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

Optimal. Leaf size=166 \[ -\frac{i \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{i \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{\tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

((1/8 + I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((I/5)*Ta
n[c + d*x]^(5/2))/(d*(a + I*a*Tan[c + d*x])^(5/2)) + Tan[c + d*x]^(3/2)/(6*a*d*(a + I*a*Tan[c + d*x])^(3/2)) -
((I/4)*Sqrt[Tan[c + d*x]])/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.287158, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3546, 3544, 205} \[ -\frac{i \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{i \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{\tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((I/5)*Ta
n[c + d*x]^(5/2))/(d*(a + I*a*Tan[c + d*x])^(5/2)) + Tan[c + d*x]^(3/2)/(6*a*d*(a + I*a*Tan[c + d*x])^(3/2)) -
((I/4)*Sqrt[Tan[c + d*x]])/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3546

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

Rule 3544

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

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 1.94971, size = 177, normalized size = 1.07 \[ \frac{i e^{-4 i (c+d x)} \sqrt{\tan (c+d x)} \left (\sqrt{-1+e^{2 i (c+d x)}} \left (11 e^{2 i (c+d x)}-23 e^{4 i (c+d x)}-3\right )+15 e^{5 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{60 \sqrt{2} a^2 d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/60)*(Sqrt[-1 + E^((2*I)*(c + d*x))]*(-3 + 11*E^((2*I)*(c + d*x)) - 23*E^((4*I)*(c + d*x))) + 15*E^((5*I)*(
c + d*x))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Tan[c + d*x]])/(Sqrt[2]*a^2*d*E^((4*I)
*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))])

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Maple [B]  time = 0.041, size = 570, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(60*I*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c
)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a-15*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(
1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a+148*(-I*a)^(1/2)
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3-60*I*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a+90*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-308*I*(-I*a)^(1/2)*(a*ta
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-15*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x
+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a+60*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-220*
tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x
+c)+I)^4/(-I*a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 3.03092, size = 1088, normalized size = 6.55 \begin{align*} -\frac{{\left (30 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{4} \,{\left (4 i \, a^{3} d \sqrt{\frac{i}{8 \, 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}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{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 ) - 30 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{4} \,{\left (-4 i \, a^{3} d \sqrt{\frac{i}{8 \, 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}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{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}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-23 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 12 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\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)^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/120*(30*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(1/4*(4*I*a^3*d*sqrt(1/8*I/(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))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(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)) - 30*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(6*I*d*x
+ 6*I*c)*log(1/4*(-4*I*a^3*d*sqrt(1/8*I/(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))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(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))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2
*I*c) + 1))*(-23*I*e^(6*I*d*x + 6*I*c) - 12*I*e^(4*I*d*x + 4*I*c) + 8*I*e^(2*I*d*x + 2*I*c) - 3*I)*e^(I*d*x +
I*c))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError