### 3.774 $$\int (-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x))^{5/2} \, dx$$

Optimal. Leaf size=146 $\frac{2}{5} (b \sinh (x)+c \cosh (x)) \left (-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}-\frac{16}{15} \sqrt{b^2-c^2} (b \sinh (x)+c \cosh (x)) \sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}+\frac{64 \left (b^2-c^2\right ) (b \sinh (x)+c \cosh (x))}{15 \sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}}$

[Out]

(64*(b^2 - c^2)*(c*Cosh[x] + b*Sinh[x]))/(15*Sqrt[-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]]) - (16*Sqrt[b^2 -
c^2]*(c*Cosh[x] + b*Sinh[x])*Sqrt[-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]])/15 + (2*(c*Cosh[x] + b*Sinh[x])*(
-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(3/2))/5

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Rubi [A]  time = 0.122441, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.071, Rules used = {3113, 3112} $\frac{2}{5} (b \sinh (x)+c \cosh (x)) \left (-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}-\frac{16}{15} \sqrt{b^2-c^2} (b \sinh (x)+c \cosh (x)) \sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}+\frac{64 \left (b^2-c^2\right ) (b \sinh (x)+c \cosh (x))}{15 \sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(64*(b^2 - c^2)*(c*Cosh[x] + b*Sinh[x]))/(15*Sqrt[-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]]) - (16*Sqrt[b^2 -
c^2]*(c*Cosh[x] + b*Sinh[x])*Sqrt[-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]])/15 + (2*(c*Cosh[x] + b*Sinh[x])*(
-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(3/2))/5

Rule 3113

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

Rule 3112

Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Simp[(-2*(c*Cos[d
+ e*x] - b*Sin[d + e*x]))/(e*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]), x] /; FreeQ[{a, b, c, d, e}, x] && E
qQ[a^2 - b^2 - c^2, 0]

Rubi steps

\begin{align*} \int \left (-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2} \, dx &=\frac{2}{5} (c \cosh (x)+b \sinh (x)) \left (-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}-\frac{1}{5} \left (8 \sqrt{b^2-c^2}\right ) \int \left (-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2} \, dx\\ &=-\frac{16}{15} \sqrt{b^2-c^2} (c \cosh (x)+b \sinh (x)) \sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}+\frac{2}{5} (c \cosh (x)+b \sinh (x)) \left (-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}+\frac{1}{15} \left (32 \left (b^2-c^2\right )\right ) \int \sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)} \, dx\\ &=\frac{64 \left (b^2-c^2\right ) (c \cosh (x)+b \sinh (x))}{15 \sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}}-\frac{16}{15} \sqrt{b^2-c^2} (c \cosh (x)+b \sinh (x)) \sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}+\frac{2}{5} (c \cosh (x)+b \sinh (x)) \left (-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}\\ \end{align*}

Mathematica [C]  time = 75.4769, size = 9943, normalized size = 68.1 $\text{Result too large to show}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.574, size = 288, normalized size = 2. \begin{align*}{ \left ( -{\frac{ \left ( \cosh \left ( x \right ) \right ) ^{3}}{3} \left ({b}^{2}-{c}^{2} \right ) ^{{\frac{3}{2}}}}- \left ( 2\,{b}^{2}-2\,{c}^{2} \right ) \sqrt{{b}^{2}-{c}^{2}}\cosh \left ( x \right ) \right ){\frac{1}{\sqrt{-{(\sinh \left ( x \right ){b}^{2}-\sinh \left ( x \right ){c}^{2}+{b}^{2}-{c}^{2}){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}}}}}+{\frac{1}{\sinh \left ( x \right ) }\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) +1 \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}} \left ({\frac{ \left ({b}^{2}-{c}^{2} \right ) ^{2}\cosh \left ( x \right ) }{2\,\sinh \left ( x \right ){b}^{2}-2\,\sinh \left ( x \right ){c}^{2}+2\,{b}^{2}-2\,{c}^{2}}\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) +1 \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}}}-{\frac{1}{2} \left ({b}^{2}-{c}^{2} \right ) ^{{\frac{3}{2}}}\arctan \left ({\cosh \left ( x \right ) \sqrt{\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) +1 \right ) }{\frac{1}{\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) +1 \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}}}}} \right ){\frac{1}{\sqrt{\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) +1 \right ) }}}} \right ){\frac{1}{\sqrt{-{(\sinh \left ( x \right ){b}^{2}-\sinh \left ( x \right ){c}^{2}+{b}^{2}-{c}^{2}){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^(5/2),x)

[Out]

1/(-(sinh(x)*b^2-sinh(x)*c^2+b^2-c^2)/(b^2-c^2)^(1/2))^(1/2)*(-1/3*(b^2-c^2)^(3/2)*cosh(x)^3-(2*b^2-2*c^2)*(b^
2-c^2)^(1/2)*cosh(x))+(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)*(1/2*(b^2-c^2)^2*cosh(x)/(sinh(x)*b^2-sin
h(x)*c^2+b^2-c^2)*(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)-1/2*(b^2-c^2)^(3/2)/((b^2-c^2)^(1/2)*(sinh(x)
+1))^(1/2)*arctan(((b^2-c^2)^(1/2)*(sinh(x)+1))^(1/2)*cosh(x)/(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)))
/sinh(x)/(-(sinh(x)*b^2-sinh(x)*c^2+b^2-c^2)/(b^2-c^2)^(1/2))^(1/2)

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Maxima [B]  time = 16.2856, size = 2415, normalized size = 16.54 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^(5/2),x, algorithm="maxima")

[Out]

1/20*sqrt(2)*(sqrt(b + c)*sqrt(b - c)*b^2 + 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2)*(-2*s
qrt(b + c)*sqrt(b - c)*e^(-x) + (b - c)*e^(-2*x) + b + c)^(5/2)*e^(5/2*x)/(sqrt(b + c)*sqrt(b - c)*b^2 + 2*sqr
t(b + c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2 - 5*(b^3 + b^2*c - b*c^2 - c^3)*e^(-x) + 10*(sqrt(b + c
)*sqrt(b - c)*b^2 - sqrt(b + c)*sqrt(b - c)*c^2)*e^(-2*x) - 10*(b^3 - b^2*c - b*c^2 + c^3)*e^(-3*x) + 5*(sqrt(
b + c)*sqrt(b - c)*b^2 - 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2)*e^(-4*x) - (b^3 - 3*b^2*
c + 3*b*c^2 - c^3)*e^(-5*x)) - 5/12*sqrt(2)*(b^3 + b^2*c - b*c^2 - c^3)*(-2*sqrt(b + c)*sqrt(b - c)*e^(-x) + (
b - c)*e^(-2*x) + b + c)^(5/2)*e^(3/2*x)/(sqrt(b + c)*sqrt(b - c)*b^2 + 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b
+ c)*sqrt(b - c)*c^2 - 5*(b^3 + b^2*c - b*c^2 - c^3)*e^(-x) + 10*(sqrt(b + c)*sqrt(b - c)*b^2 - sqrt(b + c)*s
qrt(b - c)*c^2)*e^(-2*x) - 10*(b^3 - b^2*c - b*c^2 + c^3)*e^(-3*x) + 5*(sqrt(b + c)*sqrt(b - c)*b^2 - 2*sqrt(b
+ c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2)*e^(-4*x) - (b^3 - 3*b^2*c + 3*b*c^2 - c^3)*e^(-5*x)) + 5/
2*sqrt(2)*(sqrt(b + c)*sqrt(b - c)*b^2 - sqrt(b + c)*sqrt(b - c)*c^2)*(-2*sqrt(b + c)*sqrt(b - c)*e^(-x) + (b
- c)*e^(-2*x) + b + c)^(5/2)*e^(1/2*x)/(sqrt(b + c)*sqrt(b - c)*b^2 + 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b +
c)*sqrt(b - c)*c^2 - 5*(b^3 + b^2*c - b*c^2 - c^3)*e^(-x) + 10*(sqrt(b + c)*sqrt(b - c)*b^2 - sqrt(b + c)*sqr
t(b - c)*c^2)*e^(-2*x) - 10*(b^3 - b^2*c - b*c^2 + c^3)*e^(-3*x) + 5*(sqrt(b + c)*sqrt(b - c)*b^2 - 2*sqrt(b +
c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2)*e^(-4*x) - (b^3 - 3*b^2*c + 3*b*c^2 - c^3)*e^(-5*x)) + 5/2*
sqrt(2)*(b^3 - b^2*c - b*c^2 + c^3)*(-2*sqrt(b + c)*sqrt(b - c)*e^(-x) + (b - c)*e^(-2*x) + b + c)^(5/2)*e^(-1
/2*x)/(sqrt(b + c)*sqrt(b - c)*b^2 + 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2 - 5*(b^3 + b^
2*c - b*c^2 - c^3)*e^(-x) + 10*(sqrt(b + c)*sqrt(b - c)*b^2 - sqrt(b + c)*sqrt(b - c)*c^2)*e^(-2*x) - 10*(b^3
- b^2*c - b*c^2 + c^3)*e^(-3*x) + 5*(sqrt(b + c)*sqrt(b - c)*b^2 - 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b + c)
*sqrt(b - c)*c^2)*e^(-4*x) - (b^3 - 3*b^2*c + 3*b*c^2 - c^3)*e^(-5*x)) - 5/12*sqrt(2)*(sqrt(b + c)*sqrt(b - c)
*b^2 - 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2)*(-2*sqrt(b + c)*sqrt(b - c)*e^(-x) + (b -
c)*e^(-2*x) + b + c)^(5/2)*e^(-3/2*x)/(sqrt(b + c)*sqrt(b - c)*b^2 + 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b +
c)*sqrt(b - c)*c^2 - 5*(b^3 + b^2*c - b*c^2 - c^3)*e^(-x) + 10*(sqrt(b + c)*sqrt(b - c)*b^2 - sqrt(b + c)*sqrt
(b - c)*c^2)*e^(-2*x) - 10*(b^3 - b^2*c - b*c^2 + c^3)*e^(-3*x) + 5*(sqrt(b + c)*sqrt(b - c)*b^2 - 2*sqrt(b +
c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2)*e^(-4*x) - (b^3 - 3*b^2*c + 3*b*c^2 - c^3)*e^(-5*x)) + 1/20*
sqrt(2)*(b^3 - 3*b^2*c + 3*b*c^2 - c^3)*(-2*sqrt(b + c)*sqrt(b - c)*e^(-x) + (b - c)*e^(-2*x) + b + c)^(5/2)*e
^(-5/2*x)/(sqrt(b + c)*sqrt(b - c)*b^2 + 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b + c)*sqrt(b - c)*c^2 - 5*(b^3
+ b^2*c - b*c^2 - c^3)*e^(-x) + 10*(sqrt(b + c)*sqrt(b - c)*b^2 - sqrt(b + c)*sqrt(b - c)*c^2)*e^(-2*x) - 10*(
b^3 - b^2*c - b*c^2 + c^3)*e^(-3*x) + 5*(sqrt(b + c)*sqrt(b - c)*b^2 - 2*sqrt(b + c)*sqrt(b - c)*b*c + sqrt(b
+ c)*sqrt(b - c)*c^2)*e^(-4*x) - (b^3 - 3*b^2*c + 3*b*c^2 - c^3)*e^(-5*x))

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Fricas [B]  time = 2.48302, size = 2079, normalized size = 14.24 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^(5/2),x, algorithm="fricas")

[Out]

1/30*sqrt(1/2)*(3*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^6 + 18*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)*sinh(
x)^5 + 3*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*sinh(x)^6 + 125*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)^4 + 5*(25*b^3 + 2
5*b^2*c - 25*b*c^2 - 25*c^3 + 9*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^2)*sinh(x)^4 + 20*(3*(b^3 + 3*b^2*c +
3*b*c^2 + c^3)*cosh(x)^3 + 25*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x))*sinh(x)^3 + 3*b^3 - 9*b^2*c + 9*b*c^2 - 3*c
^3 + 125*(b^3 - b^2*c - b*c^2 + c^3)*cosh(x)^2 + 5*(9*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^4 + 25*b^3 - 25*
b^2*c - 25*b*c^2 + 25*c^3 + 150*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)^2)*sinh(x)^2 + 2*(9*(b^3 + 3*b^2*c + 3*b*c
^2 + c^3)*cosh(x)^5 + 250*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)^3 + 125*(b^3 - b^2*c - b*c^2 + c^3)*cosh(x))*sin
h(x) - 2*(11*(b^2 + 2*b*c + c^2)*cosh(x)^5 + 55*(b^2 + 2*b*c + c^2)*cosh(x)*sinh(x)^4 + 11*(b^2 + 2*b*c + c^2)
*sinh(x)^5 - 150*(b^2 - c^2)*cosh(x)^3 + 10*(11*(b^2 + 2*b*c + c^2)*cosh(x)^2 - 15*b^2 + 15*c^2)*sinh(x)^3 + 1
0*(11*(b^2 + 2*b*c + c^2)*cosh(x)^3 - 45*(b^2 - c^2)*cosh(x))*sinh(x)^2 + 11*(b^2 - 2*b*c + c^2)*cosh(x) + (55
*(b^2 + 2*b*c + c^2)*cosh(x)^4 - 450*(b^2 - c^2)*cosh(x)^2 + 11*b^2 - 22*b*c + 11*c^2)*sinh(x))*sqrt(b^2 - c^2
))*sqrt(((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh(x)^2 - 2*sqrt(b^2 - c^2)*(cosh(x) + sinh
(x)) + b - c)/(cosh(x) + sinh(x)))/((b + c)*cosh(x)^4 + 4*(b + c)*cosh(x)*sinh(x)^3 + (b + c)*sinh(x)^4 - (b -
c)*cosh(x)^2 + (6*(b + c)*cosh(x)^2 - b + c)*sinh(x)^2 + 2*(2*(b + c)*cosh(x)^3 - (b - c)*cosh(x))*sinh(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((b*cosh(x)+c*sinh(x)-(b**2-c**2)**(1/2))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.46866, size = 887, normalized size = 6.08 \begin{align*} -\frac{\sqrt{2}{\left (3 \,{\left (\sqrt{b^{2} - c^{2}} b^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) + 2 \, \sqrt{b^{2} - c^{2}} b c \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) + \sqrt{b^{2} - c^{2}} c^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right )\right )} e^{\left (\frac{5}{2} \, x\right )} - 25 \,{\left (b^{3} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) + b^{2} c \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) - b c^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) - c^{3} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right )\right )} e^{\left (\frac{3}{2} \, x\right )} + 150 \,{\left (\sqrt{b^{2} - c^{2}} b^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) - \sqrt{b^{2} - c^{2}} c^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right )\right )} e^{\left (\frac{1}{2} \, x\right )} + 150 \,{\left (b^{3} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) - b^{2} c \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) - b c^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) + c^{3} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right )\right )} e^{\left (-\frac{1}{2} \, x\right )} - 25 \,{\left (\sqrt{b^{2} - c^{2}} b^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) - 2 \, \sqrt{b^{2} - c^{2}} b c \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) + \sqrt{b^{2} - c^{2}} c^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right )\right )} e^{\left (-\frac{3}{2} \, x\right )} + 3 \,{\left (b^{3} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) - 3 \, b^{2} c \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) + 3 \, b c^{2} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right ) - c^{3} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} + b - c\right )\right )} e^{\left (-\frac{5}{2} \, x\right )}\right )}}{60 \, \sqrt{b - c}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^(5/2),x, algorithm="giac")

[Out]

-1/60*sqrt(2)*(3*(sqrt(b^2 - c^2)*b^2*sgn(-sqrt(b^2 - c^2)*e^x + b - c) + 2*sqrt(b^2 - c^2)*b*c*sgn(-sqrt(b^2
- c^2)*e^x + b - c) + sqrt(b^2 - c^2)*c^2*sgn(-sqrt(b^2 - c^2)*e^x + b - c))*e^(5/2*x) - 25*(b^3*sgn(-sqrt(b^2
- c^2)*e^x + b - c) + b^2*c*sgn(-sqrt(b^2 - c^2)*e^x + b - c) - b*c^2*sgn(-sqrt(b^2 - c^2)*e^x + b - c) - c^3
*sgn(-sqrt(b^2 - c^2)*e^x + b - c))*e^(3/2*x) + 150*(sqrt(b^2 - c^2)*b^2*sgn(-sqrt(b^2 - c^2)*e^x + b - c) - s
qrt(b^2 - c^2)*c^2*sgn(-sqrt(b^2 - c^2)*e^x + b - c))*e^(1/2*x) + 150*(b^3*sgn(-sqrt(b^2 - c^2)*e^x + b - c) -
b^2*c*sgn(-sqrt(b^2 - c^2)*e^x + b - c) - b*c^2*sgn(-sqrt(b^2 - c^2)*e^x + b - c) + c^3*sgn(-sqrt(b^2 - c^2)*
e^x + b - c))*e^(-1/2*x) - 25*(sqrt(b^2 - c^2)*b^2*sgn(-sqrt(b^2 - c^2)*e^x + b - c) - 2*sqrt(b^2 - c^2)*b*c*s
gn(-sqrt(b^2 - c^2)*e^x + b - c) + sqrt(b^2 - c^2)*c^2*sgn(-sqrt(b^2 - c^2)*e^x + b - c))*e^(-3/2*x) + 3*(b^3*
sgn(-sqrt(b^2 - c^2)*e^x + b - c) - 3*b^2*c*sgn(-sqrt(b^2 - c^2)*e^x + b - c) + 3*b*c^2*sgn(-sqrt(b^2 - c^2)*e
^x + b - c) - c^3*sgn(-sqrt(b^2 - c^2)*e^x + b - c))*e^(-5/2*x))/sqrt(b - c)