3.130 \(\int \frac{\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac{1}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{1}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{1}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

-ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(4*Sqrt[2]*a^(5/2)*d) - 1/(5*d*(a + I*a*Tan[c + d*x])^(
5/2)) + 1/(6*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + 1/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.101062, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3526, 3479, 3480, 206} \[ \frac{1}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{1}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{1}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(4*Sqrt[2]*a^(5/2)*d) - 1/(5*d*(a + I*a*Tan[c + d*x])^(
5/2)) + 1/(6*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + 1/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3526

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)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(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] && LtQ[n
, 0]

Rule 3480

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 206

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

Rubi steps

\begin{align*} \int \frac{\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac{1}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{i \int \frac{1}{(a+i a \tan (c+d x))^{3/2}} \, dx}{2 a}\\ &=-\frac{1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i \int \frac{1}{\sqrt{a+i a \tan (c+d x)}} \, dx}{4 a^2}\\ &=-\frac{1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{1}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{i \int \sqrt{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=-\frac{1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{1}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 a^2 d}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{1}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.71922, size = 135, normalized size = 1.08 \[ \frac{e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt{1+e^{2 i (c+d x)}} \left (-e^{2 i (c+d x)}+17 e^{4 i (c+d x)}-3\right )-15 e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{240 a^2 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + E^((2*I)*(c + d*x)))^(3/2)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-3 - E^((2*I)*(c + d*x)) + 17*E^((4*I)*(c + d
*x))) - 15*E^((5*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])*Sec[c + d*x]^2)/(240*a^2*d*E^((6*I)*(c + d*x))*Sqrt[a
+ I*a*Tan[c + d*x]])

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Maple [A]  time = 0.019, size = 91, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{\frac{\sqrt{2}}{8}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}}-{\frac{1}{5} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}+{\frac{1}{4\,{a}^{2}}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{1}{6\,a} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/8/a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-1/5/(a+I*a*tan(d*x+c))^(5/2)+1
/4/a^2/(a+I*a*tan(d*x+c))^(1/2)+1/6/a/(a+I*a*tan(d*x+c))^(3/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31567, size = 845, normalized size = 6.76 \begin{align*} -\frac{{\left (15 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (17 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{120 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log((2*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(2
*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d
*x - I*c)) - 15*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(-(2*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2)
)*e^(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e
^(-I*d*x - I*c)) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(17*e^(6*I*d*x + 6*I*c) + 16*e^(4*I*d*x + 4*I*c)
- 4*e^(2*I*d*x + 2*I*c) - 3)*e^(I*d*x + I*c))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(c + d*x)/(a*(I*tan(c + d*x) + 1))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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