### 3.862 $$\int (a+b \cosh (c+d x) \sinh (c+d x))^{3/2} \, dx$$

Optimal. Leaf size=248 $\frac{i \left (4 a^2+b^2\right ) \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{6 \sqrt{2} d \sqrt{2 a+b \sinh (2 c+2 d x)}}+\frac{b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{6 \sqrt{2} d}-\frac{2 i \sqrt{2} a \sqrt{2 a+b \sinh (2 c+2 d x)} E\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{3 d \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}$

[Out]

(b*Cosh[2*c + 2*d*x]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/(6*Sqrt[2]*d) - (((2*I)/3)*Sqrt[2]*a*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]])/(d*Sqrt[(2*a + b*Sinh[2*c + 2*d*x
])/(2*a - I*b)]) + ((I/6)*(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)])/(Sqrt[2]*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])

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Rubi [A]  time = 0.253067, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.35, Rules used = {2666, 2656, 2752, 2663, 2661, 2655, 2653} $\frac{i \left (4 a^2+b^2\right ) \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{6 \sqrt{2} d \sqrt{2 a+b \sinh (2 c+2 d x)}}+\frac{b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{6 \sqrt{2} d}-\frac{2 i \sqrt{2} a \sqrt{2 a+b \sinh (2 c+2 d x)} E\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{3 d \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(3/2),x]

[Out]

(b*Cosh[2*c + 2*d*x]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/(6*Sqrt[2]*d) - (((2*I)/3)*Sqrt[2]*a*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]])/(d*Sqrt[(2*a + b*Sinh[2*c + 2*d*x
])/(2*a - I*b)]) + ((I/6)*(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)])/(Sqrt[2]*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 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))^{3/2} \, dx &=\int \left (a+\frac{1}{2} b \sinh (2 c+2 d x)\right )^{3/2} \, dx\\ &=\frac{b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{2}{3} \int \frac{\frac{1}{8} \left (12 a^2-b^2\right )+a b \sinh (2 c+2 d x)}{\sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}} \, dx\\ &=\frac{b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{1}{3} (4 a) \int \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)} \, dx+\frac{1}{12} \left (-4 a^2-b^2\right ) \int \frac{1}{\sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}} \, dx\\ &=\frac{b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{\left (4 a \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}{3 \sqrt{\frac{a+\frac{1}{2} b \sinh (2 c+2 d x)}{a-\frac{i b}{2}}}}+\frac{\left (\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}{12 \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}}\\ &=\frac{b \cosh (2 c+2 d x) \sqrt{2 a+b \sinh (2 c+2 d x)}}{6 \sqrt{2} d}-\frac{2 i \sqrt{2} a 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)}}{3 d \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac{i \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}}}{6 \sqrt{2} d \sqrt{2 a+b \sinh (2 c+2 d x)}}\\ \end{align*}

Mathematica [A]  time = 0.803819, size = 202, normalized size = 0.81 $\frac{-2 i \left (4 a^2+b^2\right ) \sqrt{\frac{2 a+b \sinh (2 (c+d x))}{2 a-i b}} F\left (\frac{1}{4} (-4 i c-4 i d x+\pi )|-\frac{2 i b}{2 a-i b}\right )+b (4 a \cosh (2 (c+d x))+b \sinh (4 (c+d x)))+16 a (b+2 i a) \sqrt{\frac{2 a+b \sinh (2 (c+d x))}{2 a-i b}} E\left (\frac{1}{4} (-4 i c-4 i d x+\pi )|-\frac{2 i b}{2 a-i b}\right )}{12 d \sqrt{4 a+2 b \sinh (2 (c+d x))}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(3/2),x]

[Out]

(16*a*((2*I)*a + b)*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)] - (2*I)*(4*a^2 + b^2)*EllipticF[((-4*I)*c + Pi - (4*I)*d*x)/4, ((-2*I)*b)/(2*a - I*b)]*Sq
rt[(2*a + b*Sinh[2*(c + d*x)])/(2*a - I*b)] + b*(4*a*Cosh[2*(c + d*x)] + b*Sinh[4*(c + d*x)]))/(12*d*Sqrt[4*a
+ 2*b*Sinh[2*(c + d*x)]])

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Maple [B]  time = 0.49, size = 935, normalized size = 3.8 \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))^(3/2),x)

[Out]

1/6*(4*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^2*
b+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))*b^3+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^3+6*(-(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^2-32*(-(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^3-8*(-(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)*Ellipt
icE((-(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^2+b^3*sinh(2*d*x+2*c)^3+2*a*b
^2*sinh(2*d*x+2*c)^2+b^3*sinh(2*d*x+2*c)+2*a*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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*cosh(d*x + c)*sinh(d*x + c) + a)^(3/2), x)

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

[Out]

integral((b*cosh(d*x + c)*sinh(d*x + c) + a)^(3/2), 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*cosh(d*x+c)*sinh(d*x+c))**(3/2),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{3}{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))^(3/2),x, algorithm="giac")

[Out]

integrate((b*cosh(d*x + c)*sinh(d*x + c) + a)^(3/2), x)