\(\int x (a+i a \sinh (c+d x))^{5/2} \, dx\) [132]

Optimal result

Integrand size = 19, antiderivative size = 312 \[ \int x (a+i a \sinh (c+d x))^{5/2} \, dx=-\frac {128 a^2 \sqrt {a+i a \sinh (c+d x)}}{15 d^2}-\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {32 a^2 x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {64 a^2 x \sqrt {a+i a \sinh (c+d x)} \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{15 d} \]

[Out]

-128/15*a^2*(a+I*a*sinh(d*x+c))^(1/2)/d^2-64/45*a^2*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^2*(a+I*a*sinh(d*x+c))^(1/2)/d
^2-16/25*a^2*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^4*(a+I*a*sinh(d*x+c))^(1/2)/d^2+32/15*a^2*x*cosh(1/2*c+1/4*I*Pi+1/2*
d*x)*sinh(1/2*c+1/4*I*Pi+1/2*d*x)*(a+I*a*sinh(d*x+c))^(1/2)/d+8/5*a^2*x*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^3*sinh(1/
2*c+1/4*I*Pi+1/2*d*x)*(a+I*a*sinh(d*x+c))^(1/2)/d+64/15*a^2*x*(a+I*a*sinh(d*x+c))^(1/2)*tanh(1/2*c+1/4*I*Pi+1/
2*d*x)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3400, 3391, 3377, 2718} \[ \int x (a+i a \sinh (c+d x))^{5/2} \, dx=-\frac {128 a^2 \sqrt {a+i a \sinh (c+d x)}}{15 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}-\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}+\frac {64 a^2 x \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh ^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {32 a^2 x \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d} \]

[In]

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

[Out]

(-128*a^2*Sqrt[a + I*a*Sinh[c + d*x]])/(15*d^2) - (64*a^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^2*Sqrt[a + I*a*Sinh[c
+ d*x]])/(45*d^2) - (16*a^2*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^4*Sqrt[a + I*a*Sinh[c + d*x]])/(25*d^2) + (32*a^2*
x*Cosh[c/2 + (I/4)*Pi + (d*x)/2]*Sinh[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]])/(15*d) + (8*a^2*x
*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^3*Sinh[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]])/(5*d) + (64*a^2*
x*Sqrt[a + I*a*Sinh[c + d*x]]*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(15*d)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 3400

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(2*a)^IntPart[n]
*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n])), Int[(c + d*x)^m*Sin[e/2
+ a*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps \begin{align*} \text {integral}& = \left (4 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int x \sinh ^5\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx \\ & = -\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}-\frac {1}{5} \left (16 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int x \sinh ^3\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx \\ & = -\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {32 a^2 x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {1}{15} \left (32 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int x \sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx \\ & = -\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {32 a^2 x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {64 a^2 x \sqrt {a+i a \sinh (c+d x)} \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{15 d}-\frac {\left (64 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \cosh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{15 d} \\ & = -\frac {128 a^2 \sqrt {a+i a \sinh (c+d x)}}{15 d^2}-\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {32 a^2 x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {64 a^2 x \sqrt {a+i a \sinh (c+d x)} \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.43 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.70 \[ \int x (a+i a \sinh (c+d x))^{5/2} \, dx=\frac {a^2 (-i+\sinh (c+d x))^2 \sqrt {a+i a \sinh (c+d x)} \left (2250 (2-i d x) \cosh \left (\frac {1}{2} (c+d x)\right )+(-250-375 i d x) \cosh \left (\frac {3}{2} (c+d x)\right )-18 \cosh \left (\frac {5}{2} (c+d x)\right )+45 i d x \cosh \left (\frac {5}{2} (c+d x)\right )+4500 i \sinh \left (\frac {1}{2} (c+d x)\right )-2250 d x \sinh \left (\frac {1}{2} (c+d x)\right )+250 i \sinh \left (\frac {3}{2} (c+d x)\right )+375 d x \sinh \left (\frac {3}{2} (c+d x)\right )-18 i \sinh \left (\frac {5}{2} (c+d x)\right )+45 d x \sinh \left (\frac {5}{2} (c+d x)\right )\right )}{450 d^2 \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

[In]

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

[Out]

(a^2*(-I + Sinh[c + d*x])^2*Sqrt[a + I*a*Sinh[c + d*x]]*(2250*(2 - I*d*x)*Cosh[(c + d*x)/2] + (-250 - (375*I)*
d*x)*Cosh[(3*(c + d*x))/2] - 18*Cosh[(5*(c + d*x))/2] + (45*I)*d*x*Cosh[(5*(c + d*x))/2] + (4500*I)*Sinh[(c +
d*x)/2] - 2250*d*x*Sinh[(c + d*x)/2] + (250*I)*Sinh[(3*(c + d*x))/2] + 375*d*x*Sinh[(3*(c + d*x))/2] - (18*I)*
Sinh[(5*(c + d*x))/2] + 45*d*x*Sinh[(5*(c + d*x))/2]))/(450*d^2*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^5)

Maple [F]

\[\int x \left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int x (a+i a \sinh (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F(-1)]

Timed out. \[ \int x (a+i a \sinh (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int x (a+i a \sinh (c+d x))^{5/2} \, dx=\int { {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} x \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int x (a+i a \sinh (c+d x))^{5/2} \, dx=\int { {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} x \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x (a+i a \sinh (c+d x))^{5/2} \, dx=\int x\,{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

[In]

int(x*(a + a*sinh(c + d*x)*1i)^(5/2),x)

[Out]

int(x*(a + a*sinh(c + d*x)*1i)^(5/2), x)