\(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [100]

Optimal result

Integrand size = 26, antiderivative size = 130 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {4 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d} \]

[Out]

4*I*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-4*I*a^2*(a+I*a*tan(d*x+c))^(1/2)/d
-2/3*I*a*(a+I*a*tan(d*x+c))^(3/2)/d-2/7*I*(a+I*a*tan(d*x+c))^(7/2)/a/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3624, 3559, 3561, 212} \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {4 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

((4*I)*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((4*I)*a^2*Sqrt[a + I*a*Tan[
c + d*x]])/d - (((2*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/d - (((2*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a*d)

Rule 212

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

Rule 3559

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^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*} \text {integral}& = -\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\int (a+i a \tan (c+d x))^{5/2} \, dx \\ & = -\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-(2 a) \int (a+i a \tan (c+d x))^{3/2} \, dx \\ & = -\frac {4 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\left (4 a^2\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx \\ & = -\frac {4 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}+\frac {\left (8 i a^3\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = \frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {4 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {84 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a^2 \sqrt {a+i a \tan (c+d x)} \left (-52 i+16 \tan (c+d x)+9 i \tan ^2(c+d x)-3 \tan ^3(c+d x)\right )}{21 d} \]

[In]

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

[Out]

((84*I)*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + 2*a^2*Sqrt[a + I*a*Tan[c + d*x
]]*(-52*I + 16*Tan[c + d*x] + (9*I)*Tan[c + d*x]^2 - 3*Tan[c + d*x]^3))/(21*d)

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+2 a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d a}\)	\(96\)
default	\(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+2 a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d a}\)	\(96\)

[In]

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

[Out]

2*I/d/a*(-1/7*(a+I*a*tan(d*x+c))^(7/2)-1/3*a^2*(a+I*a*tan(d*x+c))^(3/2)-2*a^3*(a+I*a*tan(d*x+c))^(1/2)+2*a^(7/
2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (95) = 190\).

Time = 0.25 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.85 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (21 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 21 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) + 2 \, \sqrt {2} {\left (40 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 77 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 70 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 21 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{21 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-2/21*(21*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) +
d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
)))*e^(-I*d*x - I*c)/a^2) - 21*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e
^(2*I*d*x + 2*I*c) + d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e
^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) + 2*sqrt(2)*(40*I*a^2*e^(7*I*d*x + 7*I*c) + 77*I*a^2*e^(5*I*d*
x + 5*I*c) + 70*I*a^2*e^(3*I*d*x + 3*I*c) + 21*I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^
(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 i \, {\left (21 \, \sqrt {2} a^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} + 7 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} + 42 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{5}\right )}}{21 \, a^{3} d} \]

[In]

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

[Out]

-2/21*I*(21*sqrt(2)*a^(11/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*t
an(d*x + c) + a))) + 3*(I*a*tan(d*x + c) + a)^(7/2)*a^2 + 7*(I*a*tan(d*x + c) + a)^(3/2)*a^4 + 42*sqrt(I*a*tan
(d*x + c) + a)*a^5)/(a^3*d)

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}\,2{}\mathrm {i}}{7\,a\,d}-\frac {a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {\sqrt {2}\,{\left (-a\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,4{}\mathrm {i}}{d} \]

[In]

int(tan(c + d*x)^2*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

(2^(1/2)*(-a)^(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*4i)/d - ((a + a*tan(c + d*x)*
1i)^(7/2)*2i)/(7*a*d) - (a*(a + a*tan(c + d*x)*1i)^(3/2)*2i)/(3*d) - (a^2*(a + a*tan(c + d*x)*1i)^(1/2)*4i)/d