\(\int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx\) [125]

Optimal result

Integrand size = 20, antiderivative size = 110 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {(3 i A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}} \]

[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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2829, 2729, 2728, 212} \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {(5 B+3 i A) \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(5 B+3 i A) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac {(-B+i A) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}} \]

[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 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 2728

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 2729

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 2829

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*} \text {integral}& = \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) \text {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) \text {arctanh}\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*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.67 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right ) \left (4 i (A+i B) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )+(3 i A+5 B) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^3+(1-i) \sqrt [4]{-1} (3 A-5 i B) \arctan \left (\frac {i+\tanh \left (\frac {x}{4}\right )}{\sqrt {2}}\right ) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^4+8 (A+i B) \sinh \left (\frac {x}{2}\right )+2 (3 i A+5 B) \sinh \left (\frac {x}{2}\right ) (-i+\sinh (x))\right )}{16 (a+i a \sinh (x))^{5/2}} \]

[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))

Maple [F]

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

[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)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (77) = 154\).

Time = 0.30 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.15 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (\frac {\sqrt {\frac {1}{2}} a^{3} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (3 i \, A + 5 \, B\right )}}{3 i \, A + 5 \, B}\right ) - \sqrt {\frac {1}{2}} {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (-\frac {\sqrt {\frac {1}{2}} a^{3} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} - \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (3 i \, A + 5 \, B\right )}}{3 i \, A + 5 \, B}\right ) + 2 \, {\left ({\left (-3 i \, A - 5 \, B\right )} e^{\left (4 \, x\right )} - {\left (11 \, A + 3 i \, B\right )} e^{\left (3 \, x\right )} + {\left (-11 i \, A + 3 \, B\right )} e^{\left (2 \, x\right )} - {\left (3 \, A - 5 i \, B\right )} e^{x}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}}}{16 \, {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )}} \]

[In]

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

[Out]

1/16*(sqrt(1/2)*(a^3*e^(4*x) - 4*I*a^3*e^(3*x) - 6*a^3*e^(2*x) + 4*I*a^3*e^x + 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)*(a^3*e^(4*x) - 4*I*a^3*e^(3*x) - 6*a^3*e^(2*x) + 4*I*a^3*e^x + a^3)*sqrt(-(9*A^2 - 3
0*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)) + 2*((-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)))/(a^3*e^(4*x) - 4*I*a^3*e^(3*x) - 6*a^3*e^(2*x) + 4*I*a^3*e^x + a^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[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)