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

Optimal. Leaf size=176 \[ \frac{2 i B \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},\frac{2 b}{b+i a}\right )}{b \sqrt{a+b \sinh (x)}}-\frac{2 \cosh (x) (A b-a B)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{2 i (A b-a B) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{b \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

[Out]

(-2*(A*b - a*B)*Cosh[x])/((a^2 + b^2)*Sqrt[a + b*Sinh[x]]) + ((2*I)*(A*b - a*B)*EllipticE[Pi/4 - (I/2)*x, (2*b
)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/(b*(a^2 + b^2)*Sqrt[(a + b*Sinh[x])/(a - I*b)]) + ((2*I)*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 = 0.226655, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \cosh (x) (A b-a B)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{2 i (A b-a B) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{b \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2 i B \sqrt{\frac{a+b \sinh (x)}{a-i b}} F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{b \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*b - a*B)*Cosh[x])/((a^2 + b^2)*Sqrt[a + b*Sinh[x]]) + ((2*I)*(A*b - a*B)*EllipticE[Pi/4 - (I/2)*x, (2*b
)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/(b*(a^2 + b^2)*Sqrt[(a + b*Sinh[x])/(a - I*b)]) + ((2*I)*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 2754

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 + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*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] && LtQ[m, -1] && 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 \frac{A+B \sinh (x)}{(a+b \sinh (x))^{3/2}} \, dx &=-\frac{2 (A b-a B) \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}-\frac{2 \int \frac{\frac{1}{2} (-a A-b B)-\frac{1}{2} (A b-a B) \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx}{a^2+b^2}\\ &=-\frac{2 (A b-a B) \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{B \int \frac{1}{\sqrt{a+b \sinh (x)}} \, dx}{b}+\frac{(A b-a B) \int \sqrt{a+b \sinh (x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{2 (A b-a B) \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{\left ((A b-a B) \sqrt{a+b \sinh (x)}\right ) \int \sqrt{\frac{a}{a-i b}+\frac{b \sinh (x)}{a-i b}} \, dx}{b \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{\left (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}{b \sqrt{a+b \sinh (x)}}\\ &=-\frac{2 (A b-a B) \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{2 i (A b-a B) E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right ) \sqrt{a+b \sinh (x)}}{b \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2 i 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}}}{b \sqrt{a+b \sinh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.618416, size = 159, normalized size = 0.9 \[ \frac{2 i B \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )+2 b \cosh (x) (a B-A b)+\frac{2 i (A b-a B) (a+b \sinh (x)) E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )}{\sqrt{\frac{a+b \sinh (x)}{a-i b}}}}{b \left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(-(A*b) + a*B)*Cosh[x] + ((2*I)*(A*b - a*B)*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*(a + b*Sinh
[x]))/Sqrt[(a + b*Sinh[x])/(a - I*b)] + (2*I)*(a^2 + b^2)*B*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*
Sqrt[(a + b*Sinh[x])/(a - I*b)])/(b*(a^2 + b^2)*Sqrt[a + b*Sinh[x]])

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Maple [B]  time = 0.163, size = 517, normalized size = 2.9 \begin{align*}{\frac{1}{\cosh \left ( x \right ) }\sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}} \left ( 2\,{\frac{B}{b\sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}}} \left ({\frac{a}{b}}-i \right ) \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}}\sqrt{{\frac{ \left ( i-\sinh \left ( x \right ) \right ) b}{ib+a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( x \right ) \right ) b}{ib-a}}}{\it EllipticF} \left ( \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}},\sqrt{{\frac{a-ib}{ib+a}}} \right ) }+{\frac{Ab-aB}{b} \left ( -2\,{\frac{b \left ( \cosh \left ( x \right ) \right ) ^{2}}{ \left ({a}^{2}+{b}^{2} \right ) \sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}}}}+2\,{\frac{a}{ \left ({a}^{2}+{b}^{2} \right ) \sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}}} \left ({\frac{a}{b}}-i \right ) \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}}\sqrt{{\frac{ \left ( i-\sinh \left ( x \right ) \right ) b}{ib+a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( x \right ) \right ) b}{ib-a}}}{\it EllipticF} \left ( \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}},\sqrt{{\frac{a-ib}{ib+a}}} \right ) }+2\,{\frac{b}{ \left ({a}^{2}+{b}^{2} \right ) \sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}}} \left ({\frac{a}{b}}-i \right ) \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}}\sqrt{{\frac{ \left ( i-\sinh \left ( x \right ) \right ) b}{ib+a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( x \right ) \right ) b}{ib-a}}} \left ( \left ( -{\frac{a}{b}}-i \right ){\it EllipticE} \left ( \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}},\sqrt{{\frac{a-ib}{ib+a}}} \right ) +i{\it EllipticF} \left ( \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}},\sqrt{{\frac{a-ib}{ib+a}}} \right ) \right ) } \right ) } \right ){\frac{1}{\sqrt{a+b\sinh \left ( x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a+b*sinh(x))*cosh(x)^2)^(1/2)*(2*B/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))+(A*b-B*a)/b*(-2*b*cosh(x)^2/(a^2+b^2)/((a+b*sinh(x))*cosh(x)^2)^(1/2)+2*a/(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)*EllipticF(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))+2*b/(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)))))/cosh(x)/(a+b*sinh(x))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

[Out]

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