3.6.67 \(\int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [567]

Optimal. Leaf size=223 \[ \frac {\left (a^2+2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\tan (c+d x)}} \]

[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/d/tan(d*x+c)^(1/2)-2/3*a^2/d/tan(d*x+c)^(3/2)

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Rubi [A]
time = 0.13, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3623, 3610, 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 {\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 {2 a^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^2/Tan[c + d*x]^(5/2),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) - (2
*a^2)/(3*d*Tan[c + d*x]^(3/2)) - (4*a*b)/(d*Sqrt[Tan[c + d*x]])

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 3610

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

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 3623

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.22, size = 77, normalized size = 0.35 \begin {gather*} -\frac {2 \left (\left (a^2-b^2\right ) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(c+d x)\right )+b \left (b+6 a \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\tan ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*((a^2 - b^2)*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2] + b*(b + 6*a*Hypergeometric2F1[-1/4, 1, 3/4,
-Tan[c + d*x]^2]*Tan[c + d*x])))/(3*d*Tan[c + d*x]^(3/2))

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Maple [A]
time = 0.05, size = 212, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {2 a^{2}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 a b}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}}{d}\)	\(212\)
default	\(\frac {-\frac {2 a^{2}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 a b}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}}{d}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2/3*a^2/tan(d*x+c)^(3/2)-4*a*b/tan(d*x+c)^(1/2)+1/4*(-a^2+b^2)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+t
an(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)*t
an(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))))

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Maxima [A]
time = 0.49, size = 185, normalized size = 0.83 \begin {gather*} -\frac {6 \, \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 ) + 6 \, \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 ) + 3 \, \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 ) - 3 \, \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 ) + \frac {8 \, {\left (6 \, a b \tan \left (d x + c\right ) + a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*sqrt(2)*(a^2 + 2*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 6*sqrt(2)*(a^2 + 2
*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + 3*sqrt(2)*(a^2 - 2*a*b - b^2)*log(sqrt(2)*
sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 3*sqrt(2)*(a^2 - 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d
*x + c) + 1) + 8*(6*a*b*tan(d*x + c) + a^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. 5149 vs. \(2 (189) = 378\).
time = 1.41, size = 5149, normalized size = 23.09 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*sqrt(2)*(d^5*cos(d*x + c)^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)*((a^2 - b^2)*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) - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*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)*(2*(a^9*b - 12*a^7*b^3 + 38*a^5*b^5 - 1
2*a^3*b^7 + a*b^9)*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) - (a^14 - 11*a^1
2*b^2 + 25*a^10*b^4 + 37*a^8*b^6 - 37*a^6*b^8 - 25*a^4*b^10 + 11*a^2*b^12 - b^14)*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)*((a^10 - 5*a^8*b^2 - 6*a^6*b^4 + 6*a^4*b^6 + 5*a^2*b^8 - b^10)*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)
- 2*(a^13*b - 2*a^11*b^3 - 17*a^9*b^5 - 28*a^7*b^7 - 17*a^5*b^9 - 2*a^3*b^11 + a*b^13)*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 + 3
67*a^8*b^16 + 12*a^6*b^18 - 30*a^4*b^20 - 4*a^2*b^22 + b^24)) + 12*sqrt(2)*(d^5*cos(d*x + c)^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)*((a^2 - b^2)*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)
- 2*(a^5*b + 2*a^3*b^3 + a*b^5)*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)*(2*(a^9*b - 12*a^7*b^3 + 38*a^5*b^5 - 12*a^3*b^7 + a*b^9)*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) - (a^14 - 11*a^12*b^2 + 25*a^10*b^4 + 37*a^8*b^6 - 37*a^6*b^8 - 25*a^4
*b^10 + 11*a^2*b^12 - b^14)*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)*((a^10 - 5*a^8*b^2
- 6*a^6*b^4 + 6*a^4*b^6 + 5*a^2*b^8 - b^10)*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) - 2*(a^13*b - 2*a^11*b^3 - 17*a^9*b^5 - 28*a^7*b^7 -
17*a^5*b^9 - 2*a^3*b^11 + a*b^13)*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*(...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\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((a+b*tan(d*x+c))**2/tan(d*x+c)**(5/2),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((a+b*tan(d*x+c))^2/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 5.07, size = 968, normalized size = 4.34 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {32\,a^4\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a\,b^3}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a^3\,b}{d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{a^6\,d^2\,16{}\mathrm {i}+32\,a^5\,b\,d^2-a^4\,b^2\,d^2\,112{}\mathrm {i}-192\,a^3\,b^3\,d^2+a^2\,b^4\,d^2\,112{}\mathrm {i}+32\,a\,b^5\,d^2-b^6\,d^2\,16{}\mathrm {i}}+\frac {32\,b^4\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a\,b^3}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a^3\,b}{d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{a^6\,d^2\,16{}\mathrm {i}+32\,a^5\,b\,d^2-a^4\,b^2\,d^2\,112{}\mathrm {i}-192\,a^3\,b^3\,d^2+a^2\,b^4\,d^2\,112{}\mathrm {i}+32\,a\,b^5\,d^2-b^6\,d^2\,16{}\mathrm {i}}-\frac {192\,a^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a\,b^3}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a^3\,b}{d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}}{a^6\,d^2\,16{}\mathrm {i}+32\,a^5\,b\,d^2-a^4\,b^2\,d^2\,112{}\mathrm {i}-192\,a^3\,b^3\,d^2+a^2\,b^4\,d^2\,112{}\mathrm {i}+32\,a\,b^5\,d^2-b^6\,d^2\,16{}\mathrm {i}}\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 {atanh}\left (\frac {32\,a^4\,d^3\,\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}}}{-a^6\,d^2\,16{}\mathrm {i}+32\,a^5\,b\,d^2+a^4\,b^2\,d^2\,112{}\mathrm {i}-192\,a^3\,b^3\,d^2-a^2\,b^4\,d^2\,112{}\mathrm {i}+32\,a\,b^5\,d^2+b^6\,d^2\,16{}\mathrm {i}}+\frac {32\,b^4\,d^3\,\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}}}{-a^6\,d^2\,16{}\mathrm {i}+32\,a^5\,b\,d^2+a^4\,b^2\,d^2\,112{}\mathrm {i}-192\,a^3\,b^3\,d^2-a^2\,b^4\,d^2\,112{}\mathrm {i}+32\,a\,b^5\,d^2+b^6\,d^2\,16{}\mathrm {i}}-\frac {192\,a^2\,b^2\,d^3\,\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}}}{-a^6\,d^2\,16{}\mathrm {i}+32\,a^5\,b\,d^2+a^4\,b^2\,d^2\,112{}\mathrm {i}-192\,a^3\,b^3\,d^2-a^2\,b^4\,d^2\,112{}\mathrm {i}+32\,a\,b^5\,d^2+b^6\,d^2\,16{}\mathrm {i}}\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}}-\frac {\frac {2\,a^2}{3}+4\,b\,\mathrm {tan}\left (c+d\,x\right )\,a}{d\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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