3.4.91 \(\int \frac {\cosh ^5(e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\) [391]

Optimal. Leaf size=134 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {(a-b) \cosh ^2(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(a-b) (3 a+2 b) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}} \]

[Out]

arctanh(sinh(f*x+e)*b^(1/2)/(a+b*sinh(f*x+e)^2)^(1/2))/b^(5/2)/f-1/3*(a-b)*cosh(f*x+e)^2*sinh(f*x+e)/a/b/f/(a+
b*sinh(f*x+e)^2)^(3/2)-1/3*(a-b)*(3*a+2*b)*sinh(f*x+e)/a^2/b^2/f/(a+b*sinh(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3269, 424, 393, 223, 212} \begin {gather*} -\frac {(a-b) (3 a+2 b) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {(a-b) \sinh (e+f x) \cosh ^2(e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]^5/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

ArcTanh[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]]/(b^(5/2)*f) - ((a - b)*Cosh[e + f*x]^2*Sinh[e + f
*x])/(3*a*b*f*(a + b*Sinh[e + f*x]^2)^(3/2)) - ((a - b)*(3*a + 2*b)*Sinh[e + f*x])/(3*a^2*b^2*f*Sqrt[a + b*Sin
h[e + f*x]^2])

Rule 212

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
 + 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
 /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 424

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
 FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 3269

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac {(a-b) \cosh ^2(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {a+2 b+3 a x^2}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 a b f}\\ &=-\frac {(a-b) \cosh ^2(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(a-b) (3 a+2 b) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{b^2 f}\\ &=-\frac {(a-b) \cosh ^2(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(a-b) (3 a+2 b) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{b^2 f}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {(a-b) \cosh ^2(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(a-b) (3 a+2 b) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.64, size = 126, normalized size = 0.94 \begin {gather*} \frac {\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sinh (e+f x)}{\sqrt {2 a-b+b \cosh (2 (e+f x))}}\right )}{b^{5/2}}+\frac {2 \sqrt {2} (-a+b) \left (3 a^2+a b-b^2+b (2 a+b) \cosh (2 (e+f x))\right ) \sinh (e+f x)}{3 a^2 b^2 (2 a-b+b \cosh (2 (e+f x)))^{3/2}}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cosh[e + f*x]^5/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

(ArcTanh[(Sqrt[2]*Sqrt[b]*Sinh[e + f*x])/Sqrt[2*a - b + b*Cosh[2*(e + f*x)]]]/b^(5/2) + (2*Sqrt[2]*(-a + b)*(3
*a^2 + a*b - b^2 + b*(2*a + b)*Cosh[2*(e + f*x)])*Sinh[e + f*x])/(3*a^2*b^2*(2*a - b + b*Cosh[2*(e + f*x)])^(3
/2)))/f

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.40, size = 65, normalized size = 0.49

method result size
default \(\frac {\mathit {`\,int/indef0`\,}\left (\frac {\cosh ^{4}\left (f x +e \right )}{\left (b^{2} \left (\sinh ^{4}\left (f x +e \right )\right )+2 a b \left (\sinh ^{2}\left (f x +e \right )\right )+a^{2}\right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\) \(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

`int/indef0`(cosh(f*x+e)^4/(b^2*sinh(f*x+e)^4+2*a*b*sinh(f*x+e)^2+a^2)/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/
f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(cosh(f*x + e)^5/(b*sinh(f*x + e)^2 + a)^(5/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2932 vs. \(2 (120) = 240\).
time = 0.69, size = 6774, normalized size = 50.55 \begin {gather*} \text {too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(a^2*b^2*cosh(f*x + e)^8 + 8*a^2*b^2*cosh(f*x + e)*sinh(f*x + e)^7 + a^2*b^2*sinh(f*x + e)^8 + 4*(2*a
^3*b - a^2*b^2)*cosh(f*x + e)^6 + 4*(7*a^2*b^2*cosh(f*x + e)^2 + 2*a^3*b - a^2*b^2)*sinh(f*x + e)^6 + 8*(7*a^2
*b^2*cosh(f*x + e)^3 + 3*(2*a^3*b - a^2*b^2)*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(8*a^4 - 8*a^3*b + 3*a^2*b^2)*
cosh(f*x + e)^4 + 2*(35*a^2*b^2*cosh(f*x + e)^4 + 8*a^4 - 8*a^3*b + 3*a^2*b^2 + 30*(2*a^3*b - a^2*b^2)*cosh(f*
x + e)^2)*sinh(f*x + e)^4 + a^2*b^2 + 8*(7*a^2*b^2*cosh(f*x + e)^5 + 10*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^3 +
(8*a^4 - 8*a^3*b + 3*a^2*b^2)*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^2 + 4*(7*a^
2*b^2*cosh(f*x + e)^6 + 15*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^4 + 2*a^3*b - a^2*b^2 + 3*(8*a^4 - 8*a^3*b + 3*a^
2*b^2)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 8*(a^2*b^2*cosh(f*x + e)^7 + 3*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^5 +
 (8*a^4 - 8*a^3*b + 3*a^2*b^2)*cosh(f*x + e)^3 + (2*a^3*b - a^2*b^2)*cosh(f*x + e))*sinh(f*x + e))*sqrt(b)*log
(-((a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^8 + 8*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)*sinh(f*x + e)^7 + (a^2*b
- 2*a*b^2 + b^3)*sinh(f*x + e)^8 + 2*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^6 + 2*(a^3 - 4*a^2*b + 5*
a*b^2 - 2*b^3 + 14*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^2)*sinh(f*x + e)^6 + 4*(14*(a^2*b - 2*a*b^2 + b^3)*co
sh(f*x + e)^3 + 3*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e))*sinh(f*x + e)^5 + (9*a^2*b - 14*a*b^2 + 6*b
^3)*cosh(f*x + e)^4 + (70*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^4 + 9*a^2*b - 14*a*b^2 + 6*b^3 + 30*(a^3 - 4*a
^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^2)*sinh(f*x + e)^4 + 4*(14*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^5 + 10*
(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^3 + (9*a^2*b - 14*a*b^2 + 6*b^3)*cosh(f*x + e))*sinh(f*x + e)^
3 + b^3 + 2*(3*a*b^2 - 2*b^3)*cosh(f*x + e)^2 + 2*(14*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^6 + 15*(a^3 - 4*a^
2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^4 + 3*a*b^2 - 2*b^3 + 3*(9*a^2*b - 14*a*b^2 + 6*b^3)*cosh(f*x + e)^2)*sin
h(f*x + e)^2 + sqrt(2)*((a^2 - 2*a*b + b^2)*cosh(f*x + e)^6 + 6*(a^2 - 2*a*b + b^2)*cosh(f*x + e)*sinh(f*x + e
)^5 + (a^2 - 2*a*b + b^2)*sinh(f*x + e)^6 - 3*(a^2 - 2*a*b + b^2)*cosh(f*x + e)^4 + 3*(5*(a^2 - 2*a*b + b^2)*c
osh(f*x + e)^2 - a^2 + 2*a*b - b^2)*sinh(f*x + e)^4 + 4*(5*(a^2 - 2*a*b + b^2)*cosh(f*x + e)^3 - 3*(a^2 - 2*a*
b + b^2)*cosh(f*x + e))*sinh(f*x + e)^3 - (4*a*b - 3*b^2)*cosh(f*x + e)^2 + (15*(a^2 - 2*a*b + b^2)*cosh(f*x +
 e)^4 - 18*(a^2 - 2*a*b + b^2)*cosh(f*x + e)^2 - 4*a*b + 3*b^2)*sinh(f*x + e)^2 - b^2 + 2*(3*(a^2 - 2*a*b + b^
2)*cosh(f*x + e)^5 - 6*(a^2 - 2*a*b + b^2)*cosh(f*x + e)^3 - (4*a*b - 3*b^2)*cosh(f*x + e))*sinh(f*x + e))*sqr
t(b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) +
 sinh(f*x + e)^2)) + 4*(2*(a^2*b - 2*a*b^2 + b^3)*cosh(f*x + e)^7 + 3*(a^3 - 4*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f
*x + e)^5 + (9*a^2*b - 14*a*b^2 + 6*b^3)*cosh(f*x + e)^3 + (3*a*b^2 - 2*b^3)*cosh(f*x + e))*sinh(f*x + e))/(co
sh(f*x + e)^6 + 6*cosh(f*x + e)^5*sinh(f*x + e) + 15*cosh(f*x + e)^4*sinh(f*x + e)^2 + 20*cosh(f*x + e)^3*sinh
(f*x + e)^3 + 15*cosh(f*x + e)^2*sinh(f*x + e)^4 + 6*cosh(f*x + e)*sinh(f*x + e)^5 + sinh(f*x + e)^6)) + 3*(a^
2*b^2*cosh(f*x + e)^8 + 8*a^2*b^2*cosh(f*x + e)*sinh(f*x + e)^7 + a^2*b^2*sinh(f*x + e)^8 + 4*(2*a^3*b - a^2*b
^2)*cosh(f*x + e)^6 + 4*(7*a^2*b^2*cosh(f*x + e)^2 + 2*a^3*b - a^2*b^2)*sinh(f*x + e)^6 + 8*(7*a^2*b^2*cosh(f*
x + e)^3 + 3*(2*a^3*b - a^2*b^2)*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(8*a^4 - 8*a^3*b + 3*a^2*b^2)*cosh(f*x + e
)^4 + 2*(35*a^2*b^2*cosh(f*x + e)^4 + 8*a^4 - 8*a^3*b + 3*a^2*b^2 + 30*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^2)*si
nh(f*x + e)^4 + a^2*b^2 + 8*(7*a^2*b^2*cosh(f*x + e)^5 + 10*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^3 + (8*a^4 - 8*a
^3*b + 3*a^2*b^2)*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^2 + 4*(7*a^2*b^2*cosh(f
*x + e)^6 + 15*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^4 + 2*a^3*b - a^2*b^2 + 3*(8*a^4 - 8*a^3*b + 3*a^2*b^2)*cosh(
f*x + e)^2)*sinh(f*x + e)^2 + 8*(a^2*b^2*cosh(f*x + e)^7 + 3*(2*a^3*b - a^2*b^2)*cosh(f*x + e)^5 + (8*a^4 - 8*
a^3*b + 3*a^2*b^2)*cosh(f*x + e)^3 + (2*a^3*b - a^2*b^2)*cosh(f*x + e))*sinh(f*x + e))*sqrt(b)*log((b*cosh(f*x
 + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*a*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2
 + a)*sinh(f*x + e)^2 + sqrt(2)*(cosh(f*x + e)^2 + 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2 + 1)*sqrt(b
)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + si
nh(f*x + e)^2)) + 4*(b*cosh(f*x + e)^3 + a*cosh(f*x + e))*sinh(f*x + e) + b)/(cosh(f*x + e)^2 + 2*cosh(f*x + e
)*sinh(f*x + e) + sinh(f*x + e)^2)) - 8*sqrt(2)*((2*a^2*b^2 - a*b^3 - b^4)*cosh(f*x + e)^6 + 6*(2*a^2*b^2 - a*
b^3 - b^4)*cosh(f*x + e)*sinh(f*x + e)^5 + (2*a^2*b^2 - a*b^3 - b^4)*sinh(f*x + e)^6 + 3*(2*a^3*b - 2*a^2*b^2
- a*b^3 + b^4)*cosh(f*x + e)^4 + 3*(2*a^3*b - 2*a^2*b^2 - a*b^3 + b^4 + 5*(2*a^2*b^2 - a*b^3 - b^4)*cosh(f*x +
 e)^2)*sinh(f*x + e)^4 - 2*a^2*b^2 + a*b^3 + b^...

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)**5/(a+b*sinh(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.97Error: Bad Argument Type

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {cosh}\left (e+f\,x\right )}^5}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(e + f*x)^5/(a + b*sinh(e + f*x)^2)^(5/2),x)

[Out]

int(cosh(e + f*x)^5/(a + b*sinh(e + f*x)^2)^(5/2), x)

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