3.228 \(\int \frac{\tan ^{\frac{3}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=164 \[ \frac{\sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

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

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Rubi [A]  time = 0.282123, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3547, 3546, 3544, 205} \[ \frac{\sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

Rule 3547

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

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

Mathematica [A]  time = 1.77519, size = 171, normalized size = 1.04 \[ -\frac{i e^{-6 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-4 e^{2 i (c+d x)}-16 e^{4 i (c+d x)}+17 e^{6 i (c+d x)}-15 e^{5 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+3\right )}{60 \sqrt{2} a^3 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(3/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-I/60)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(3 - 4*E^((2*I)*(c + d*x)) - 16*E^((4*I)*(c +
d*x)) + 17*E^((6*I)*(c + d*x)) - 15*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x)
)/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*a^3*d*E^((6*I)*(c + d*x))*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.041, size = 571, normalized size = 3.5 \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)^(3/2)/(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*(15*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)^4*a-90*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)^2*a+60*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-212*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-52*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan
(d*x+c)))^(1/2)*tan(d*x+c)^3+15*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))*a-60*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+220*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan
(d*x+c)+60*(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)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 3.12825, size = 1076, normalized size = 6.56 \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 \, 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 \, 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 (17 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 18 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, 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)^(3/2)/(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*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*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))*(17*e^(6*I*d*x + 6*I*c) + 18*e^(4*I*d*x + 4*I*c) - 2*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(-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)**(3/2)/(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)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError