3.1147 \(\int \frac{(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=173 \[ -\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f}+\frac{i (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{(d+i c) \sqrt{c+d \tan (e+f x)}}{2 a f \sqrt{a+i a \tan (e+f x)}} \]

[Out]

((-I/2)*(c - I*d)^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e +
f*x]])])/(Sqrt[2]*a^(3/2)*f) + ((I*c + d)*Sqrt[c + d*Tan[e + f*x]])/(2*a*f*Sqrt[a + I*a*Tan[e + f*x]]) + ((I/
3)*(c + d*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^(3/2))

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Rubi [A]  time = 0.34049, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3546, 3544, 208} \[ -\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f}+\frac{i (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{(d+i c) \sqrt{c+d \tan (e+f x)}}{2 a f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(3/2),x]

[Out]

((-I/2)*(c - I*d)^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e +
f*x]])])/(Sqrt[2]*a^(3/2)*f) + ((I*c + d)*Sqrt[c + d*Tan[e + f*x]])/(2*a*f*Sqrt[a + I*a*Tan[e + f*x]]) + ((I/
3)*(c + d*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^(3/2))

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 208

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

Rubi steps

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

Mathematica [A]  time = 3.9387, size = 240, normalized size = 1.39 \[ \frac{\sec ^{\frac{3}{2}}(e+f x) \left (-i \sqrt{2} (c-i d)^{3/2} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (1+e^{2 i (e+f x)}\right )^{3/2} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )-\frac{2 ((3 c-5 i d) \tan (e+f x)-5 i c-3 d) \sqrt{c+d \tan (e+f x)}}{3 \sec ^{\frac{3}{2}}(e+f x)}\right )}{4 f (a+i a \tan (e+f x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sec[e + f*x]^(3/2)*((-I)*Sqrt[2]*(c - I*d)^(3/2)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(3/2)*(1 + E^((2
*I)*(e + f*x)))^(3/2)*Log[2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-1 +
E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])] - (2*((-5*I)*c - 3*d + (3*c - (5*I)*d)*Tan[e + f*x])*Sqrt[
c + d*Tan[e + f*x]])/(3*Sec[e + f*x]^(3/2))))/(4*f*(a + I*a*Tan[e + f*x])^(3/2))

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Maple [B]  time = 0.052, size = 1275, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/2),x)

[Out]

1/24/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2*(-20*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2
)*tan(f*x+e)^2*d^2-12*I*tan(f*x+e)^2*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-3*2^(1/2)*(-a*(I*d-c))^(1
/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*ta
n(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)^3*c^2-3*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a
*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*
tan(f*x+e)^3*d^2-8*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c*d-32*tan(f*x+e)*c^2*(a*(c+d*tan(
f*x+e))*(1+I*tan(f*x+e)))^(1/2)-3*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)
*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2-3*I*2^(1/2)*(
-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f
*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*d^2+9*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-
I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I
))*tan(f*x+e)^2*d^2+9*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)
*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)^2*c^2+9*2^(1/2)*(-
a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*
x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*tan(f*x+e)*c^2+9*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f
*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(
f*x+e)+I))*tan(f*x+e)*d^2+12*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^2+20*I*c^2*(a*(c+d*tan(f*x+e))*(1
+I*tan(f*x+e)))^(1/2)-32*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*d^2-8*(a*(c+d*tan(f*x+e))*(1+I
*tan(f*x+e)))^(1/2)*c*d)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(I*c-d)/(-tan(f*x+e)+I)^3

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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((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.80498, size = 1405, normalized size = 8.12 \begin{align*} -\frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{2}{\left ({\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c + d\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{i \, c + d}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} - \sqrt{2}{\left ({\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c + d\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{i \, c + d}\right ) - \sqrt{2}{\left ({\left (4 i \, c + 4 \, d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (5 i \, c + 3 \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{12 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(1/2)*a^2*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^3*f^2))*e^(4*I*f*x + 4*I*e)*log((2*sqrt(
1/2)*a^2*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^3*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(2)*((I*c + d)*e^(2*
I*f*x + 2*I*e) + I*c + d)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^
(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e))*e^(-I*f*x - I*e)/(I*c + d)) - 3*sqrt(1/2)*a^2*f*sqrt(-(c^3 - 3*I*c^2*
d - 3*c*d^2 + I*d^3)/(a^3*f^2))*e^(4*I*f*x + 4*I*e)*log(-(2*sqrt(1/2)*a^2*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 +
I*d^3)/(a^3*f^2))*e^(2*I*f*x + 2*I*e) - sqrt(2)*((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c + d)*sqrt(((c - I*d)*e^(
2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e))*e^(-
I*f*x - I*e)/(I*c + d)) - sqrt(2)*((4*I*c + 4*d)*e^(4*I*f*x + 4*I*e) + (5*I*c + 3*d)*e^(2*I*f*x + 2*I*e) + I*c
- d)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) +
1))*e^(I*f*x + I*e))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**(3/2),x)

[Out]

Exception raised: AttributeError

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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((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError