3.105 \(\int (a+b \sinh (x))^{5/2} \, dx\)

Optimal. Leaf size=179 \[ -\frac {16 i a \left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{15 \sqrt {a+b \sinh (x)}}+\frac {2 i \left (23 a^2-9 b^2\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{15 \sqrt {\frac {a+b \sinh (x)}{a-i b}}}+\frac {2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac {16}{15} a b \cosh (x) \sqrt {a+b \sinh (x)} \]

[Out]

2/5*b*cosh(x)*(a+b*sinh(x))^(3/2)+16/15*a*b*cosh(x)*(a+b*sinh(x))^(1/2)+2/15*I*(23*a^2-9*b^2)*(sin(1/4*Pi+1/2*
I*x)^2)^(1/2)/sin(1/4*Pi+1/2*I*x)*EllipticE(cos(1/4*Pi+1/2*I*x),2^(1/2)*(b/(I*a+b))^(1/2))*(a+b*sinh(x))^(1/2)
/((a+b*sinh(x))/(a-I*b))^(1/2)-16/15*I*a*(a^2+b^2)*(sin(1/4*Pi+1/2*I*x)^2)^(1/2)/sin(1/4*Pi+1/2*I*x)*EllipticF
(cos(1/4*Pi+1/2*I*x),2^(1/2)*(b/(I*a+b))^(1/2))*((a+b*sinh(x))/(a-I*b))^(1/2)/(a+b*sinh(x))^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {2656, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {16 i a \left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{15 \sqrt {a+b \sinh (x)}}+\frac {2 i \left (23 a^2-9 b^2\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{15 \sqrt {\frac {a+b \sinh (x)}{a-i b}}}+\frac {2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac {16}{15} a b \cosh (x) \sqrt {a+b \sinh (x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[x])^(5/2),x]

[Out]

(16*a*b*Cosh[x]*Sqrt[a + b*Sinh[x]])/15 + (2*b*Cosh[x]*(a + b*Sinh[x])^(3/2))/5 + (((2*I)/15)*(23*a^2 - 9*b^2)
*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/Sqrt[(a + b*Sinh[x])/(a - I*b)] - (((16*I)/15
)*a*(a^2 + b^2)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[(a + b*Sinh[x])/(a - I*b)])/Sqrt[a + b*Sinh[x]
]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

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

Rule 2656

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[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*
x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

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

Rule 2752

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

Rule 2753

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

Rubi steps

\begin {align*} \int (a+b \sinh (x))^{5/2} \, dx &=\frac {2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2}{5} \int \sqrt {a+b \sinh (x)} \left (\frac {1}{2} \left (5 a^2-3 b^2\right )+4 a b \sinh (x)\right ) \, dx\\ &=\frac {16}{15} a b \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac {4}{15} \int \frac {\frac {1}{4} a \left (15 a^2-17 b^2\right )+\frac {1}{4} b \left (23 a^2-9 b^2\right ) \sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx\\ &=\frac {16}{15} a b \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{15} \left (23 a^2-9 b^2\right ) \int \sqrt {a+b \sinh (x)} \, dx-\frac {1}{15} \left (8 a \left (a^2+b^2\right )\right ) \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx\\ &=\frac {16}{15} a b \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac {\left (\left (23 a^2-9 b^2\right ) \sqrt {a+b \sinh (x)}\right ) \int \sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}} \, dx}{15 \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (8 a \left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}} \, dx}{15 \sqrt {a+b \sinh (x)}}\\ &=\frac {16}{15} a b \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2 i \left (23 a^2-9 b^2\right ) E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {a+b \sinh (x)}}{15 \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {16 i a \left (a^2+b^2\right ) F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{15 \sqrt {a+b \sinh (x)}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 178, normalized size = 0.99 \[ \frac {b \cosh (x) \left (22 a^2+28 a b \sinh (x)+3 b^2 \cosh (2 x)-3 b^2\right )-16 i a \left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} F\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )+2 \left (23 i a^3+23 a^2 b-9 i a b^2-9 b^3\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} E\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )}{15 \sqrt {a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[x])^(5/2),x]

[Out]

(2*((23*I)*a^3 + 23*a^2*b - (9*I)*a*b^2 - 9*b^3)*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a + b
*Sinh[x])/(a - I*b)] - (16*I)*a*(a^2 + b^2)*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a + b*Sinh
[x])/(a - I*b)] + b*Cosh[x]*(22*a^2 - 3*b^2 + 3*b^2*Cosh[2*x] + 28*a*b*Sinh[x]))/(15*Sqrt[a + b*Sinh[x]])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \sinh \relax (x)^{2} + 2 \, a b \sinh \relax (x) + a^{2}\right )} \sqrt {b \sinh \relax (x) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*sinh(x)^2 + 2*a*b*sinh(x) + a^2)*sqrt(b*sinh(x) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(x) + a)^(5/2), x)

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maple [B]  time = 0.20, size = 917, normalized size = 5.12 \[ \frac {\frac {16 i \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \EllipticF \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a^{3} b}{15}+\frac {16 i \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \EllipticF \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a \,b^{3}}{15}+2 \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \EllipticF \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a^{4}+\frac {4 \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \EllipticF \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a^{2} b^{2}}{5}-\frac {6 \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \EllipticF \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) b^{4}}{5}-\frac {46 \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \EllipticE \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a^{4}}{15}-\frac {28 \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \EllipticE \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a^{2} b^{2}}{15}+\frac {6 \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \EllipticE \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) b^{4}}{5}+\frac {2 b^{4} \left (\sinh ^{4}\relax (x )\right )}{5}+\frac {28 a \,b^{3} \left (\sinh ^{3}\relax (x )\right )}{15}+\frac {22 a^{2} b^{2} \left (\sinh ^{2}\relax (x )\right )}{15}+\frac {2 b^{4} \left (\sinh ^{2}\relax (x )\right )}{5}+\frac {28 a \,b^{3} \sinh \relax (x )}{15}+\frac {22 a^{2} b^{2}}{15}}{b \cosh \relax (x ) \sqrt {a +b \sinh \relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sinh(x))^(5/2),x)

[Out]

2/15*(8*I*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticF
((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^3*b+8*I*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x)
)*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(
1/2))*a*b^3+15*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*Elli
pticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^4+6*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x
))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^
(1/2))*a^2*b^2-9*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*El
lipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*b^4-23*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sin
h(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticE((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a
))^(1/2))*a^4-14*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*El
lipticE((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^2*b^2+9*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-
sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticE((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*
b+a))^(1/2))*b^4+3*b^4*sinh(x)^4+14*a*b^3*sinh(x)^3+11*a^2*b^2*sinh(x)^2+3*b^4*sinh(x)^2+14*a*b^3*sinh(x)+11*a
^2*b^2)/b/cosh(x)/(a+b*sinh(x))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(x) + a)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(x))^(5/2),x)

[Out]

int((a + b*sinh(x))^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(x))**(5/2),x)

[Out]

Integral((a + b*sinh(x))**(5/2), x)

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