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

Optimal. Leaf size=259 \[ -\frac{2 i \left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},\frac{2 b}{b+i a}\right )}{105 b \sqrt{a+b \sinh (x)}}+\frac{2}{105} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt{a+b \sinh (x)}+\frac{2 i \left (161 a^2 A b+15 a^3 B-145 a b^2 B-63 A b^3\right ) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{105 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2}{35} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac{2}{7} B \cosh (x) (a+b \sinh (x))^{5/2} \]

[Out]

(2*(56*a*A*b + 15*a^2*B - 25*b^2*B)*Cosh[x]*Sqrt[a + b*Sinh[x]])/105 + (2*(7*A*b + 5*a*B)*Cosh[x]*(a + b*Sinh[
x])^(3/2))/35 + (2*B*Cosh[x]*(a + b*Sinh[x])^(5/2))/7 + (((2*I)/105)*(161*a^2*A*b - 63*A*b^3 + 15*a^3*B - 145*
a*b^2*B)*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/(b*Sqrt[(a + b*Sinh[x])/(a - I*b)]) -
(((2*I)/105)*(a^2 + b^2)*(56*a*A*b + 15*a^2*B - 25*b^2*B)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[(a
+ b*Sinh[x])/(a - I*b)])/(b*Sqrt[a + b*Sinh[x]])

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Rubi [A]  time = 0.446324, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2}{105} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt{a+b \sinh (x)}-\frac{2 i \left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{105 b \sqrt{a+b \sinh (x)}}+\frac{2 i \left (161 a^2 A b+15 a^3 B-145 a b^2 B-63 A b^3\right ) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{105 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2}{35} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac{2}{7} B \cosh (x) (a+b \sinh (x))^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(56*a*A*b + 15*a^2*B - 25*b^2*B)*Cosh[x]*Sqrt[a + b*Sinh[x]])/105 + (2*(7*A*b + 5*a*B)*Cosh[x]*(a + b*Sinh[
x])^(3/2))/35 + (2*B*Cosh[x]*(a + b*Sinh[x])^(5/2))/7 + (((2*I)/105)*(161*a^2*A*b - 63*A*b^3 + 15*a^3*B - 145*
a*b^2*B)*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/(b*Sqrt[(a + b*Sinh[x])/(a - I*b)]) -
(((2*I)/105)*(a^2 + b^2)*(56*a*A*b + 15*a^2*B - 25*b^2*B)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[(a
+ b*Sinh[x])/(a - I*b)])/(b*Sqrt[a + b*Sinh[x]])

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]

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 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 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 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 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]

Rubi steps

\begin{align*} \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx &=\frac{2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac{2}{7} \int (a+b \sinh (x))^{3/2} \left (\frac{1}{2} (7 a A-5 b B)+\frac{1}{2} (7 A b+5 a B) \sinh (x)\right ) \, dx\\ &=\frac{2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac{2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac{4}{35} \int \sqrt{a+b \sinh (x)} \left (\frac{1}{4} \left (35 a^2 A-21 A b^2-40 a b B\right )+\frac{1}{4} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \sinh (x)\right ) \, dx\\ &=\frac{2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac{2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac{8}{105} \int \frac{\frac{1}{8} \left (105 a^3 A-119 a A b^2-135 a^2 b B+25 b^3 B\right )+\frac{1}{8} \left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx\\ &=\frac{2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac{2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}-\frac{\left (\left (a^2+b^2\right ) \left (56 a A b+15 a^2 B-25 b^2 B\right )\right ) \int \frac{1}{\sqrt{a+b \sinh (x)}} \, dx}{105 b}+\frac{\left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) \int \sqrt{a+b \sinh (x)} \, dx}{105 b}\\ &=\frac{2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac{2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac{\left (\left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) \sqrt{a+b \sinh (x)}\right ) \int \sqrt{\frac{a}{a-i b}+\frac{b \sinh (x)}{a-i b}} \, dx}{105 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{\left (\left (a^2+b^2\right ) \left (56 a A b+15 a^2 B-25 b^2 B\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}{105 b \sqrt{a+b \sinh (x)}}\\ &=\frac{2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac{2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac{2 i \left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right ) \sqrt{a+b \sinh (x)}}{105 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{2 i \left (a^2+b^2\right ) \left (56 a A b+15 a^2 B-25 b^2 B\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}}}{105 b \sqrt{a+b \sinh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.771175, size = 241, normalized size = 0.93 \[ \frac{\cosh (x) (a+b \sinh (x)) \left (90 a^2 B+6 b \sinh (x) (15 a B+7 A b)+154 a A b+15 b^2 B \cosh (2 x)-65 b^2 B\right )+\frac{2 i \sqrt{\frac{a+b \sinh (x)}{a-i b}} \left (b \left (105 a^3 A-135 a^2 b B-119 a A b^2+25 b^3 B\right ) \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )+\left (161 a^2 A b+15 a^3 B-145 a b^2 B-63 A b^3\right ) \left ((a-i b) E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )-a \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )\right )\right )}{b}}{105 \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)*(b*(105*a^3*A - 119*a*A*b^2 - 135*a^2*b*B + 25*b^3*B)*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)
] + (161*a^2*A*b - 63*A*b^3 + 15*a^3*B - 145*a*b^2*B)*((a - I*b)*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I
*b)] - a*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]))*Sqrt[(a + b*Sinh[x])/(a - I*b)])/b + Cosh[x]*(a +
b*Sinh[x])*(154*a*A*b + 90*a^2*B - 65*b^2*B + 15*b^2*B*Cosh[2*x] + 6*b*(7*A*b + 15*a*B)*Sinh[x]))/(105*Sqrt[a
+ b*Sinh[x]])

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Maple [B]  time = 0.108, size = 1893, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(56*I*A*(-(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)*Ellip
ticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a*b^4+63*A*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-si
nh(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^5-25*I*B*(-(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))*b^5+98*A*a*b^4*sinh(x)+90*B*a^2*b^3*sinh
(x)+60*B*a*b^4*sinh(x)^4+98*A*a*b^4*sinh(x)^3+90*B*a^2*b^3*sinh(x)^3+77*A*a^2*b^3*sinh(x)^2+45*B*a^3*b^2*sinh(
x)^2+35*B*a*b^4*sinh(x)^2-15*B*(-(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))*a^5-63*A*(-(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))*b^5-10*I*B*(-(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^3+56*I*A*(-(a+b*sin
h(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^2+77*A*a^2*b^3+45*B*a^3*b^2-25*B*a*b^4+15*I*B*(-(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^4*b+15*B*b^5*sinh(x)^5+21*A*b^5*sinh(x)^4-10*B*b^5*sinh(x)^3+21*A*b^5*sinh(x)^2-
25*B*b^5*sinh(x)+42*A*(-(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^3-98*A*(-(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))*a^2*b^3-120*B*(-(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^2-120*B*(-(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^4+130*B*(-(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))*a^3*b^2+145*B*(
-(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*sin
h(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a*b^4+105*A*(-(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^4
*b-161*A*(-(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))*a^4*b)/b^2/cosh(x)/(a+b*sinh(x))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sinh \left (x\right ) + A\right )}{\left (b \sinh \left (x\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{2} \sinh \left (x\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \sinh \left (x\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sinh \left (x\right )\right )} \sqrt{b \sinh \left (x\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*sinh(x)^3 + A*a^2 + (2*B*a*b + A*b^2)*sinh(x)^2 + (B*a^2 + 2*A*a*b)*sinh(x))*sqrt(b*sinh(x) +
a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sinh \left (x\right ) + A\right )}{\left (b \sinh \left (x\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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