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

Optimal. Leaf size=232 \[ \frac{4 i a (a-b) (2 a-b) \sqrt{\frac{b \sinh ^2(c+d x)}{a}+1} \text{EllipticF}\left (i c+i d x,\frac{b}{a}\right )}{15 d \sqrt{a+b \sinh ^2(c+d x)}}-\frac{i \left (23 a^2-23 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(c+d x)} E\left (i c+i d x\left |\frac{b}{a}\right .\right )}{15 d \sqrt{\frac{b \sinh ^2(c+d x)}{a}+1}}+\frac{b \sinh (c+d x) \cosh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{4 b (2 a-b) \sinh (c+d x) \cosh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d} \]

[Out]

(4*(2*a - b)*b*Cosh[c + d*x]*Sinh[c + d*x]*Sqrt[a + b*Sinh[c + d*x]^2])/(15*d) + (b*Cosh[c + d*x]*Sinh[c + d*x
]*(a + b*Sinh[c + d*x]^2)^(3/2))/(5*d) - ((I/15)*(23*a^2 - 23*a*b + 8*b^2)*EllipticE[I*c + I*d*x, b/a]*Sqrt[a
+ b*Sinh[c + d*x]^2])/(d*Sqrt[1 + (b*Sinh[c + d*x]^2)/a]) + (((4*I)/15)*a*(a - b)*(2*a - b)*EllipticF[I*c + I*
d*x, b/a]*Sqrt[1 + (b*Sinh[c + d*x]^2)/a])/(d*Sqrt[a + b*Sinh[c + d*x]^2])

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Rubi [A]  time = 0.303784, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3180, 3170, 3172, 3178, 3177, 3183, 3182} \[ -\frac{i \left (23 a^2-23 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(c+d x)} E\left (i c+i d x\left |\frac{b}{a}\right .\right )}{15 d \sqrt{\frac{b \sinh ^2(c+d x)}{a}+1}}+\frac{b \sinh (c+d x) \cosh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{4 b (2 a-b) \sinh (c+d x) \cosh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{4 i a (a-b) (2 a-b) \sqrt{\frac{b \sinh ^2(c+d x)}{a}+1} F\left (i c+i d x\left |\frac{b}{a}\right .\right )}{15 d \sqrt{a+b \sinh ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x]^2)^(5/2),x]

[Out]

(4*(2*a - b)*b*Cosh[c + d*x]*Sinh[c + d*x]*Sqrt[a + b*Sinh[c + d*x]^2])/(15*d) + (b*Cosh[c + d*x]*Sinh[c + d*x
]*(a + b*Sinh[c + d*x]^2)^(3/2))/(5*d) - ((I/15)*(23*a^2 - 23*a*b + 8*b^2)*EllipticE[I*c + I*d*x, b/a]*Sqrt[a
+ b*Sinh[c + d*x]^2])/(d*Sqrt[1 + (b*Sinh[c + d*x]^2)/a]) + (((4*I)/15)*a*(a - b)*(2*a - b)*EllipticF[I*c + I*
d*x, b/a]*Sqrt[1 + (b*Sinh[c + d*x]^2)/a])/(d*Sqrt[a + b*Sinh[c + d*x]^2])

Rule 3180

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

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 \left (a+b \sinh ^2(c+d x)\right )^{5/2} \, dx &=\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{1}{5} \int \sqrt{a+b \sinh ^2(c+d x)} \left (a (5 a-b)+4 (2 a-b) b \sinh ^2(c+d x)\right ) \, dx\\ &=\frac{4 (2 a-b) b \cosh (c+d x) \sinh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{1}{15} \int \frac{a \left (15 a^2-11 a b+4 b^2\right )+b \left (23 a^2-23 a b+8 b^2\right ) \sinh ^2(c+d x)}{\sqrt{a+b \sinh ^2(c+d x)}} \, dx\\ &=\frac{4 (2 a-b) b \cosh (c+d x) \sinh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}-\frac{1}{15} (4 a (a-b) (2 a-b)) \int \frac{1}{\sqrt{a+b \sinh ^2(c+d x)}} \, dx+\frac{1}{15} \left (23 a^2-23 a b+8 b^2\right ) \int \sqrt{a+b \sinh ^2(c+d x)} \, dx\\ &=\frac{4 (2 a-b) b \cosh (c+d x) \sinh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{\left (\left (23 a^2-23 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(c+d x)}\right ) \int \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}} \, dx}{15 \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}}-\frac{\left (4 a (a-b) (2 a-b) \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}\right ) \int \frac{1}{\sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}} \, dx}{15 \sqrt{a+b \sinh ^2(c+d x)}}\\ &=\frac{4 (2 a-b) b \cosh (c+d x) \sinh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}-\frac{i \left (23 a^2-23 a b+8 b^2\right ) E\left (i c+i d x\left |\frac{b}{a}\right .\right ) \sqrt{a+b \sinh ^2(c+d x)}}{15 d \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}}+\frac{4 i a (a-b) (2 a-b) F\left (i c+i d x\left |\frac{b}{a}\right .\right ) \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}}{15 d \sqrt{a+b \sinh ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.39443, size = 208, normalized size = 0.9 \[ \frac{64 i a \left (2 a^2-3 a b+b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (c+d x))-b}{a}} \text{EllipticF}\left (i (c+d x),\frac{b}{a}\right )+\sqrt{2} b \sinh (2 (c+d x)) \left (88 a^2+28 b (2 a-b) \cosh (2 (c+d x))-88 a b+3 b^2 \cosh (4 (c+d x))+25 b^2\right )-16 i a \left (23 a^2-23 a b+8 b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (c+d x))-b}{a}} E\left (i (c+d x)\left |\frac{b}{a}\right .\right )}{240 d \sqrt{2 a+b \cosh (2 (c+d x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*I)*a*(23*a^2 - 23*a*b + 8*b^2)*Sqrt[(2*a - b + b*Cosh[2*(c + d*x)])/a]*EllipticE[I*(c + d*x), b/a] + (64
*I)*a*(2*a^2 - 3*a*b + b^2)*Sqrt[(2*a - b + b*Cosh[2*(c + d*x)])/a]*EllipticF[I*(c + d*x), b/a] + Sqrt[2]*b*(8
8*a^2 - 88*a*b + 25*b^2 + 28*(2*a - b)*b*Cosh[2*(c + d*x)] + 3*b^2*Cosh[4*(c + d*x)])*Sinh[2*(c + d*x)])/(240*
d*Sqrt[2*a - b + b*Cosh[2*(c + d*x)]])

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Maple [B]  time = 0.085, size = 609, normalized size = 2.6 \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(d*x+c)^2)^(5/2),x)

[Out]

1/15*(3*(-1/a*b)^(1/2)*b^3*sinh(d*x+c)*cosh(d*x+c)^6+(14*(-1/a*b)^(1/2)*a*b^2-10*(-1/a*b)^(1/2)*b^3)*cosh(d*x+
c)^4*sinh(d*x+c)+(11*(-1/a*b)^(1/2)*a^2*b-18*(-1/a*b)^(1/2)*a*b^2+7*(-1/a*b)^(1/2)*b^3)*cosh(d*x+c)^2*sinh(d*x
+c)+15*a^3*(b/a*cosh(d*x+c)^2+(a-b)/a)^(1/2)*(cosh(d*x+c)^2)^(1/2)*EllipticF(sinh(d*x+c)*(-1/a*b)^(1/2),(a/b)^
(1/2))-34*a^2*b*(b/a*cosh(d*x+c)^2+(a-b)/a)^(1/2)*(cosh(d*x+c)^2)^(1/2)*EllipticF(sinh(d*x+c)*(-1/a*b)^(1/2),(
a/b)^(1/2))+27*(b/a*cosh(d*x+c)^2+(a-b)/a)^(1/2)*(cosh(d*x+c)^2)^(1/2)*EllipticF(sinh(d*x+c)*(-1/a*b)^(1/2),(a
/b)^(1/2))*a*b^2-8*(b/a*cosh(d*x+c)^2+(a-b)/a)^(1/2)*(cosh(d*x+c)^2)^(1/2)*EllipticF(sinh(d*x+c)*(-1/a*b)^(1/2
),(a/b)^(1/2))*b^3+23*a^2*b*(b/a*cosh(d*x+c)^2+(a-b)/a)^(1/2)*(cosh(d*x+c)^2)^(1/2)*EllipticE(sinh(d*x+c)*(-1/
a*b)^(1/2),(a/b)^(1/2))-23*(b/a*cosh(d*x+c)^2+(a-b)/a)^(1/2)*(cosh(d*x+c)^2)^(1/2)*EllipticE(sinh(d*x+c)*(-1/a
*b)^(1/2),(a/b)^(1/2))*a*b^2+8*(b/a*cosh(d*x+c)^2+(a-b)/a)^(1/2)*(cosh(d*x+c)^2)^(1/2)*EllipticE(sinh(d*x+c)*(
-1/a*b)^(1/2),(a/b)^(1/2))*b^3)/(-1/a*b)^(1/2)/cosh(d*x+c)/(a+b*sinh(d*x+c)^2)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*sinh(d*x + c)^4 + 2*a*b*sinh(d*x + c)^2 + a^2)*sqrt(b*sinh(d*x + c)^2 + 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(d*x+c)**2)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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