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

Optimal result

Integrand size = 23, antiderivative size = 223 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\tan (c+d x)}} \]

[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)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3623, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\tan (c+d x)}} \]

[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*} \text {integral}& = -\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 ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \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)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.35 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \left (\left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\tan ^2(c+d x)\right )+b \left (b+6 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\tan ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[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))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}-\frac {2 a^{2}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 a b}{\sqrt {\tan \left (d x +c \right )}}}{d}\)	\(212\)
default	\(\frac {\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}-\frac {2 a^{2}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 a b}{\sqrt {\tan \left (d x +c \right )}}}{d}\)	\(212\)
parts	\(\frac {a^{2} \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {b^{2} \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {2 a b \left (-\frac {2}{\sqrt {\tan \left (d x +c \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(299\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1050 vs. \(2 (189) = 378\).

Time = 0.25 (sec) , antiderivative size = 1050, normalized size of antiderivative = 4.71 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {3 \, d \sqrt {-\frac {4 \, a^{3} b - 4 \, a b^{3} + d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} \log \left ({\left (2 \, a b d^{3} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}} + {\left (a^{6} - 7 \, a^{4} b^{2} + 7 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sqrt {-\frac {4 \, a^{3} b - 4 \, a b^{3} + d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} + {\left (a^{8} - 4 \, a^{6} b^{2} - 10 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} - 3 \, d \sqrt {-\frac {4 \, a^{3} b - 4 \, a b^{3} + d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} \log \left (-{\left (2 \, a b d^{3} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}} + {\left (a^{6} - 7 \, a^{4} b^{2} + 7 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sqrt {-\frac {4 \, a^{3} b - 4 \, a b^{3} + d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} + {\left (a^{8} - 4 \, a^{6} b^{2} - 10 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} - 3 \, d \sqrt {-\frac {4 \, a^{3} b - 4 \, a b^{3} - d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} \log \left ({\left (2 \, a b d^{3} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}} - {\left (a^{6} - 7 \, a^{4} b^{2} + 7 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sqrt {-\frac {4 \, a^{3} b - 4 \, a b^{3} - d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} + {\left (a^{8} - 4 \, a^{6} b^{2} - 10 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} + 3 \, d \sqrt {-\frac {4 \, a^{3} b - 4 \, a b^{3} - d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} \log \left (-{\left (2 \, a b d^{3} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}} - {\left (a^{6} - 7 \, a^{4} b^{2} + 7 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sqrt {-\frac {4 \, a^{3} b - 4 \, a b^{3} - d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} + {\left (a^{8} - 4 \, a^{6} b^{2} - 10 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} + 4 \, {\left (6 \, a b \tan \left (d x + c\right ) + a^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{6 \, d \tan \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-1/6*(3*d*sqrt(-(4*a^3*b - 4*a*b^3 + d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2)*l
og((2*a*b*d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^6 - 7*a^4*b^2 + 7*a^2*b^4 - b
^6)*d)*sqrt(-(4*a^3*b - 4*a*b^3 + d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2) + (a
^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 - 3*d*sqrt(-(4*a^3*b - 4*a*b
^3 + d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2)*log(-(2*a*b*d^3*sqrt(-(a^8 - 12*a
^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^6 - 7*a^4*b^2 + 7*a^2*b^4 - b^6)*d)*sqrt(-(4*a^3*b - 4*a*b^3
+ d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2) + (a^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4
*a^2*b^6 + b^8)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 - 3*d*sqrt(-(4*a^3*b - 4*a*b^3 - d^2*sqrt(-(a^8 - 12*a^6*b^
2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2)*log((2*a*b*d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6
+ b^8)/d^4) - (a^6 - 7*a^4*b^2 + 7*a^2*b^4 - b^6)*d)*sqrt(-(4*a^3*b - 4*a*b^3 - d^2*sqrt(-(a^8 - 12*a^6*b^2 +
38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2) + (a^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(tan(d*x + c
)))*tan(d*x + c)^2 + 3*d*sqrt(-(4*a^3*b - 4*a*b^3 - d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^
8)/d^4))/d^2)*log(-(2*a*b*d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) - (a^6 - 7*a^4*b^2
+ 7*a^2*b^4 - b^6)*d)*sqrt(-(4*a^3*b - 4*a*b^3 - d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)
/d^4))/d^2) + (a^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 + 4*(6*a*b*t
an(d*x + c) + a^2)*sqrt(tan(d*x + c)))/(d*tan(d*x + c)^2)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {6 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, {\left (6 \, a b \tan \left (d x + c\right ) + a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]

[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

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 6.16 (sec) , antiderivative size = 968, normalized size of antiderivative = 4.34 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=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}} \]

[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))