3.1.67 \(\int (a+i a \sinh (c+d x))^{3/2} \, dx\) [67]

Optimal. Leaf size=69 \[ \frac {8 i a^2 \cosh (c+d x)}{3 d \sqrt {a+i a \sinh (c+d x)}}+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d} \]

[Out]

8/3*I*a^2*cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))^(1/2)+2/3*I*a*cosh(d*x+c)*(a+I*a*sinh(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2726, 2725} \begin {gather*} \frac {8 i a^2 \cosh (c+d x)}{3 d \sqrt {a+i a \sinh (c+d x)}}+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Sinh[c + d*x])^(3/2),x]

[Out]

(((8*I)/3)*a^2*Cosh[c + d*x])/(d*Sqrt[a + I*a*Sinh[c + d*x]]) + (((2*I)/3)*a*Cosh[c + d*x]*Sqrt[a + I*a*Sinh[c
+ d*x]])/d

Rule 2725

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x
]])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2726

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[a*((2*n - 1)/n), Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rubi steps

\begin {align*} \int (a+i a \sinh (c+d x))^{3/2} \, dx &=\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d}+\frac {1}{3} (4 a) \int \sqrt {a+i a \sinh (c+d x)} \, dx\\ &=\frac {8 i a^2 \cosh (c+d x)}{3 d \sqrt {a+i a \sinh (c+d x)}}+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 113, normalized size = 1.64 \begin {gather*} -\frac {a (-i+\sinh (c+d x)) \sqrt {a+i a \sinh (c+d x)} \left (9 \cosh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {3}{2} (c+d x)\right )-9 i \sinh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {3}{2} (c+d x)\right )\right )}{3 d \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*(-I + Sinh[c + d*x])*Sqrt[a + I*a*Sinh[c + d*x]]*(9*Cosh[(c + d*x)/2] + Cosh[(3*(c + d*x))/2] - (9*I)*
Sinh[(c + d*x)/2] + I*Sinh[(3*(c + d*x))/2]))/(d*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^3)

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Maple [F]
time = 1.84, size = 0, normalized size = 0.00 \[\int \left (a +i a \sinh \left (d x +c \right )\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*sinh(d*x+c))^(3/2),x)

[Out]

int((a+I*a*sinh(d*x+c))^(3/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.43, size = 63, normalized size = 0.91 \begin {gather*} \frac {{\left (i \, a e^{\left (3 \, d x + 3 \, c\right )} + 9 \, a e^{\left (2 \, d x + 2 \, c\right )} + 9 i \, a e^{\left (d x + c\right )} + a\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}} e^{\left (-d x - c\right )}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(I*a*e^(3*d*x + 3*c) + 9*a*e^(2*d*x + 2*c) + 9*I*a*e^(d*x + c) + a)*sqrt(1/2*I*a*e^(-d*x - c))*e^(-d*x - c
)/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \sinh {\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*sinh(d*x+c))**(3/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sinh(c + d*x)*1i)^(3/2),x)

[Out]

int((a + a*sinh(c + d*x)*1i)^(3/2), x)

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