3.1.95 \(\int (a+i a \tan (c+d x))^{3/2} \, dx\) [95]

Optimal. Leaf size=72 \[ -\frac {2 i \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3559, 3561, 212} \begin {gather*} \frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.53, size = 79, normalized size = 1.10 \begin {gather*} \frac {2 i a e^{-i (c+d x)} \left (e^{i (c+d x)}-\sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*a*(E^(I*(c + d*x)) - Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])*Sqrt[a + I*a*Tan[c + d*x]]
)/(d*E^(I*(c + d*x)))

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Maple [A]
time = 0.15, size = 54, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {2 i a \left (\sqrt {a +i a \tan \left (d x +c \right )}-\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(54\)
default	\(\frac {2 i a \left (\sqrt {a +i a \tan \left (d x +c \right )}-\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 83, normalized size = 1.15 \begin {gather*} \frac {i \, {\left (\sqrt {2} a^{\frac {5}{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 ) + 2 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}\right )}}{a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(sqrt(2)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c
) + a))) + 2*sqrt(I*a*tan(d*x + c) + a)*a^2)/(a*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (53) = 106\).
time = 0.47, size = 215, normalized size = 2.99 \begin {gather*} -\frac {-2 i \, \sqrt {2} a \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {-\frac {a^{3}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} + {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) + \sqrt {2} \sqrt {-\frac {a^{3}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} + {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-2*I*sqrt(2)*a*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - sqrt(2)*sqrt(-a^3/d^2)*d*log(4*(a^2*e^(I*
d*x + I*c) + (I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c
)/a) + sqrt(2)*sqrt(-a^3/d^2)*d*log(4*(a^2*e^(I*d*x + I*c) + (-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(-a^3/d^2)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*tan(c + d*x) + a)**(3/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [B]
time = 0.18, size = 61, normalized size = 0.85 \begin {gather*} \frac {a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{d}+\frac {\sqrt {2}\,{\left (-a\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,2{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

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

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