3.87 \(\int (a+a \cosh (x))^{5/2} (A+B \cosh (x)) \, dx\)

Optimal. Leaf size=94 \[ \frac{64 a^3 (7 A+5 B) \sinh (x)}{105 \sqrt{a \cosh (x)+a}}+\frac{16}{105} a^2 (7 A+5 B) \sinh (x) \sqrt{a \cosh (x)+a}+\frac{2}{35} a (7 A+5 B) \sinh (x) (a \cosh (x)+a)^{3/2}+\frac{2}{7} B \sinh (x) (a \cosh (x)+a)^{5/2} \]

[Out]

(64*a^3*(7*A + 5*B)*Sinh[x])/(105*Sqrt[a + a*Cosh[x]]) + (16*a^2*(7*A + 5*B)*Sqrt[a + a*Cosh[x]]*Sinh[x])/105
+ (2*a*(7*A + 5*B)*(a + a*Cosh[x])^(3/2)*Sinh[x])/35 + (2*B*(a + a*Cosh[x])^(5/2)*Sinh[x])/7

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Rubi [A]  time = 0.0922257, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2751, 2647, 2646} \[ \frac{64 a^3 (7 A+5 B) \sinh (x)}{105 \sqrt{a \cosh (x)+a}}+\frac{16}{105} a^2 (7 A+5 B) \sinh (x) \sqrt{a \cosh (x)+a}+\frac{2}{35} a (7 A+5 B) \sinh (x) (a \cosh (x)+a)^{3/2}+\frac{2}{7} B \sinh (x) (a \cosh (x)+a)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cosh[x])^(5/2)*(A + B*Cosh[x]),x]

[Out]

(64*a^3*(7*A + 5*B)*Sinh[x])/(105*Sqrt[a + a*Cosh[x]]) + (16*a^2*(7*A + 5*B)*Sqrt[a + a*Cosh[x]]*Sinh[x])/105
+ (2*a*(7*A + 5*B)*(a + a*Cosh[x])^(3/2)*Sinh[x])/35 + (2*B*(a + a*Cosh[x])^(5/2)*Sinh[x])/7

Rule 2751

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2647

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[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2646

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

Rubi steps

\begin{align*} \int (a+a \cosh (x))^{5/2} (A+B \cosh (x)) \, dx &=\frac{2}{7} B (a+a \cosh (x))^{5/2} \sinh (x)+\frac{1}{7} (7 A+5 B) \int (a+a \cosh (x))^{5/2} \, dx\\ &=\frac{2}{35} a (7 A+5 B) (a+a \cosh (x))^{3/2} \sinh (x)+\frac{2}{7} B (a+a \cosh (x))^{5/2} \sinh (x)+\frac{1}{35} (8 a (7 A+5 B)) \int (a+a \cosh (x))^{3/2} \, dx\\ &=\frac{16}{105} a^2 (7 A+5 B) \sqrt{a+a \cosh (x)} \sinh (x)+\frac{2}{35} a (7 A+5 B) (a+a \cosh (x))^{3/2} \sinh (x)+\frac{2}{7} B (a+a \cosh (x))^{5/2} \sinh (x)+\frac{1}{105} \left (32 a^2 (7 A+5 B)\right ) \int \sqrt{a+a \cosh (x)} \, dx\\ &=\frac{64 a^3 (7 A+5 B) \sinh (x)}{105 \sqrt{a+a \cosh (x)}}+\frac{16}{105} a^2 (7 A+5 B) \sqrt{a+a \cosh (x)} \sinh (x)+\frac{2}{35} a (7 A+5 B) (a+a \cosh (x))^{3/2} \sinh (x)+\frac{2}{7} B (a+a \cosh (x))^{5/2} \sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.119413, size = 60, normalized size = 0.64 \[ \frac{1}{210} a^2 \tanh \left (\frac{x}{2}\right ) \sqrt{a (\cosh (x)+1)} ((392 A+505 B) \cosh (x)+6 (7 A+20 B) \cosh (2 x)+1246 A+15 B \cosh (3 x)+1040 B) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cosh[x])^(5/2)*(A + B*Cosh[x]),x]

[Out]

(a^2*Sqrt[a*(1 + Cosh[x])]*(1246*A + 1040*B + (392*A + 505*B)*Cosh[x] + 6*(7*A + 20*B)*Cosh[2*x] + 15*B*Cosh[3
*x])*Tanh[x/2])/210

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Maple [A]  time = 0.046, size = 71, normalized size = 0.8 \begin{align*}{\frac{8\,{a}^{3}\sqrt{2}}{105}\cosh \left ({\frac{x}{2}} \right ) \sinh \left ({\frac{x}{2}} \right ) \left ( 30\,B \left ( \sinh \left ( x/2 \right ) \right ) ^{6}+ \left ( 21\,A+105\,B \right ) \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{4}+ \left ( 70\,A+140\,B \right ) \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{2}+105\,A+105\,B \right ){\frac{1}{\sqrt{a \left ( \cosh \left ({\frac{x}{2}} \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cosh(x))^(5/2)*(A+B*cosh(x)),x)

[Out]

8/105*cosh(1/2*x)*a^3*sinh(1/2*x)*(30*B*sinh(1/2*x)^6+(21*A+105*B)*sinh(1/2*x)^4+(70*A+140*B)*sinh(1/2*x)^2+10
5*A+105*B)*2^(1/2)/(a*cosh(1/2*x)^2)^(1/2)

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Maxima [B]  time = 1.6449, size = 320, normalized size = 3.4 \begin{align*} \frac{1}{60} \,{\left (3 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (\frac{5}{2} \, x\right )} + 25 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (\frac{3}{2} \, x\right )} + 150 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (\frac{1}{2} \, x\right )} - 150 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-\frac{1}{2} \, x\right )} - 25 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-\frac{3}{2} \, x\right )} - 3 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-\frac{5}{2} \, x\right )}\right )} A + \frac{1}{168} \,{\left ({\left (3 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-x\right )} + 21 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-2 \, x\right )} + 70 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-3 \, x\right )} + 210 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-4 \, x\right )} - 105 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-5 \, x\right )} - 7 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-6 \, x\right )}\right )} e^{\left (\frac{9}{2} \, x\right )} +{\left (7 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-x\right )} + 105 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-2 \, x\right )} - 210 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-3 \, x\right )} - 70 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-4 \, x\right )} - 21 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-5 \, x\right )} - 3 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-6 \, x\right )}\right )} e^{\left (\frac{5}{2} \, x\right )}\right )} B \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cosh(x))^(5/2)*(A+B*cosh(x)),x, algorithm="maxima")

[Out]

1/60*(3*sqrt(2)*a^(5/2)*e^(5/2*x) + 25*sqrt(2)*a^(5/2)*e^(3/2*x) + 150*sqrt(2)*a^(5/2)*e^(1/2*x) - 150*sqrt(2)
*a^(5/2)*e^(-1/2*x) - 25*sqrt(2)*a^(5/2)*e^(-3/2*x) - 3*sqrt(2)*a^(5/2)*e^(-5/2*x))*A + 1/168*((3*sqrt(2)*a^(5
/2)*e^(-x) + 21*sqrt(2)*a^(5/2)*e^(-2*x) + 70*sqrt(2)*a^(5/2)*e^(-3*x) + 210*sqrt(2)*a^(5/2)*e^(-4*x) - 105*sq
rt(2)*a^(5/2)*e^(-5*x) - 7*sqrt(2)*a^(5/2)*e^(-6*x))*e^(9/2*x) + (7*sqrt(2)*a^(5/2)*e^(-x) + 105*sqrt(2)*a^(5/
2)*e^(-2*x) - 210*sqrt(2)*a^(5/2)*e^(-3*x) - 70*sqrt(2)*a^(5/2)*e^(-4*x) - 21*sqrt(2)*a^(5/2)*e^(-5*x) - 3*sqr
t(2)*a^(5/2)*e^(-6*x))*e^(5/2*x))*B

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Fricas [B]  time = 2.26143, size = 1553, normalized size = 16.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cosh(x))^(5/2)*(A+B*cosh(x)),x, algorithm="fricas")

[Out]

1/420*sqrt(1/2)*(15*B*a^2*cosh(x)^7 + 15*B*a^2*sinh(x)^7 + 21*(2*A + 5*B)*a^2*cosh(x)^6 + 35*(10*A + 11*B)*a^2
*cosh(x)^5 + 525*(4*A + 3*B)*a^2*cosh(x)^4 + 21*(5*B*a^2*cosh(x) + (2*A + 5*B)*a^2)*sinh(x)^6 - 525*(4*A + 3*B
)*a^2*cosh(x)^3 + 7*(45*B*a^2*cosh(x)^2 + 18*(2*A + 5*B)*a^2*cosh(x) + 5*(10*A + 11*B)*a^2)*sinh(x)^5 - 35*(10
*A + 11*B)*a^2*cosh(x)^2 + 35*(15*B*a^2*cosh(x)^3 + 9*(2*A + 5*B)*a^2*cosh(x)^2 + 5*(10*A + 11*B)*a^2*cosh(x)
+ 15*(4*A + 3*B)*a^2)*sinh(x)^4 - 21*(2*A + 5*B)*a^2*cosh(x) + 35*(15*B*a^2*cosh(x)^4 + 12*(2*A + 5*B)*a^2*cos
h(x)^3 + 10*(10*A + 11*B)*a^2*cosh(x)^2 + 60*(4*A + 3*B)*a^2*cosh(x) - 15*(4*A + 3*B)*a^2)*sinh(x)^3 - 15*B*a^
2 + 35*(9*B*a^2*cosh(x)^5 + 9*(2*A + 5*B)*a^2*cosh(x)^4 + 10*(10*A + 11*B)*a^2*cosh(x)^3 + 90*(4*A + 3*B)*a^2*
cosh(x)^2 - 45*(4*A + 3*B)*a^2*cosh(x) - (10*A + 11*B)*a^2)*sinh(x)^2 + 7*(15*B*a^2*cosh(x)^6 + 18*(2*A + 5*B)
*a^2*cosh(x)^5 + 25*(10*A + 11*B)*a^2*cosh(x)^4 + 300*(4*A + 3*B)*a^2*cosh(x)^3 - 225*(4*A + 3*B)*a^2*cosh(x)^
2 - 10*(10*A + 11*B)*a^2*cosh(x) - 3*(2*A + 5*B)*a^2)*sinh(x))*sqrt(a/(cosh(x) + sinh(x)))/(cosh(x)^3 + 3*cosh
(x)^2*sinh(x) + 3*cosh(x)*sinh(x)^2 + sinh(x)^3)

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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+a*cosh(x))**(5/2)*(A+B*cosh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.24837, size = 207, normalized size = 2.2 \begin{align*} -\frac{1}{840} \, \sqrt{2}{\left (\frac{{\left (2100 \, A a^{6} e^{\left (3 \, x\right )} + 1575 \, B a^{6} e^{\left (3 \, x\right )} + 350 \, A a^{6} e^{\left (2 \, x\right )} + 385 \, B a^{6} e^{\left (2 \, x\right )} + 42 \, A a^{6} e^{x} + 105 \, B a^{6} e^{x} + 15 \, B a^{6}\right )} e^{\left (-\frac{7}{2} \, x\right )}}{a^{\frac{7}{2}}} - \frac{15 \, B a^{\frac{19}{2}} e^{\left (\frac{7}{2} \, x\right )} + 42 \, A a^{\frac{19}{2}} e^{\left (\frac{5}{2} \, x\right )} + 105 \, B a^{\frac{19}{2}} e^{\left (\frac{5}{2} \, x\right )} + 350 \, A a^{\frac{19}{2}} e^{\left (\frac{3}{2} \, x\right )} + 385 \, B a^{\frac{19}{2}} e^{\left (\frac{3}{2} \, x\right )} + 2100 \, A a^{\frac{19}{2}} e^{\left (\frac{1}{2} \, x\right )} + 1575 \, B a^{\frac{19}{2}} e^{\left (\frac{1}{2} \, x\right )}}{a^{7}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cosh(x))^(5/2)*(A+B*cosh(x)),x, algorithm="giac")

[Out]

-1/840*sqrt(2)*((2100*A*a^6*e^(3*x) + 1575*B*a^6*e^(3*x) + 350*A*a^6*e^(2*x) + 385*B*a^6*e^(2*x) + 42*A*a^6*e^
x + 105*B*a^6*e^x + 15*B*a^6)*e^(-7/2*x)/a^(7/2) - (15*B*a^(19/2)*e^(7/2*x) + 42*A*a^(19/2)*e^(5/2*x) + 105*B*
a^(19/2)*e^(5/2*x) + 350*A*a^(19/2)*e^(3/2*x) + 385*B*a^(19/2)*e^(3/2*x) + 2100*A*a^(19/2)*e^(1/2*x) + 1575*B*
a^(19/2)*e^(1/2*x))/a^7)