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

Optimal. Leaf size=94 \[ -\frac{(3 A-5 B) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{2} \sqrt{a-a \cosh (x)}}\right )}{16 \sqrt{2} a^{5/2}}-\frac{(3 A-5 B) \sinh (x)}{16 a (a-a \cosh (x))^{3/2}}-\frac{(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}} \]

[Out]

-((3*A - 5*B)*ArcTan[(Sqrt[a]*Sinh[x])/(Sqrt[2]*Sqrt[a - a*Cosh[x]])])/(16*Sqrt[2]*a^(5/2)) - ((A + B)*Sinh[x]
)/(4*(a - a*Cosh[x])^(5/2)) - ((3*A - 5*B)*Sinh[x])/(16*a*(a - a*Cosh[x])^(3/2))

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Rubi [A]  time = 0.095537, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2750, 2650, 2649, 206} \[ -\frac{(3 A-5 B) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{2} \sqrt{a-a \cosh (x)}}\right )}{16 \sqrt{2} a^{5/2}}-\frac{(3 A-5 B) \sinh (x)}{16 a (a-a \cosh (x))^{3/2}}-\frac{(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*A - 5*B)*ArcTan[(Sqrt[a]*Sinh[x])/(Sqrt[2]*Sqrt[a - a*Cosh[x]])])/(16*Sqrt[2]*a^(5/2)) - ((A + B)*Sinh[x]
)/(4*(a - a*Cosh[x])^(5/2)) - ((3*A - 5*B)*Sinh[x])/(16*a*(a - a*Cosh[x])^(3/2))

Rule 2750

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

Rule 2650

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx &=-\frac{(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}+\frac{(3 A-5 B) \int \frac{1}{(a-a \cosh (x))^{3/2}} \, dx}{8 a}\\ &=-\frac{(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}-\frac{(3 A-5 B) \sinh (x)}{16 a (a-a \cosh (x))^{3/2}}+\frac{(3 A-5 B) \int \frac{1}{\sqrt{a-a \cosh (x)}} \, dx}{32 a^2}\\ &=-\frac{(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}-\frac{(3 A-5 B) \sinh (x)}{16 a (a-a \cosh (x))^{3/2}}+\frac{(i (3 A-5 B)) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{i a \sinh (x)}{\sqrt{a-a \cosh (x)}}\right )}{16 a^2}\\ &=-\frac{(3 A-5 B) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{2} \sqrt{a-a \cosh (x)}}\right )}{16 \sqrt{2} a^{5/2}}-\frac{(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}-\frac{(3 A-5 B) \sinh (x)}{16 a (a-a \cosh (x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.360171, size = 108, normalized size = 1.15 \[ \frac{\sinh ^5\left (\frac{x}{2}\right ) \left (-(A+B) \text{csch}^4\left (\frac{x}{4}\right )+2 (3 A-5 B) \text{csch}^2\left (\frac{x}{4}\right )+(A+B) \text{sech}^4\left (\frac{x}{4}\right )+2 (3 A-5 B) \text{sech}^2\left (\frac{x}{4}\right )+8 (3 A-5 B) \log \left (\tanh \left (\frac{x}{4}\right )\right )\right )}{32 a^2 (\cosh (x)-1)^2 \sqrt{a-a \cosh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(3*A - 5*B)*Csch[x/4]^2 - (A + B)*Csch[x/4]^4 + 8*(3*A - 5*B)*Log[Tanh[x/4]] + 2*(3*A - 5*B)*Sech[x/4]^2 +
(A + B)*Sech[x/4]^4)*Sinh[x/2]^5)/(32*a^2*(-1 + Cosh[x])^2*Sqrt[a - a*Cosh[x]])

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Maple [A]  time = 0.07, size = 118, normalized size = 1.3 \begin{align*} -{\frac{1}{32\,{a}^{2}} \left ( \left ( -6\,A+10\,B \right ) \cosh \left ({\frac{x}{2}} \right ) \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{2}+ \left ( 4\,A+4\,B \right ) \cosh \left ({\frac{x}{2}} \right ) + \left ( 3\,\ln \left ( 1+\cosh \left ( x/2 \right ) \right ) A-3\,\ln \left ( -1+\cosh \left ( x/2 \right ) \right ) A-5\,B\ln \left ( 1+\cosh \left ( x/2 \right ) \right ) +5\,B\ln \left ( -1+\cosh \left ( x/2 \right ) \right ) \right ) \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{4} \right ) \left ( 1+\cosh \left ({\frac{x}{2}} \right ) \right ) ^{-1} \left ( -1+\cosh \left ({\frac{x}{2}} \right ) \right ) ^{-1} \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sinh \left ( x/2 \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/a^2*((-6*A+10*B)*cosh(1/2*x)*sinh(1/2*x)^2+(4*A+4*B)*cosh(1/2*x)+(3*ln(1+cosh(1/2*x))*A-3*ln(-1+cosh(1/2
*x))*A-5*B*ln(1+cosh(1/2*x))+5*B*ln(-1+cosh(1/2*x)))*sinh(1/2*x)^4)/(1+cosh(1/2*x))/(-1+cosh(1/2*x))/sinh(1/2*
x)/(-2*sinh(1/2*x)^2*a)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cosh \left (x\right ) + A}{{\left (-a \cosh \left (x\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(x))/(a-a*cosh(x))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.24865, 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+B*cosh(x))/(a-a*cosh(x))^(5/2),x, algorithm="fricas")

[Out]

1/32*(sqrt(2)*((3*A - 5*B)*cosh(x)^4 + (3*A - 5*B)*sinh(x)^4 - 4*(3*A - 5*B)*cosh(x)^3 + 4*((3*A - 5*B)*cosh(x
) - 3*A + 5*B)*sinh(x)^3 + 6*(3*A - 5*B)*cosh(x)^2 + 6*((3*A - 5*B)*cosh(x)^2 - 2*(3*A - 5*B)*cosh(x) + 3*A -
5*B)*sinh(x)^2 - 4*(3*A - 5*B)*cosh(x) + 4*((3*A - 5*B)*cosh(x)^3 - 3*(3*A - 5*B)*cosh(x)^2 + 3*(3*A - 5*B)*co
sh(x) - 3*A + 5*B)*sinh(x) + 3*A - 5*B)*sqrt(-a)*log((2*sqrt(2)*sqrt(1/2)*sqrt(-a)*sqrt(-a/(cosh(x) + sinh(x))
)*(cosh(x) + sinh(x)) - a*cosh(x) - a*sinh(x) - a)/(cosh(x) + sinh(x) - 1)) - 4*sqrt(1/2)*((3*A - 5*B)*cosh(x)
^4 + (3*A - 5*B)*sinh(x)^4 - (11*A + 3*B)*cosh(x)^3 + (4*(3*A - 5*B)*cosh(x) - 11*A - 3*B)*sinh(x)^3 - (11*A +
3*B)*cosh(x)^2 + (6*(3*A - 5*B)*cosh(x)^2 - 3*(11*A + 3*B)*cosh(x) - 11*A - 3*B)*sinh(x)^2 + (3*A - 5*B)*cosh
(x) + (4*(3*A - 5*B)*cosh(x)^3 - 3*(11*A + 3*B)*cosh(x)^2 - 2*(11*A + 3*B)*cosh(x) + 3*A - 5*B)*sinh(x))*sqrt(
-a/(cosh(x) + sinh(x))))/(a^3*cosh(x)^4 + a^3*sinh(x)^4 - 4*a^3*cosh(x)^3 + 6*a^3*cosh(x)^2 - 4*a^3*cosh(x) +
4*(a^3*cosh(x) - a^3)*sinh(x)^3 + a^3 + 6*(a^3*cosh(x)^2 - 2*a^3*cosh(x) + a^3)*sinh(x)^2 + 4*(a^3*cosh(x)^3 -
3*a^3*cosh(x)^2 + 3*a^3*cosh(x) - a^3)*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((A+B*cosh(x))/(a-a*cosh(x))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.33991, size = 255, normalized size = 2.71 \begin{align*} -\frac{\sqrt{2}{\left (3 \, A - 5 \, B\right )} \arctan \left (\frac{\sqrt{-a e^{x}}}{\sqrt{a}}\right )}{16 \, a^{\frac{5}{2}} \mathrm{sgn}\left (-e^{x} + 1\right )} + \frac{\sqrt{2}{\left (3 \, \sqrt{-a e^{x}} A a^{3} e^{\left (3 \, x\right )} - 5 \, \sqrt{-a e^{x}} B a^{3} e^{\left (3 \, x\right )} - 11 \, \sqrt{-a e^{x}} A a^{3} e^{\left (2 \, x\right )} - 3 \, \sqrt{-a e^{x}} B a^{3} e^{\left (2 \, x\right )} - 11 \, \sqrt{-a e^{x}} A a^{3} e^{x} - 3 \, \sqrt{-a e^{x}} B a^{3} e^{x} + 3 \, \sqrt{-a e^{x}} A a^{3} - 5 \, \sqrt{-a e^{x}} B a^{3}\right )}}{16 \,{\left (a e^{x} - a\right )}^{4} a^{2} \mathrm{sgn}\left (-e^{x} + 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(2)*(3*A - 5*B)*arctan(sqrt(-a*e^x)/sqrt(a))/(a^(5/2)*sgn(-e^x + 1)) + 1/16*sqrt(2)*(3*sqrt(-a*e^x)*
A*a^3*e^(3*x) - 5*sqrt(-a*e^x)*B*a^3*e^(3*x) - 11*sqrt(-a*e^x)*A*a^3*e^(2*x) - 3*sqrt(-a*e^x)*B*a^3*e^(2*x) -
11*sqrt(-a*e^x)*A*a^3*e^x - 3*sqrt(-a*e^x)*B*a^3*e^x + 3*sqrt(-a*e^x)*A*a^3 - 5*sqrt(-a*e^x)*B*a^3)/((a*e^x -
a)^4*a^2*sgn(-e^x + 1))