3.6.62 \(\int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx\) [562]

Optimal. Leaf size=268 \[ \frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d} \]

[Out]

-1/2*(a^2-2*a*b-b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*(a^2-2*a*b-b^2)*arctan(1+2^(1/2)*tan(d*
x+c)^(1/2))/d*2^(1/2)-1/4*(a^2+2*a*b-b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+1/4*(a^2+2*a*b-b
^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-4*a*b*tan(d*x+c)^(1/2)/d+2/3*(a^2-b^2)*tan(d*x+c)^(3/2
)/d+4/5*a*b*tan(d*x+c)^(5/2)/d+2/7*b^2*tan(d*x+c)^(7/2)/d

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Rubi [A]
time = 0.18, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3624, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^2,x]

[Out]

((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a^2 - 2*a*b - b^2)*ArcTan[1 + Sqr
t[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a^2 + 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]
])/(2*Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (4
*a*b*Sqrt[Tan[c + d*x]])/d + (2*(a^2 - b^2)*Tan[c + d*x]^(3/2))/(3*d) + (4*a*b*Tan[c + d*x]^(5/2))/(5*d) + (2*
b^2*Tan[c + d*x]^(7/2))/(7*d)

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3624

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

\begin {align*} \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\int \tan ^{\frac {5}{2}}(c+d x) \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\int \tan ^{\frac {3}{2}}(c+d x) \left (-2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\int \sqrt {\tan (c+d x)} \left (-a^2+b^2-2 a b \tan (c+d x)\right ) \, dx\\ &=-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\int \frac {2 a b-\left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\frac {2 \text {Subst}\left (\int \frac {2 a b+\left (-a^2+b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}-\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}\\ &=-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}\\ &=\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.13, size = 133, normalized size = 0.50 \begin {gather*} \frac {-105 (-1)^{3/4} (a-i b)^2 \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+105 (-1)^{3/4} (a+i b)^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {\tan (c+d x)} \left (-210 a b+35 \left (a^2-b^2\right ) \tan (c+d x)+42 a b \tan ^2(c+d x)+15 b^2 \tan ^3(c+d x)\right )}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^2,x]

[Out]

(-105*(-1)^(3/4)*(a - I*b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 105*(-1)^(3/4)*(a + I*b)^2*ArcTanh[(-1)^(
3/4)*Sqrt[Tan[c + d*x]]] + 2*Sqrt[Tan[c + d*x]]*(-210*a*b + 35*(a^2 - b^2)*Tan[c + d*x] + 42*a*b*Tan[c + d*x]^
2 + 15*b^2*Tan[c + d*x]^3))/(105*d)

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Maple [A]
time = 0.07, size = 250, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {\frac {2 b^{2} \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {4 a b \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 a^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {2 b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-4 a b \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(250\)
default	\(\frac {\frac {2 b^{2} \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {4 a b \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 a^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {2 b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-4 a b \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(250\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/7*b^2*tan(d*x+c)^(7/2)+4/5*a*b*tan(d*x+c)^(5/2)+2/3*a^2*tan(d*x+c)^(3/2)-2/3*b^2*tan(d*x+c)^(3/2)-4*a*b
*tan(d*x+c)^(1/2)+1/2*a*b*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(
d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-a^2+b^2)*2^(1/2)*(l
n((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+
c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))

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Maxima [A]
time = 0.49, size = 217, normalized size = 0.81 \begin {gather*} \frac {120 \, b^{2} \tan \left (d x + c\right )^{\frac {7}{2}} + 336 \, a b \tan \left (d x + c\right )^{\frac {5}{2}} - 210 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 210 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 105 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 105 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 1680 \, a b \sqrt {\tan \left (d x + c\right )} + 280 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{420 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/420*(120*b^2*tan(d*x + c)^(7/2) + 336*a*b*tan(d*x + c)^(5/2) - 210*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sq
rt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) - 210*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqr
t(tan(d*x + c)))) + 105*sqrt(2)*(a^2 + 2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 105*s
qrt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 1680*a*b*sqrt(tan(d*x + c)) +
280*(a^2 - b^2)*tan(d*x + c)^(3/2))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 5080 vs. \(2 (226) = 452\).
time = 1.75, size = 5080, normalized size = 18.96 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/420*(420*sqrt(2)*d^5*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 +
4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*
a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4)
*arctan(-((a^16 - 20*a^12*b^4 - 64*a^10*b^6 - 90*a^8*b^8 - 64*a^6*b^10 - 20*a^4*b^12 + b^16)*d^4*sqrt((a^8 + 4
*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) - sq
rt(2)*(2*a*b*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4
- 12*a^2*b^6 + b^8)/d^4) + (a^6 + a^4*b^2 - a^2*b^4 - b^6)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b
^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*
b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(((a^12 - 10*
a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 + b^12)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 +
4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + sqrt(2)*((a^10 - 13*a^8*b^2 + 50*a^6*b^4 - 50*a^4*b^6 + 13*a^2*b^8 - b^10
)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + 2*(a^13*b - 10*a^11*b^3 + 15*a^
9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*
a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6
*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b
^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^10*b^6 + 134*a^8*b^8 + 72*a^6*b^10 - 4*a^4*b^12
- 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4) +
sqrt(2)*(2*(a^9*b - 4*a^7*b^3 - 10*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2
*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^14 - 3*a^12*b^2 - 15*a^10*b
^4 - 11*a^8*b^6 + 11*a^6*b^8 + 15*a^4*b^10 + 3*a^2*b^12 - b^14)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a
^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*
a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x
+ c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4))/(a^24 - 4*a^22*b^2 - 30*a^20*b
^4 + 12*a^18*b^6 + 367*a^16*b^8 + 1016*a^14*b^10 + 1372*a^12*b^12 + 1016*a^10*b^14 + 367*a^8*b^16 + 12*a^6*b^1
8 - 30*a^4*b^20 - 4*a^2*b^22 + b^24))*cos(d*x + c)^3 + 420*sqrt(2)*d^5*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a
^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*
b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4)*sqrt((a^8 -
12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4)*arctan(((a^16 - 20*a^12*b^4 - 64*a^10*b^6 - 90*a^8*b^8 - 64*a
^6*b^10 - 20*a^4*b^12 + b^16)*d^4*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6
*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + sqrt(2)*(2*a*b*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6
+ b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^6 + a^4*b^2 - a^2*b^4 - b^6)*d^5
*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 +
b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38
*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 +
b^12)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) - sqrt(2)*((a^10 - 13*a^8*b^
2 + 50*a^6*b^4 - 50*a^4*b^6 + 13*a^2*b^8 - b^10)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)
*cos(d*x + c) + 2*(a^13*b - 10*a^11*b^3 + 15*a^9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d
*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6
*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*
x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^1
0*b^6 + 134*a^8*b^8 + 72*a^6*b^10 - 4*a^4*b^12 - 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c))*((a^8 + 4*a^6*
b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4) - sqrt(2)*(2*(a^9*b - 4*a^7*b^3 - 10*a^5*b^5 - 4*a^3*b^7 + a*b^9
)*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b
^6 + b^8)/d^4) + (a^14 - 3*a^12*b^2 - 15*a^10*b^4 - 11*a^8*b^6 + 11*a^6*b^8 + 15*a^4*b^10 + 3*a^2*b^12 - b^14)
*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b
^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(5/2)*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*tan(c + d*x)**(5/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 6.99, size = 995, normalized size = 3.71 \begin {gather*} {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (\frac {2\,a^2}{3\,d}-\frac {2\,b^2}{3\,d}\right )+\frac {2\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{7\,d}-\frac {4\,a\,b\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {4\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\mathrm {atan}\left (\frac {a^4\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,32{}\mathrm {i}}{\frac {16\,a^6}{d}-\frac {16\,b^6}{d}+\frac {112\,a^2\,b^4}{d}-\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}+\frac {b^4\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,32{}\mathrm {i}}{\frac {16\,a^6}{d}-\frac {16\,b^6}{d}+\frac {112\,a^2\,b^4}{d}-\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}-\frac {a^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,192{}\mathrm {i}}{\frac {16\,a^6}{d}-\frac {16\,b^6}{d}+\frac {112\,a^2\,b^4}{d}-\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}\right )\,\sqrt {-\frac {a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a^4\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,32{}\mathrm {i}}{\frac {16\,b^6}{d}-\frac {16\,a^6}{d}-\frac {112\,a^2\,b^4}{d}+\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}+\frac {b^4\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,32{}\mathrm {i}}{\frac {16\,b^6}{d}-\frac {16\,a^6}{d}-\frac {112\,a^2\,b^4}{d}+\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}-\frac {a^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,192{}\mathrm {i}}{\frac {16\,b^6}{d}-\frac {16\,a^6}{d}-\frac {112\,a^2\,b^4}{d}+\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}\right )\,\sqrt {\frac {a^4\,1{}\mathrm {i}+4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}-4\,a\,b^3+b^4\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))^2,x)

[Out]

tan(c + d*x)^(3/2)*((2*a^2)/(3*d) - (2*b^2)/(3*d)) - atan((a^4*tan(c + d*x)^(1/2)*((a^3*b)/d^2 - (b^4*1i)/(4*d
^2) - (a*b^3)/d^2 - (a^4*1i)/(4*d^2) + (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*a^6)/d - (16*b^6)/d + (a*b^5*32i)
/d + (a^5*b*32i)/d + (112*a^2*b^4)/d - (a^3*b^3*192i)/d - (112*a^4*b^2)/d) + (b^4*tan(c + d*x)^(1/2)*((a^3*b)/
d^2 - (b^4*1i)/(4*d^2) - (a*b^3)/d^2 - (a^4*1i)/(4*d^2) + (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*a^6)/d - (16*b
^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d + (112*a^2*b^4)/d - (a^3*b^3*192i)/d - (112*a^4*b^2)/d) - (a^2*b^2*tan(c
+ d*x)^(1/2)*((a^3*b)/d^2 - (b^4*1i)/(4*d^2) - (a*b^3)/d^2 - (a^4*1i)/(4*d^2) + (a^2*b^2*3i)/(2*d^2))^(1/2)*1
92i)/((16*a^6)/d - (16*b^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d + (112*a^2*b^4)/d - (a^3*b^3*192i)/d - (112*a^4*
b^2)/d))*(-(4*a*b^3 - 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i)/(4*d^2))^(1/2)*2i + atan((a^4*tan(c + d*x)^(1/2)
*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d^2) - (a*b^3)/d^2 + (a^3*b)/d^2 - (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*b^6)
/d - (16*a^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d - (112*a^2*b^4)/d - (a^3*b^3*192i)/d + (112*a^4*b^2)/d) + (b^4
*tan(c + d*x)^(1/2)*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d^2) - (a*b^3)/d^2 + (a^3*b)/d^2 - (a^2*b^2*3i)/(2*d^2))^(
1/2)*32i)/((16*b^6)/d - (16*a^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d - (112*a^2*b^4)/d - (a^3*b^3*192i)/d + (112
*a^4*b^2)/d) - (a^2*b^2*tan(c + d*x)^(1/2)*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d^2) - (a*b^3)/d^2 + (a^3*b)/d^2 -
(a^2*b^2*3i)/(2*d^2))^(1/2)*192i)/((16*b^6)/d - (16*a^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d - (112*a^2*b^4)/d -
(a^3*b^3*192i)/d + (112*a^4*b^2)/d))*((4*a^3*b - 4*a*b^3 + a^4*1i + b^4*1i - a^2*b^2*6i)/(4*d^2))^(1/2)*2i +
(2*b^2*tan(c + d*x)^(7/2))/(7*d) - (4*a*b*tan(c + d*x)^(1/2))/d + (4*a*b*tan(c + d*x)^(5/2))/(5*d)

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