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

Optimal. Leaf size=93 \[ \frac{(3 A+5 B) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{2} \sqrt{a \cosh (x)+a}}\right )}{16 \sqrt{2} a^{5/2}}+\frac{(3 A+5 B) \sinh (x)}{16 a (a \cosh (x)+a)^{3/2}}+\frac{(A-B) \sinh (x)}{4 (a \cosh (x)+a)^{5/2}} \]

[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 = 0.0885073, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2750, 2650, 2649, 206} \[ \frac{(3 A+5 B) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{2} \sqrt{a \cosh (x)+a}}\right )}{16 \sqrt{2} a^{5/2}}+\frac{(3 A+5 B) \sinh (x)}{16 a (a \cosh (x)+a)^{3/2}}+\frac{(A-B) \sinh (x)}{4 (a \cosh (x)+a)^{5/2}} \]

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 = 0.153856, size = 57, normalized size = 0.61 \[ \frac{\sinh (x) ((3 A+5 B) \cosh (x)+7 A+B)+4 (3 A+5 B) \cosh ^5\left (\frac{x}{2}\right ) \tan ^{-1}\left (\sinh \left (\frac{x}{2}\right )\right )}{16 (a (\cosh (x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(3*A + 5*B)*ArcTan[Sinh[x/2]]*Cosh[x/2]^5 + (7*A + B + (3*A + 5*B)*Cosh[x])*Sinh[x])/(16*(a*(1 + Cosh[x]))^
(5/2))

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Maple [B]  time = 0.065, size = 209, normalized size = 2.3 \begin{align*} -{\frac{\sqrt{2}}{32\,{a}^{3}}\sqrt{ \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{2}a} \left ( 3\,A\ln \left ( 2\,{\frac{\sqrt{ \left ( \sinh \left ( x/2 \right ) \right ) ^{2}a}\sqrt{-a}-a}{\cosh \left ( x/2 \right ) }} \right ) \left ( \cosh \left ( x/2 \right ) \right ) ^{4}a+5\,B\ln \left ( 2\,{\frac{\sqrt{ \left ( \sinh \left ( x/2 \right ) \right ) ^{2}a}\sqrt{-a}-a}{\cosh \left ( x/2 \right ) }} \right ) \left ( \cosh \left ( x/2 \right ) \right ) ^{4}a-3\,A\sqrt{ \left ( \sinh \left ( x/2 \right ) \right ) ^{2}a}\sqrt{-a} \left ( \cosh \left ( x/2 \right ) \right ) ^{2}-5\,B\sqrt{ \left ( \sinh \left ( x/2 \right ) \right ) ^{2}a}\sqrt{-a} \left ( \cosh \left ( x/2 \right ) \right ) ^{2}-2\,A\sqrt{-a}\sqrt{ \left ( \sinh \left ( x/2 \right ) \right ) ^{2}a}+2\,B\sqrt{ \left ( \sinh \left ( x/2 \right ) \right ) ^{2}a}\sqrt{-a} \right ) \left ( \cosh \left ({\frac{x}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{-a}}} \left ( \sinh \left ({\frac{x}{2}} \right ) \right ) ^{-1}{\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+B*cosh(x))/(a+a*cosh(x))^(5/2),x)

[Out]

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

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Maxima [B]  time = 2.0715, size = 576, normalized size = 6.19 \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="maxima")

[Out]

1/80*(sqrt(2)*((15*e^(9/2*x) + 70*e^(7/2*x) + 128*e^(5/2*x) - 70*e^(3/2*x) - 15*e^(1/2*x))/(a^(5/2)*e^(5*x) +
5*a^(5/2)*e^(4*x) + 10*a^(5/2)*e^(3*x) + 10*a^(5/2)*e^(2*x) + 5*a^(5/2)*e^x + a^(5/2)) + 15*arctan(e^(1/2*x))/
a^(5/2)) - 128*sqrt(2)*e^(5/2*x)/(a^(5/2)*e^(5*x) + 5*a^(5/2)*e^(4*x) + 10*a^(5/2)*e^(3*x) + 10*a^(5/2)*e^(2*x
) + 5*a^(5/2)*e^x + a^(5/2)))*A + 1/672*(sqrt(2)*((105*e^(9/2*x) + 490*e^(7/2*x) + 896*e^(5/2*x) + 790*e^(3/2*
x) - 105*e^(1/2*x))/(a^(5/2)*e^(5*x) + 5*a^(5/2)*e^(4*x) + 10*a^(5/2)*e^(3*x) + 10*a^(5/2)*e^(2*x) + 5*a^(5/2)
*e^x + a^(5/2)) + 105*arctan(e^(1/2*x))/a^(5/2)) + 7*sqrt(2)*((15*e^(9/2*x) + 70*e^(7/2*x) - 128*e^(5/2*x) - 7
0*e^(3/2*x) - 15*e^(1/2*x))/(a^(5/2)*e^(5*x) + 5*a^(5/2)*e^(4*x) + 10*a^(5/2)*e^(3*x) + 10*a^(5/2)*e^(2*x) + 5
*a^(5/2)*e^x + a^(5/2)) + 15*arctan(e^(1/2*x))/a^(5/2)) - 128*(7*sqrt(2)*sqrt(a)*e^(7/2*x) + 3*sqrt(2)*sqrt(a)
*e^(3/2*x))/(a^3*e^(5*x) + 5*a^3*e^(4*x) + 10*a^3*e^(3*x) + 10*a^3*e^(2*x) + 5*a^3*e^x + a^3))*B

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Fricas [B]  time = 2.24973, size = 1453, normalized size = 15.62 \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]

-1/16*(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)*c
osh(x) + 3*A + 5*B)*sinh(x) + 3*A + 5*B)*sqrt(a)*arctan(sqrt(2)*sqrt(1/2)*sqrt(a)*sqrt(a/(cosh(x) + sinh(x)))/
a) - 2*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 [A]  time = 1.41516, size = 159, normalized size = 1.71 \begin{align*} \frac{\sqrt{2}{\left (3 \, A + 5 \, B\right )} \arctan \left (e^{\left (\frac{1}{2} \, x\right )}\right )}{16 \, a^{\frac{5}{2}}} + \frac{\sqrt{2}{\left (3 \, A a^{\frac{7}{2}} e^{\left (\frac{7}{2} \, x\right )} + 5 \, B a^{\frac{7}{2}} e^{\left (\frac{7}{2} \, x\right )} + 11 \, A a^{\frac{7}{2}} e^{\left (\frac{5}{2} \, x\right )} - 3 \, B a^{\frac{7}{2}} e^{\left (\frac{5}{2} \, x\right )} - 11 \, A a^{\frac{7}{2}} e^{\left (\frac{3}{2} \, x\right )} + 3 \, B a^{\frac{7}{2}} e^{\left (\frac{3}{2} \, x\right )} - 3 \, A a^{\frac{7}{2}} e^{\left (\frac{1}{2} \, x\right )} - 5 \, B a^{\frac{7}{2}} e^{\left (\frac{1}{2} \, x\right )}\right )}}{16 \,{\left (a e^{x} + a\right )}^{4} a^{2}} \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(e^(1/2*x))/a^(5/2) + 1/16*sqrt(2)*(3*A*a^(7/2)*e^(7/2*x) + 5*B*a^(7/2)*e^(7/2*
x) + 11*A*a^(7/2)*e^(5/2*x) - 3*B*a^(7/2)*e^(5/2*x) - 11*A*a^(7/2)*e^(3/2*x) + 3*B*a^(7/2)*e^(3/2*x) - 3*A*a^(
7/2)*e^(1/2*x) - 5*B*a^(7/2)*e^(1/2*x))/((a*e^x + a)^4*a^2)