### 3.861 $$\int (a+b \cosh (c+d x) \sinh (c+d x))^{5/2} \, dx$$

Optimal. Leaf size=301 $\frac{2 i \sqrt{2} a \left (4 a^2+b^2\right ) \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{15 d \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{i \left (92 a^2-9 b^2\right ) \sqrt{2 a+b \sinh (2 c+2 d x)} E\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{60 \sqrt{2} d \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac{b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{2 \sqrt{2} a b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{15 d}$

[Out]

(2*Sqrt[2]*a*b*Cosh[2*c + 2*d*x]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/(15*d) + (b*Cosh[2*c + 2*d*x]*(2*a + b*Sinh[
2*c + 2*d*x])^(3/2))/(20*Sqrt[2]*d) - ((I/60)*(92*a^2 - 9*b^2)*EllipticE[((2*I)*c - Pi/2 + (2*I)*d*x)/2, (2*b)
/((2*I)*a + b)]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/(Sqrt[2]*d*Sqrt[(2*a + b*Sinh[2*c + 2*d*x])/(2*a - I*b)]) + (
((2*I)/15)*Sqrt[2]*a*(4*a^2 + b^2)*EllipticF[((2*I)*c - Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*a + b)]*Sqrt[(2*a +
b*Sinh[2*c + 2*d*x])/(2*a - I*b)])/(d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])

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Rubi [A]  time = 0.39189, antiderivative size = 301, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.4, Rules used = {2666, 2656, 2753, 2752, 2663, 2661, 2655, 2653} $\frac{2 i \sqrt{2} a \left (4 a^2+b^2\right ) \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{15 d \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{i \left (92 a^2-9 b^2\right ) \sqrt{2 a+b \sinh (2 c+2 d x)} E\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{60 \sqrt{2} d \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac{b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{2 \sqrt{2} a b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{15 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[2]*a*b*Cosh[2*c + 2*d*x]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/(15*d) + (b*Cosh[2*c + 2*d*x]*(2*a + b*Sinh[
2*c + 2*d*x])^(3/2))/(20*Sqrt[2]*d) - ((I/60)*(92*a^2 - 9*b^2)*EllipticE[((2*I)*c - Pi/2 + (2*I)*d*x)/2, (2*b)
/((2*I)*a + b)]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/(Sqrt[2]*d*Sqrt[(2*a + b*Sinh[2*c + 2*d*x])/(2*a - I*b)]) + (
((2*I)/15)*Sqrt[2]*a*(4*a^2 + b^2)*EllipticF[((2*I)*c - Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*a + b)]*Sqrt[(2*a +
b*Sinh[2*c + 2*d*x])/(2*a - I*b)])/(d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2656

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*
x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

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 \cosh (c+d x) \sinh (c+d x))^{5/2} \, dx &=\int \left (a+\frac{1}{2} b \sinh (2 c+2 d x)\right )^{5/2} \, dx\\ &=\frac{b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{2}{5} \int \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)} \left (\frac{1}{8} \left (20 a^2-3 b^2\right )+2 a b \sinh (2 c+2 d x)\right ) \, dx\\ &=\frac{2 \sqrt{2} a b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{15 d}+\frac{b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{4}{15} \int \frac{\frac{1}{16} a \left (60 a^2-17 b^2\right )+\frac{1}{32} b \left (92 a^2-9 b^2\right ) \sinh (2 c+2 d x)}{\sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}} \, dx\\ &=\frac{2 \sqrt{2} a b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{15 d}+\frac{b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{1}{60} \left (92 a^2-9 b^2\right ) \int \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)} \, dx-\frac{1}{15} \left (2 a \left (4 a^2+b^2\right )\right ) \int \frac{1}{\sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}} \, dx\\ &=\frac{2 \sqrt{2} a b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{15 d}+\frac{b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{\left (\left (92 a^2-9 b^2\right ) \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}\right ) \int \sqrt{\frac{a}{a-\frac{i b}{2}}+\frac{b \sinh (2 c+2 d x)}{2 \left (a-\frac{i b}{2}\right )}} \, dx}{60 \sqrt{\frac{a+\frac{1}{2} b \sinh (2 c+2 d x)}{a-\frac{i b}{2}}}}-\frac{\left (2 a \left (4 a^2+b^2\right ) \sqrt{\frac{a+\frac{1}{2} b \sinh (2 c+2 d x)}{a-\frac{i b}{2}}}\right ) \int \frac{1}{\sqrt{\frac{a}{a-\frac{i b}{2}}+\frac{b \sinh (2 c+2 d x)}{2 \left (a-\frac{i b}{2}\right )}}} \, dx}{15 \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}}\\ &=\frac{2 \sqrt{2} a b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{15 d}+\frac{b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}-\frac{i \left (92 a^2-9 b^2\right ) E\left (\frac{1}{2} \left (2 i c-\frac{\pi }{2}+2 i d x\right )|\frac{2 b}{2 i a+b}\right ) \sqrt{2 a+b \sinh (2 c+2 d x)}}{60 \sqrt{2} d \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac{2 i \sqrt{2} a \left (4 a^2+b^2\right ) F\left (\frac{1}{2} \left (2 i c-\frac{\pi }{2}+2 i d x\right )|\frac{2 b}{2 i a+b}\right ) \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}{15 d \sqrt{2 a+b \sinh (2 c+2 d x)}}\\ \end{align*}

Mathematica [A]  time = 1.37497, size = 239, normalized size = 0.79 $\frac{-32 i a \left (4 a^2+b^2\right ) \sqrt{\frac{2 a+b \sinh (2 (c+d x))}{2 a-i b}} F\left (\frac{1}{4} (-4 i c-4 i d x+\pi )|-\frac{2 i b}{2 a-i b}\right )+2 \left (92 a^2 b+184 i a^3-18 i a b^2-9 b^3\right ) \sqrt{\frac{2 a+b \sinh (2 (c+d x))}{2 a-i b}} E\left (\frac{1}{4} (-4 i c-4 i d x+\pi )|-\frac{2 i b}{2 a-i b}\right )+b \left (88 a^2 \cosh (2 (c+d x))+b \sinh (4 (c+d x)) (28 a+3 b \sinh (2 (c+d x)))\right )}{120 d \sqrt{4 a+2 b \sinh (2 (c+d x))}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*((184*I)*a^3 + 92*a^2*b - (18*I)*a*b^2 - 9*b^3)*EllipticE[((-4*I)*c + Pi - (4*I)*d*x)/4, ((-2*I)*b)/(2*a -
I*b)]*Sqrt[(2*a + b*Sinh[2*(c + d*x)])/(2*a - I*b)] - (32*I)*a*(4*a^2 + b^2)*EllipticF[((-4*I)*c + Pi - (4*I)*
d*x)/4, ((-2*I)*b)/(2*a - I*b)]*Sqrt[(2*a + b*Sinh[2*(c + d*x)])/(2*a - I*b)] + b*(88*a^2*Cosh[2*(c + d*x)] +
b*(28*a + 3*b*Sinh[2*(c + d*x)])*Sinh[4*(c + d*x)]))/(120*d*Sqrt[4*a + 2*b*Sinh[2*(c + d*x)]])

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Maple [B]  time = 0.542, size = 1260, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(64*I*(-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+
2*c))*b/(I*b-2*a))^(1/2)*EllipticF((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*a^
3*b+16*I*(-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+2*
c))*b/(I*b-2*a))^(1/2)*EllipticF((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*a*b^
3+240*(-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+2*c))
*b/(I*b-2*a))^(1/2)*EllipticF((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*a^4+24*
(-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+2*c))*b/(I*
b-2*a))^(1/2)*EllipticF((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*a^2*b^2-9*(-(
2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+2*c))*b/(I*b-2
*a))^(1/2)*EllipticF((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*b^4-368*(-(2*a+b
*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+2*c))*b/(I*b-2*a))^
(1/2)*EllipticE((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*a^4-56*(-(2*a+b*sinh(
2*d*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+2*c))*b/(I*b-2*a))^(1/2)*
EllipticE((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*a^2*b^2+9*(-(2*a+b*sinh(2*d
*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+2*c))*b/(I*b-2*a))^(1/2)*Ell
ipticE((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*b^4+3*b^4*sinh(2*d*x+2*c)^4+28
*a*b^3*sinh(2*d*x+2*c)^3+44*a^2*b^2*sinh(2*d*x+2*c)^2+3*b^4*sinh(2*d*x+2*c)^2+28*a*b^3*sinh(2*d*x+2*c)+44*a^2*
b^2)/b/cosh(2*d*x+2*c)/(4*a+2*b*sinh(2*d*x+2*c))^(1/2)/d

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*cosh(d*x + c)*sinh(d*x + c) + 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} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2}\right )} \sqrt{b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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