3.127 \(\int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx\)

Optimal. Leaf size=207 \[ -\frac{2 i \left (a^2+b^2\right ) (3 a B+5 A b) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},\frac{2 b}{b+i a}\right )}{15 b \sqrt{a+b \sinh (x)}}+\frac{2 i \left (3 a^2 B+20 a A b-9 b^2 B\right ) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{15 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2}{15} \cosh (x) (3 a B+5 A b) \sqrt{a+b \sinh (x)}+\frac{2}{5} B \cosh (x) (a+b \sinh (x))^{3/2} \]

[Out]

(2*(5*A*b + 3*a*B)*Cosh[x]*Sqrt[a + b*Sinh[x]])/15 + (2*B*Cosh[x]*(a + b*Sinh[x])^(3/2))/5 + (((2*I)/15)*(20*a
*A*b + 3*a^2*B - 9*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)/15)*(a^2 + b^2)*(5*A*b + 3*a*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 = 0.311112, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{2 i \left (a^2+b^2\right ) (3 a B+5 A b) \sqrt{\frac{a+b \sinh (x)}{a-i b}} F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{15 b \sqrt{a+b \sinh (x)}}+\frac{2 i \left (3 a^2 B+20 a A b-9 b^2 B\right ) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{15 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2}{15} \cosh (x) (3 a B+5 A b) \sqrt{a+b \sinh (x)}+\frac{2}{5} B \cosh (x) (a+b \sinh (x))^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[x])^(3/2)*(A + B*Sinh[x]),x]

[Out]

(2*(5*A*b + 3*a*B)*Cosh[x]*Sqrt[a + b*Sinh[x]])/15 + (2*B*Cosh[x]*(a + b*Sinh[x])^(3/2))/5 + (((2*I)/15)*(20*a
*A*b + 3*a^2*B - 9*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)/15)*(a^2 + b^2)*(5*A*b + 3*a*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))^{3/2} (A+B \sinh (x)) \, 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} (5 a A-3 b B)+\frac{1}{2} (5 A b+3 a B) \sinh (x)\right ) \, dx\\ &=\frac{2}{15} (5 A b+3 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} \left (15 a^2 A-5 A b^2-12 a b B\right )+\frac{1}{4} \left (20 a A b+3 a^2 B-9 b^2 B\right ) \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx\\ &=\frac{2}{15} (5 A b+3 a B) \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}-\frac{\left (\left (a^2+b^2\right ) (5 A b+3 a B)\right ) \int \frac{1}{\sqrt{a+b \sinh (x)}} \, dx}{15 b}+\frac{\left (20 a A b+3 a^2 B-9 b^2 B\right ) \int \sqrt{a+b \sinh (x)} \, dx}{15 b}\\ &=\frac{2}{15} (5 A b+3 a B) \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac{\left (\left (20 a A b+3 a^2 B-9 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}{15 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{\left (\left (a^2+b^2\right ) (5 A b+3 a B) \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 b \sqrt{a+b \sinh (x)}}\\ &=\frac{2}{15} (5 A b+3 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 (20 a A b+3 a^2 B-9 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)}}{15 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{2 i \left (a^2+b^2\right ) (5 A b+3 a B) 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 b \sqrt{a+b \sinh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.549874, size = 196, normalized size = 0.95 \[ \frac{2 \left (\cosh (x) (a+b \sinh (x)) (6 a B+5 A b+3 b B \sinh (x))+\frac{i \sqrt{\frac{a+b \sinh (x)}{a-i b}} \left (b \left (15 a^2 A-12 a b B-5 A b^2\right ) \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )+\left (3 a^2 B+20 a A b-9 b^2 B\right ) \left ((a-i b) E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )-a \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )\right )\right )}{b}\right )}{15 \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[x])^(3/2)*(A + B*Sinh[x]),x]

[Out]

(2*((I*(b*(15*a^2*A - 5*A*b^2 - 12*a*b*B)*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)] + (20*a*A*b + 3*a^
2*B - 9*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])*(5*A*b + 6*a*B + 3*b*B*Sinh
[x])))/(15*Sqrt[a + b*Sinh[x]])

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Maple [B]  time = 0.157, size = 1037, normalized size = 5. \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))^(3/2)*(A+B*sinh(x)),x)

[Out]

((a+b*sinh(x))*cosh(x)^2)^(1/2)*(B*b^2*(2/5/b*sinh(x)*((a+b*sinh(x))*cosh(x)^2)^(1/2)-8/15*a/b^2*((a+b*sinh(x)
)*cosh(x)^2)^(1/2)-4/15*a/b*(a/b-I)*((-b*sinh(x)-a)/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*
b/(I*b-a))^(1/2)/((a+b*sinh(x))*cosh(x)^2)^(1/2)*EllipticF(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1
/2))+2*(-3/5+8/15*a^2/b^2)*(a/b-I)*((-b*sinh(x)-a)/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b
/(I*b-a))^(1/2)/((a+b*sinh(x))*cosh(x)^2)^(1/2)*((-a/b-I)*EllipticE(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I
*b+a))^(1/2))+I*EllipticF(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))))+(A*b^2+2*B*a*b)*(2/3/b*((a
+b*sinh(x))*cosh(x)^2)^(1/2)-2/3*(a/b-I)*((-b*sinh(x)-a)/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh
(x))*b/(I*b-a))^(1/2)/((a+b*sinh(x))*cosh(x)^2)^(1/2)*EllipticF(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a
))^(1/2))-4/3*a/b*(a/b-I)*((-b*sinh(x)-a)/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))
^(1/2)/((a+b*sinh(x))*cosh(x)^2)^(1/2)*((-a/b-I)*EllipticE(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1
/2))+I*EllipticF(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))))+2*(2*A*a*b+B*a^2)*(a/b-I)*((-b*sinh
(x)-a)/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)/((a+b*sinh(x))*cosh(x)^2)^(1
/2)*((-a/b-I)*EllipticE(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))+I*EllipticF(((-b*sinh(x)-a)/(I
*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2)))+2*a^2*A*(a/b-I)*((-b*sinh(x)-a)/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^
(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)/((a+b*sinh(x))*cosh(x)^2)^(1/2)*EllipticF(((-b*sinh(x)-a)/(I*b-a))^(1/2),(
(a-I*b)/(I*b+a))^(1/2)))/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{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(x))^(3/2)*(A+B*sinh(x)),x, algorithm="maxima")

[Out]

integrate((B*sinh(x) + A)*(b*sinh(x) + a)^(3/2), x)

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sinh{\left (x \right )}\right ) \left (a + b \sinh{\left (x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(x))**(3/2)*(A+B*sinh(x)),x)

[Out]

Integral((A + B*sinh(x))*(a + b*sinh(x))**(3/2), x)

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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{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(x))^(3/2)*(A+B*sinh(x)),x, algorithm="giac")

[Out]

integrate((B*sinh(x) + A)*(b*sinh(x) + a)^(3/2), x)