3.83 \(\int \sinh ^2(e+f x) (a+b \sinh ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=236 \[ \frac{i a (3 a-4 b) (a-b) \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1} \text{EllipticF}\left (i e+i f x,\frac{b}{a}\right )}{15 b f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{i \left (3 a^2-13 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{15 b f \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}}+\frac{\sinh (e+f x) \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 f}+\frac{(3 a-4 b) \sinh (e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{15 f} \]

[Out]

((3*a - 4*b)*Cosh[e + f*x]*Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(15*f) + (Cosh[e + f*x]*Sinh[e + f*x]*(a
+ b*Sinh[e + f*x]^2)^(3/2))/(5*f) - ((I/15)*(3*a^2 - 13*a*b + 8*b^2)*EllipticE[I*e + I*f*x, b/a]*Sqrt[a + b*S
inh[e + f*x]^2])/(b*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) + ((I/15)*a*(3*a - 4*b)*(a - b)*EllipticF[I*e + I*f*x,
b/a]*Sqrt[1 + (b*Sinh[e + f*x]^2)/a])/(b*f*Sqrt[a + b*Sinh[e + f*x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.325431, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3170, 3172, 3178, 3177, 3183, 3182} \[ -\frac{i \left (3 a^2-13 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{15 b f \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}}+\frac{\sinh (e+f x) \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 f}+\frac{(3 a-4 b) \sinh (e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{15 f}+\frac{i a (3 a-4 b) (a-b) \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1} F\left (i e+i f x\left |\frac{b}{a}\right .\right )}{15 b f \sqrt{a+b \sinh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a - 4*b)*Cosh[e + f*x]*Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(15*f) + (Cosh[e + f*x]*Sinh[e + f*x]*(a
+ b*Sinh[e + f*x]^2)^(3/2))/(5*f) - ((I/15)*(3*a^2 - 13*a*b + 8*b^2)*EllipticE[I*e + I*f*x, b/a]*Sqrt[a + b*S
inh[e + f*x]^2])/(b*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) + ((I/15)*a*(3*a - 4*b)*(a - b)*EllipticF[I*e + I*f*x,
b/a]*Sqrt[1 + (b*Sinh[e + f*x]^2)/a])/(b*f*Sqrt[a + b*Sinh[e + f*x]^2])

Rule 3170

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[(B*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(2*f*(p + 1)), x] + Dist[1/(2*(p + 1)), Int[(a + b*Si
n[e + f*x]^2)^(p - 1)*Simp[a*B + 2*a*A*(p + 1) + (2*A*b*(p + 1) + B*(b + 2*a*p + 2*b*p))*Sin[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, e, f, A, B}, x] && GtQ[p, 0]

Rule 3172

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[
B/b, Int[Sqrt[a + b*Sin[e + f*x]^2], x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /;
FreeQ[{a, b, e, f, A, B}, x]

Rule 3178

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

Rule 3177

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[e + f*x, -(b/a)])/f, x]
/; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3183

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

Rule 3182

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1*EllipticF[e + f*x, -(b/a)])/(Sqrt[a]*
f), x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \sinh ^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx &=\frac{\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 f}-\frac{1}{5} \int \left (a-(3 a-4 b) \sinh ^2(e+f x)\right ) \sqrt{a+b \sinh ^2(e+f x)} \, dx\\ &=\frac{(3 a-4 b) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{15 f}+\frac{\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 f}-\frac{1}{15} \int \frac{2 a (3 a-2 b)-\left (3 a^2-13 a b+8 b^2\right ) \sinh ^2(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\\ &=\frac{(3 a-4 b) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{15 f}+\frac{\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 f}-\frac{(a (3 a-4 b) (a-b)) \int \frac{1}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx}{15 b}+\frac{\left (3 a^2-13 a b+8 b^2\right ) \int \sqrt{a+b \sinh ^2(e+f x)} \, dx}{15 b}\\ &=\frac{(3 a-4 b) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{15 f}+\frac{\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 f}+\frac{\left (\left (3 a^2-13 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(e+f x)}\right ) \int \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}} \, dx}{15 b \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}}-\frac{\left (a (3 a-4 b) (a-b) \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}\right ) \int \frac{1}{\sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}} \, dx}{15 b \sqrt{a+b \sinh ^2(e+f x)}}\\ &=\frac{(3 a-4 b) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{15 f}+\frac{\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 f}-\frac{i \left (3 a^2-13 a b+8 b^2\right ) E\left (i e+i f x\left |\frac{b}{a}\right .\right ) \sqrt{a+b \sinh ^2(e+f x)}}{15 b f \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}}+\frac{i a (3 a-4 b) (a-b) F\left (i e+i f x\left |\frac{b}{a}\right .\right ) \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}}{15 b f \sqrt{a+b \sinh ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.36271, size = 213, normalized size = 0.9 \[ \frac{16 i a \left (3 a^2-7 a b+4 b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+\sqrt{2} b \sinh (2 (e+f x)) \left (48 a^2+4 b (9 a-7 b) \cosh (2 (e+f x))-68 a b+3 b^2 \cosh (4 (e+f x))+25 b^2\right )-16 i a \left (3 a^2-13 a b+8 b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{240 b f \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*I)*a*(3*a^2 - 13*a*b + 8*b^2)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] + (16*
I)*a*(3*a^2 - 7*a*b + 4*b^2)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] + Sqrt[2]*b*(
48*a^2 - 68*a*b + 25*b^2 + 4*(9*a - 7*b)*b*Cosh[2*(e + f*x)] + 3*b^2*Cosh[4*(e + f*x)])*Sinh[2*(e + f*x)])/(24
0*b*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

________________________________________________________________________________________

Maple [A]  time = 0.071, size = 535, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(f*x+e)^2*(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

-1/15*(-3*(-1/a*b)^(1/2)*b^2*sinh(f*x+e)*cosh(f*x+e)^6+(-9*(-1/a*b)^(1/2)*a*b+10*(-1/a*b)^(1/2)*b^2)*cosh(f*x+
e)^4*sinh(f*x+e)+(-6*(-1/a*b)^(1/2)*a^2+13*(-1/a*b)^(1/2)*a*b-7*(-1/a*b)^(1/2)*b^2)*cosh(f*x+e)^2*sinh(f*x+e)+
9*a^2*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2)
)-17*a*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2
))*b+8*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2
))*b^2-3*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1
/2))*a^2+13*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)
^(1/2))*a*b-8*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/
b)^(1/2))*b^2)/(-1/a*b)^(1/2)/cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)/f

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(f*x + e)^2 + a)^(3/2)*sinh(f*x + e)^2, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sinh \left (f x + e\right )^{4} + a \sinh \left (f x + e\right )^{2}\right )} \sqrt{b \sinh \left (f x + e\right )^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sinh(f*x + e)^4 + a*sinh(f*x + e)^2)*sqrt(b*sinh(f*x + e)^2 + a), x)

________________________________________________________________________________________

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(sinh(f*x+e)**2*(a+b*sinh(f*x+e)**2)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sinh \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(f*x + e)^2 + a)^(3/2)*sinh(f*x + e)^2, x)