3.125 \(\int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx\)

Optimal. Leaf size=110 \[ \frac {(5 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(5 B+3 i A) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac {(-B+i A) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}} \]

[Out]

1/4*(I*A-B)*cosh(x)/(a+I*a*sinh(x))^(5/2)+1/16*(3*I*A+5*B)*cosh(x)/a/(a+I*a*sinh(x))^(3/2)+1/32*(3*I*A+5*B)*ar
ctanh(1/2*cosh(x)*a^(1/2)*2^(1/2)/(a+I*a*sinh(x))^(1/2))/a^(5/2)*2^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2750, 2650, 2649, 206} \[ \frac {(5 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(5 B+3 i A) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac {(-B+i A) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)*A + 5*B)*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[x]])])/(16*Sqrt[2]*a^(5/2)) + ((I*A - B)
*Cosh[x])/(4*(a + I*a*Sinh[x])^(5/2)) + (((3*I)*A + 5*B)*Cosh[x])/(16*a*(a + I*a*Sinh[x])^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

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

Rule 2650

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

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx &=\frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 A-5 i B) \int \frac {1}{(a+i a \sinh (x))^{3/2}} \, dx}{8 a}\\ &=\frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac {(3 A-5 i B) \int \frac {1}{\sqrt {a+i a \sinh (x)}} \, dx}{32 a^2}\\ &=\frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac {(3 i A+5 B) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cosh (x)}{\sqrt {a+i a \sinh (x)}}\right )}{16 a^2}\\ &=\frac {(3 i A+5 B) \tanh ^{-1}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 184, normalized size = 1.67 \[ \frac {\left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right ) \left (8 (A+i B) \sinh \left (\frac {x}{2}\right )+2 (5 B+3 i A) \sinh \left (\frac {x}{2}\right ) (\sinh (x)-i)+(5 B+3 i A) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^3+4 i (A+i B) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )+(1-i) \sqrt [4]{-1} (3 A-5 i B) \tan ^{-1}\left (\frac {\tanh \left (\frac {x}{4}\right )+i}{\sqrt {2}}\right ) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^4\right )}{16 (a+i a \sinh (x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cosh[x/2] + I*Sinh[x/2])*((4*I)*(A + I*B)*(Cosh[x/2] + I*Sinh[x/2]) + ((3*I)*A + 5*B)*(Cosh[x/2] + I*Sinh[x/
2])^3 + (1 - I)*(-1)^(1/4)*(3*A - (5*I)*B)*ArcTan[(I + Tanh[x/4])/Sqrt[2]]*(Cosh[x/2] + I*Sinh[x/2])^4 + 8*(A
+ I*B)*Sinh[x/2] + 2*((3*I)*A + 5*B)*Sinh[x/2]*(-I + Sinh[x])))/(16*(a + I*a*Sinh[x])^(5/2))

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fricas [B]  time = 0.48, size = 356, normalized size = 3.24 \[ \frac {\sqrt {\frac {1}{2}} {\left (4 \, a^{3} e^{\left (4 \, x\right )} - 16 i \, a^{3} e^{\left (3 \, x\right )} - 24 \, a^{3} e^{\left (2 \, x\right )} + 16 i \, a^{3} e^{x} + 4 \, a^{3}\right )} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (\frac {\sqrt {\frac {1}{2}} a^{3} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (3 i \, A + 5 \, B\right )}}{3 i \, A + 5 \, B}\right ) - \sqrt {\frac {1}{2}} {\left (4 \, a^{3} e^{\left (4 \, x\right )} - 16 i \, a^{3} e^{\left (3 \, x\right )} - 24 \, a^{3} e^{\left (2 \, x\right )} + 16 i \, a^{3} e^{x} + 4 \, a^{3}\right )} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (-\frac {\sqrt {\frac {1}{2}} a^{3} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} - \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (3 i \, A + 5 \, B\right )}}{3 i \, A + 5 \, B}\right ) + 8 \, {\left ({\left (-3 i \, A - 5 \, B\right )} e^{\left (4 \, x\right )} - {\left (11 \, A + 3 i \, B\right )} e^{\left (3 \, x\right )} + {\left (-11 i \, A + 3 \, B\right )} e^{\left (2 \, x\right )} - {\left (3 \, A - 5 i \, B\right )} e^{x}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}}}{8 \, {\left (8 \, a^{3} e^{\left (4 \, x\right )} - 32 i \, a^{3} e^{\left (3 \, x\right )} - 48 \, a^{3} e^{\left (2 \, x\right )} + 32 i \, a^{3} e^{x} + 8 \, a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(1/2)*(4*a^3*e^(4*x) - 16*I*a^3*e^(3*x) - 24*a^3*e^(2*x) + 16*I*a^3*e^x + 4*a^3)*sqrt(-(9*A^2 - 30*I*
A*B - 25*B^2)/a^5)*log((sqrt(1/2)*a^3*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5) + sqrt(1/2*I*a*e^(-x))*(3*I*A + 5
*B))/(3*I*A + 5*B)) - sqrt(1/2)*(4*a^3*e^(4*x) - 16*I*a^3*e^(3*x) - 24*a^3*e^(2*x) + 16*I*a^3*e^x + 4*a^3)*sqr
t(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5)*log(-(sqrt(1/2)*a^3*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5) - sqrt(1/2*I*a*
e^(-x))*(3*I*A + 5*B))/(3*I*A + 5*B)) + 8*((-3*I*A - 5*B)*e^(4*x) - (11*A + 3*I*B)*e^(3*x) + (-11*I*A + 3*B)*e
^(2*x) - (3*A - 5*I*B)*e^x)*sqrt(1/2*I*a*e^(-x)))/(8*a^3*e^(4*x) - 32*I*a^3*e^(3*x) - 48*a^3*e^(2*x) + 32*I*a^
3*e^x + 8*a^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sinh \relax (x) + A}{{\left (i \, a \sinh \relax (x) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sinh(x) + A)/(I*a*sinh(x) + a)^(5/2), x)

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \frac {A +B \sinh \relax (x )}{\left (a +i a \sinh \relax (x )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x)

[Out]

int((A+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sinh \relax (x) + A}{{\left (i \, a \sinh \relax (x) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sinh(x) + A)/(I*a*sinh(x) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\mathrm {sinh}\relax (x)}{{\left (a+a\,\mathrm {sinh}\relax (x)\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sinh(x))/(a + a*sinh(x)*1i)^(5/2),x)

[Out]

int((A + B*sinh(x))/(a + a*sinh(x)*1i)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(a+I*a*sinh(x))**(5/2),x)

[Out]

Timed out

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