3.4.98 \(\int \frac {1}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\) [398]
Optimal. Leaf size=251 \[ -\frac {b \cosh (e+f x) \sinh (e+f x)}{3 a (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (2 a-b) b \cosh (e+f x) \sinh (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {2 i (2 a-b) E\left (i e+i f x\left |\frac {b}{a}\right .\right ) \sqrt {a+b \sinh ^2(e+f x)}}{3 a^2 (a-b)^2 f \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}+\frac {i F\left (i e+i f x\left |\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}{3 a (a-b) f \sqrt {a+b \sinh ^2(e+f x)}} \]
[Out]
-1/3*b*cosh(f*x+e)*sinh(f*x+e)/a/(a-b)/f/(a+b*sinh(f*x+e)^2)^(3/2)-2/3*(2*a-b)*b*cosh(f*x+e)*sinh(f*x+e)/a^2/(
a-b)^2/f/(a+b*sinh(f*x+e)^2)^(1/2)-2/3*I*(2*a-b)*(cos(I*e+I*f*x)^2)^(1/2)/cos(I*e+I*f*x)*EllipticE(sin(I*e+I*f
*x),(b/a)^(1/2))*(a+b*sinh(f*x+e)^2)^(1/2)/a^2/(a-b)^2/f/(1+b*sinh(f*x+e)^2/a)^(1/2)+1/3*I*(cos(I*e+I*f*x)^2)^
(1/2)/cos(I*e+I*f*x)*EllipticF(sin(I*e+I*f*x),(b/a)^(1/2))*(1+b*sinh(f*x+e)^2/a)^(1/2)/a/(a-b)/f/(a+b*sinh(f*x
+e)^2)^(1/2)
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Rubi [A]
time = 0.20, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3263, 3252,
3251, 3257, 3256, 3262, 3261} \begin {gather*} -\frac {2 b (2 a-b) \sinh (e+f x) \cosh (e+f x)}{3 a^2 f (a-b)^2 \sqrt {a+b \sinh ^2(e+f x)}}-\frac {2 i (2 a-b) \sqrt {a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac {b}{a}\right .\right )}{3 a^2 f (a-b)^2 \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}}-\frac {b \sinh (e+f x) \cosh (e+f x)}{3 a f (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {i \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1} F\left (i e+i f x\left |\frac {b}{a}\right .\right )}{3 a f (a-b) \sqrt {a+b \sinh ^2(e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*Sinh[e + f*x]^2)^(-5/2),x]
[Out]
-1/3*(b*Cosh[e + f*x]*Sinh[e + f*x])/(a*(a - b)*f*(a + b*Sinh[e + f*x]^2)^(3/2)) - (2*(2*a - b)*b*Cosh[e + f*x
]*Sinh[e + f*x])/(3*a^2*(a - b)^2*f*Sqrt[a + b*Sinh[e + f*x]^2]) - (((2*I)/3)*(2*a - b)*EllipticE[I*e + I*f*x,
b/a]*Sqrt[a + b*Sinh[e + f*x]^2])/(a^2*(a - b)^2*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) + ((I/3)*EllipticF[I*e +
I*f*x, b/a]*Sqrt[1 + (b*Sinh[e + f*x]^2)/a])/(a*(a - b)*f*Sqrt[a + b*Sinh[e + f*x]^2])
Rule 3251
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[
B/b, Int[Sqrt[a + b*Sin[e + f*x]^2], x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /;
FreeQ[{a, b, e, f, A, B}, x]
Rule 3252
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Dist[
1/(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b
- a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]
Rule 3256
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]/f)*EllipticE[e + f*x, -b/a], x] /
; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
Rule 3257
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[1 + b*(Sin
[e + f*x]^2/a)], Int[Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] && !GtQ[a, 0]
Rule 3261
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(Sqrt[a]*f))*EllipticF[e + f*x, -b/a]
, x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
Rule 3262
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] && !GtQ[a, 0]
Rule 3263
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^
(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && N
eQ[a + b, 0] && LtQ[p, -1]
Rubi steps
\begin {align*} \int \frac {1}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=-\frac {b \cosh (e+f x) \sinh (e+f x)}{3 a (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\int \frac {-3 a+2 b+b \sinh ^2(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx}{3 a (a-b)}\\ &=-\frac {b \cosh (e+f x) \sinh (e+f x)}{3 a (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (2 a-b) b \cosh (e+f x) \sinh (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {\int \frac {-a (3 a-b)-2 (2 a-b) b \sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx}{3 a^2 (a-b)^2}\\ &=-\frac {b \cosh (e+f x) \sinh (e+f x)}{3 a (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (2 a-b) b \cosh (e+f x) \sinh (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {\int \frac {1}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx}{3 a (a-b)}+\frac {(2 (2 a-b)) \int \sqrt {a+b \sinh ^2(e+f x)} \, dx}{3 a^2 (a-b)^2}\\ &=-\frac {b \cosh (e+f x) \sinh (e+f x)}{3 a (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (2 a-b) b \cosh (e+f x) \sinh (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\left (2 (2 a-b) \sqrt {a+b \sinh ^2(e+f x)}\right ) \int \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}} \, dx}{3 a^2 (a-b)^2 \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}-\frac {\sqrt {1+\frac {b \sinh ^2(e+f x)}{a}} \int \frac {1}{\sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}} \, dx}{3 a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}\\ &=-\frac {b \cosh (e+f x) \sinh (e+f x)}{3 a (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (2 a-b) b \cosh (e+f x) \sinh (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {2 i (2 a-b) E\left (i e+i f x\left |\frac {b}{a}\right .\right ) \sqrt {a+b \sinh ^2(e+f x)}}{3 a^2 (a-b)^2 f \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}+\frac {i F\left (i e+i f x\left |\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}{3 a (a-b) f \sqrt {a+b \sinh ^2(e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 1.08, size = 190, normalized size = 0.76 \begin {gather*} \frac {-2 i a^2 (2 a-b) \left (\frac {2 a-b+b \cosh (2 (e+f x))}{a}\right )^{3/2} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )+i a^2 (a-b) \left (\frac {2 a-b+b \cosh (2 (e+f x))}{a}\right )^{3/2} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )+\sqrt {2} b \left (-5 a^2+5 a b-b^2+b (-2 a+b) \cosh (2 (e+f x))\right ) \sinh (2 (e+f x))}{3 a^2 (a-b)^2 f (2 a-b+b \cosh (2 (e+f x)))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Sinh[e + f*x]^2)^(-5/2),x]
[Out]
((-2*I)*a^2*(2*a - b)*((2*a - b + b*Cosh[2*(e + f*x)])/a)^(3/2)*EllipticE[I*(e + f*x), b/a] + I*a^2*(a - b)*((
2*a - b + b*Cosh[2*(e + f*x)])/a)^(3/2)*EllipticF[I*(e + f*x), b/a] + Sqrt[2]*b*(-5*a^2 + 5*a*b - b^2 + b*(-2*
a + b)*Cosh[2*(e + f*x)])*Sinh[2*(e + f*x)])/(3*a^2*(a - b)^2*f*(2*a - b + b*Cosh[2*(e + f*x)])^(3/2))
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Maple [A]
time = 1.76, size = 406, normalized size = 1.62
| | |
| method |
result |
size |
| | |
| default |
\(\frac {\sqrt {\left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right ) \left (\cosh ^{2}\left (f x +e \right )\right )}\, \left (-\frac {\sinh \left (f x +e \right ) \sqrt {\left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right ) \left (\cosh ^{2}\left (f x +e \right )\right )}}{3 a b \left (a -b \right ) \left (\sinh ^{2}\left (f x +e \right )+\frac {a}{b}\right )^{2}}-\frac {2 b \left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right ) \left (-b +2 a \right )}{3 a^{2} \left (a -b \right )^{2} \sqrt {\left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right ) \left (\cosh ^{2}\left (f x +e \right )\right )}}+\frac {\left (3 a -b \right ) \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )}{\left (3 a^{3}-6 a^{2} b +3 a \,b^{2}\right ) \sqrt {-\frac {b}{a}}\, \sqrt {\left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right ) \left (\cosh ^{2}\left (f x +e \right )\right )}}-\frac {2 b \left (-b +2 a \right ) \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \left (\EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-\EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )\right )}{3 a^{2} \left (a -b \right )^{2} \sqrt {-\frac {b}{a}}\, \sqrt {\left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right ) \left (\cosh ^{2}\left (f x +e \right )\right )}}\right )}{\cosh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f}\) |
\(406\) |
| risch |
\(\text {Expression too large to display}\) |
\(16744\) |
| | |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*sinh(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
[Out]
((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2)*(-1/3/a/b/(a-b)*sinh(f*x+e)*((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2
)/(sinh(f*x+e)^2+a/b)^2-2/3*b*cosh(f*x+e)^2/a^2/(a-b)^2*sinh(f*x+e)*(-b+2*a)/((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^
2)^(1/2)+(3*a-b)/(3*a^3-6*a^2*b+3*a*b^2)/(-1/a*b)^(1/2)*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)/((
a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))-2/3*b*(-b+2*a)/a^2/(
a-b)^2/(-1/a*b)^(1/2)*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)/((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^
(1/2)*(EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))-EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))))/c
osh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)/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(1/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")
[Out]
integrate((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. 5442 vs.
\(2 (259) = 518\).
time = 0.19, size = 5442, normalized size = 21.68 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="fricas")
[Out]
2/3*(((4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)^8 + 8*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)*sinh(f*x + e)^
7 + (4*a^2*b^3 - 4*a*b^4 + b^5)*sinh(f*x + e)^8 + 4*(8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^5)*cosh(f*x + e)^6 +
4*(8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^5 + 7*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)^2)*sinh(f*x + e)^6 +
8*(7*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)^3 + 3*(8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^5)*cosh(f*x + e))*s
inh(f*x + e)^5 + 4*a^2*b^3 - 4*a*b^4 + b^5 + 2*(32*a^4*b - 64*a^3*b^2 + 52*a^2*b^3 - 20*a*b^4 + 3*b^5)*cosh(f*
x + e)^4 + 2*(32*a^4*b - 64*a^3*b^2 + 52*a^2*b^3 - 20*a*b^4 + 3*b^5 + 35*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x
+ e)^4 + 30*(8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^5)*cosh(f*x + e)^2)*sinh(f*x + e)^4 + 8*(7*(4*a^2*b^3 - 4*a*
b^4 + b^5)*cosh(f*x + e)^5 + 10*(8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^5)*cosh(f*x + e)^3 + (32*a^4*b - 64*a^3*
b^2 + 52*a^2*b^3 - 20*a*b^4 + 3*b^5)*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^
5)*cosh(f*x + e)^2 + 4*(7*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)^6 + 8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^5
+ 15*(8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^5)*cosh(f*x + e)^4 + 3*(32*a^4*b - 64*a^3*b^2 + 52*a^2*b^3 - 20*a*
b^4 + 3*b^5)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 8*((4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)^7 + 3*(8*a^3*b^2
- 12*a^2*b^3 + 6*a*b^4 - b^5)*cosh(f*x + e)^5 + (32*a^4*b - 64*a^3*b^2 + 52*a^2*b^3 - 20*a*b^4 + 3*b^5)*cosh(f
*x + e)^3 + (8*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 - b^5)*cosh(f*x + e))*sinh(f*x + e) - 2*((2*a*b^4 - b^5)*cosh(f*
x + e)^8 + 8*(2*a*b^4 - b^5)*cosh(f*x + e)*sinh(f*x + e)^7 + (2*a*b^4 - b^5)*sinh(f*x + e)^8 + 4*(4*a^2*b^3 -
4*a*b^4 + b^5)*cosh(f*x + e)^6 + 4*(4*a^2*b^3 - 4*a*b^4 + b^5 + 7*(2*a*b^4 - b^5)*cosh(f*x + e)^2)*sinh(f*x +
e)^6 + 8*(7*(2*a*b^4 - b^5)*cosh(f*x + e)^3 + 3*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e))*sinh(f*x + e)^5 + 2
*a*b^4 - b^5 + 2*(16*a^3*b^2 - 24*a^2*b^3 + 14*a*b^4 - 3*b^5)*cosh(f*x + e)^4 + 2*(16*a^3*b^2 - 24*a^2*b^3 + 1
4*a*b^4 - 3*b^5 + 35*(2*a*b^4 - b^5)*cosh(f*x + e)^4 + 30*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)^2)*sinh(f*
x + e)^4 + 8*(7*(2*a*b^4 - b^5)*cosh(f*x + e)^5 + 10*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)^3 + (16*a^3*b^2
- 24*a^2*b^3 + 14*a*b^4 - 3*b^5)*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)
^2 + 4*(7*(2*a*b^4 - b^5)*cosh(f*x + e)^6 + 4*a^2*b^3 - 4*a*b^4 + b^5 + 15*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*
x + e)^4 + 3*(16*a^3*b^2 - 24*a^2*b^3 + 14*a*b^4 - 3*b^5)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 8*((2*a*b^4 - b^5
)*cosh(f*x + e)^7 + 3*(4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e)^5 + (16*a^3*b^2 - 24*a^2*b^3 + 14*a*b^4 - 3*b^
5)*cosh(f*x + e)^3 + (4*a^2*b^3 - 4*a*b^4 + b^5)*cosh(f*x + e))*sinh(f*x + e))*sqrt((a^2 - a*b)/b^2))*sqrt(b)*
sqrt((2*b*sqrt((a^2 - a*b)/b^2) - 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 - a*b)/b^2) - 2*a + b)/b)*
(cosh(f*x + e) + sinh(f*x + e))), (8*a^2 - 8*a*b + b^2 + 4*(2*a*b - b^2)*sqrt((a^2 - a*b)/b^2))/b^2) - ((6*a^3
*b^2 - 5*a^2*b^3 + a*b^4)*cosh(f*x + e)^8 + 8*(6*a^3*b^2 - 5*a^2*b^3 + a*b^4)*cosh(f*x + e)*sinh(f*x + e)^7 +
(6*a^3*b^2 - 5*a^2*b^3 + a*b^4)*sinh(f*x + e)^8 + 4*(12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4)*cosh(f*x + e)^
6 + 4*(12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4 + 7*(6*a^3*b^2 - 5*a^2*b^3 + a*b^4)*cosh(f*x + e)^2)*sinh(f*x
+ e)^6 + 8*(7*(6*a^3*b^2 - 5*a^2*b^3 + a*b^4)*cosh(f*x + e)^3 + 3*(12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4)
*cosh(f*x + e))*sinh(f*x + e)^5 + 6*a^3*b^2 - 5*a^2*b^3 + a*b^4 + 2*(48*a^5 - 88*a^4*b + 66*a^3*b^2 - 23*a^2*b
^3 + 3*a*b^4)*cosh(f*x + e)^4 + 2*(48*a^5 - 88*a^4*b + 66*a^3*b^2 - 23*a^2*b^3 + 3*a*b^4 + 35*(6*a^3*b^2 - 5*a
^2*b^3 + a*b^4)*cosh(f*x + e)^4 + 30*(12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4)*cosh(f*x + e)^2)*sinh(f*x + e
)^4 + 8*(7*(6*a^3*b^2 - 5*a^2*b^3 + a*b^4)*cosh(f*x + e)^5 + 10*(12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4)*co
sh(f*x + e)^3 + (48*a^5 - 88*a^4*b + 66*a^3*b^2 - 23*a^2*b^3 + 3*a*b^4)*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(12
*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4)*cosh(f*x + e)^2 + 4*(7*(6*a^3*b^2 - 5*a^2*b^3 + a*b^4)*cosh(f*x + e)^
6 + 12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4 + 15*(12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4)*cosh(f*x + e)^4
+ 3*(48*a^5 - 88*a^4*b + 66*a^3*b^2 - 23*a^2*b^3 + 3*a*b^4)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 8*((6*a^3*b^2
- 5*a^2*b^3 + a*b^4)*cosh(f*x + e)^7 + 3*(12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4)*cosh(f*x + e)^5 + (48*a^5
- 88*a^4*b + 66*a^3*b^2 - 23*a^2*b^3 + 3*a*b^4)*cosh(f*x + e)^3 + (12*a^4*b - 16*a^3*b^2 + 7*a^2*b^3 - a*b^4)
*cosh(f*x + e))*sinh(f*x + e) + 2*((3*a^2*b^3 - 5*a*b^4 + 2*b^5)*cosh(f*x + e)^8 + 8*(3*a^2*b^3 - 5*a*b^4 + 2*
b^5)*cosh(f*x + e)*sinh(f*x + e)^7 + (3*a^2*b^3 - 5*a*b^4 + 2*b^5)*sinh(f*x + e)^8 + 4*(6*a^3*b^2 - 13*a^2*b^3
+ 9*a*b^4 - 2*b^5)*cosh(f*x + e)^6 + 4*(6*a^3*b^2 - 13*a^2*b^3 + 9*a*b^4 - 2*b^5 + 7*(3*a^2*b^3 - 5*a*b^4 + 2
*b^5)*cosh(f*x + e)^2)*sinh(f*x + e)^6 + 8*(7*(3*a^2*b^3 - 5*a*b^4 + 2*b^5)*cosh(f*x + e)^3 + 3*(6*a^3*b^2 - 1
3*a^2*b^3 + 9*a*b^4 - 2*b^5)*cosh(f*x + e))*sin...
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sinh(f*x+e)**2)**(5/2),x)
[Out]
Integral((a + b*sinh(e + f*x)**2)**(-5/2), x)
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="giac")
[Out]
Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Evaluation time:
0.48Error: Bad Argument Type
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\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(1/(a + b*sinh(e + f*x)^2)^(5/2),x)
[Out]
int(1/(a + b*sinh(e + f*x)^2)^(5/2), x)
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