3.325 \(\int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx\)

Optimal. Leaf size=234 \[ \frac {2 f^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {2 f^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}+\frac {2 f (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {2 f (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))} \]

[Out]

-(f*x+e)^2/b/d/(a+b*sinh(d*x+c))+2*f*(f*x+e)*ln(1+b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b/d^2/(a^2+b^2)^(1/2)-2*f*
(f*x+e)*ln(1+b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/d^2/(a^2+b^2)^(1/2)+2*f^2*polylog(2,-b*exp(d*x+c)/(a-(a^2+b^2
)^(1/2)))/b/d^3/(a^2+b^2)^(1/2)-2*f^2*polylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/d^3/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5464, 3322, 2264, 2190, 2279, 2391} \[ \frac {2 f^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {2 f^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b d^3 \sqrt {a^2+b^2}}+\frac {2 f (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {2 f (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*f*(e + f*x)*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(b*Sqrt[a^2 + b^2]*d^2) - (2*f*(e + f*x)*Log[1
+ (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/(b*Sqrt[a^2 + b^2]*d^2) + (2*f^2*PolyLog[2, -((b*E^(c + d*x))/(a - S
qrt[a^2 + b^2]))])/(b*Sqrt[a^2 + b^2]*d^3) - (2*f^2*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(b*S
qrt[a^2 + b^2]*d^3) - (e + f*x)^2/(b*d*(a + b*Sinh[c + d*x]))

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 5464

Int[Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)])^(n_.), x_Symbo
l] :> Simp[((e + f*x)^m*(a + b*Sinh[c + d*x])^(n + 1))/(b*d*(n + 1)), x] - Dist[(f*m)/(b*d*(n + 1)), Int[(e +
f*x)^(m - 1)*(a + b*Sinh[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n,
-1]

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx &=-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))}+\frac {(2 f) \int \frac {e+f x}{a+b \sinh (c+d x)} \, dx}{b d}\\ &=-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))}+\frac {(4 f) \int \frac {e^{c+d x} (e+f x)}{-b+2 a e^{c+d x}+b e^{2 (c+d x)}} \, dx}{b d}\\ &=-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))}+\frac {(4 f) \int \frac {e^{c+d x} (e+f x)}{2 a-2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{\sqrt {a^2+b^2} d}-\frac {(4 f) \int \frac {e^{c+d x} (e+f x)}{2 a+2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{\sqrt {a^2+b^2} d}\\ &=\frac {2 f (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {2 f (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))}-\frac {\left (2 f^2\right ) \int \log \left (1+\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2+b^2}}\right ) \, dx}{b \sqrt {a^2+b^2} d^2}+\frac {\left (2 f^2\right ) \int \log \left (1+\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2+b^2}}\right ) \, dx}{b \sqrt {a^2+b^2} d^2}\\ &=\frac {2 f (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {2 f (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))}-\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 b x}{2 a-2 \sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b \sqrt {a^2+b^2} d^3}+\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 b x}{2 a+2 \sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b \sqrt {a^2+b^2} d^3}\\ &=\frac {2 f (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {2 f (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {2 f^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {2 f^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 175, normalized size = 0.75 \[ \frac {2 f \left (d (e+f x) \left (\log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )-\log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )\right )+f \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*f*(d*(e + f*x)*(Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^
2])]) + f*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] - f*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 +
b^2]))]))/(b*Sqrt[a^2 + b^2]*d^3) - (e + f*x)^2/(b*d*(a + b*Sinh[c + d*x]))

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fricas [B]  time = 0.51, size = 1378, normalized size = 5.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((b^2*f^2*cosh(d*x + c)^2 + b^2*f^2*sinh(d*x + c)^2 + 2*a*b*f^2*cosh(d*x + c) - b^2*f^2 + 2*(b^2*f^2*cosh(d*
x + c) + a*b*f^2)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x
+ c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - (b^2*f^2*cosh(d*x + c)^2 + b^2*f^2*sinh(d*x + c)^2
+ 2*a*b*f^2*cosh(d*x + c) - b^2*f^2 + 2*(b^2*f^2*cosh(d*x + c) + a*b*f^2)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2
)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b
+ 1) + (b^2*d*e*f - b^2*c*f^2 - (b^2*d*e*f - b^2*c*f^2)*cosh(d*x + c)^2 - (b^2*d*e*f - b^2*c*f^2)*sinh(d*x + c
)^2 - 2*(a*b*d*e*f - a*b*c*f^2)*cosh(d*x + c) - 2*(a*b*d*e*f - a*b*c*f^2 + (b^2*d*e*f - b^2*c*f^2)*cosh(d*x +
c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 + b^2)/b^2)
+ 2*a) - (b^2*d*e*f - b^2*c*f^2 - (b^2*d*e*f - b^2*c*f^2)*cosh(d*x + c)^2 - (b^2*d*e*f - b^2*c*f^2)*sinh(d*x
+ c)^2 - 2*(a*b*d*e*f - a*b*c*f^2)*cosh(d*x + c) - 2*(a*b*d*e*f - a*b*c*f^2 + (b^2*d*e*f - b^2*c*f^2)*cosh(d*x
+ c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 + b^2)/b
^2) + 2*a) - (b^2*d*f^2*x + b^2*c*f^2 - (b^2*d*f^2*x + b^2*c*f^2)*cosh(d*x + c)^2 - (b^2*d*f^2*x + b^2*c*f^2)*
sinh(d*x + c)^2 - 2*(a*b*d*f^2*x + a*b*c*f^2)*cosh(d*x + c) - 2*(a*b*d*f^2*x + a*b*c*f^2 + (b^2*d*f^2*x + b^2*
c*f^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d
*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) + (b^2*d*f^2*x + b^2*c*f^2 - (b^2*d*f^2*x + b^2*c*f^2
)*cosh(d*x + c)^2 - (b^2*d*f^2*x + b^2*c*f^2)*sinh(d*x + c)^2 - 2*(a*b*d*f^2*x + a*b*c*f^2)*cosh(d*x + c) - 2*
(a*b*d*f^2*x + a*b*c*f^2 + (b^2*d*f^2*x + b^2*c*f^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*log(-
(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) - ((a^2
+ b^2)*d^2*f^2*x^2 + 2*(a^2 + b^2)*d^2*e*f*x + (a^2 + b^2)*d^2*e^2)*cosh(d*x + c) - ((a^2 + b^2)*d^2*f^2*x^2
+ 2*(a^2 + b^2)*d^2*e*f*x + (a^2 + b^2)*d^2*e^2)*sinh(d*x + c))/((a^2*b^2 + b^4)*d^3*cosh(d*x + c)^2 + (a^2*b^
2 + b^4)*d^3*sinh(d*x + c)^2 + 2*(a^3*b + a*b^3)*d^3*cosh(d*x + c) - (a^2*b^2 + b^4)*d^3 + 2*((a^2*b^2 + b^4)*
d^3*cosh(d*x + c) + (a^3*b + a*b^3)*d^3)*sinh(d*x + c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} \cosh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.00, size = 491, normalized size = 2.10 \[ -\frac {2 \left (x^{2} f^{2}+2 e f x +e^{2}\right ) {\mathrm e}^{d x +c}}{b d \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}-\frac {4 f e \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {a^{2}+b^{2}}}+\frac {2 f^{2} \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d^{2} b \sqrt {a^{2}+b^{2}}}+\frac {2 f^{2} \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{3} b \sqrt {a^{2}+b^{2}}}-\frac {2 f^{2} \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d^{2} b \sqrt {a^{2}+b^{2}}}-\frac {2 f^{2} \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{3} b \sqrt {a^{2}+b^{2}}}+\frac {2 f^{2} \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{3} b \sqrt {a^{2}+b^{2}}}-\frac {2 f^{2} \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{3} b \sqrt {a^{2}+b^{2}}}+\frac {4 f^{2} c \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{3} b \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(f^2*x^2+2*e*f*x+e^2)/b/d*exp(d*x+c)/(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b)-4/d^2/b*f*e/(a^2+b^2)^(1/2)*arctan
h(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))+2/d^2/b*f^2/(a^2+b^2)^(1/2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(
-a+(a^2+b^2)^(1/2)))*x+2/d^3/b*f^2/(a^2+b^2)^(1/2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*
c-2/d^2/b*f^2/(a^2+b^2)^(1/2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*x-2/d^3/b*f^2/(a^2+b^2)
^(1/2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*c+2/d^3/b*f^2/(a^2+b^2)^(1/2)*dilog((-b*exp(d*
x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))-2/d^3/b*f^2/(a^2+b^2)^(1/2)*dilog((b*exp(d*x+c)+(a^2+b^2)^(1/2)+
a)/(a+(a^2+b^2)^(1/2)))+4/d^3/b*f^2*c/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -2 \, {\left (\frac {x^{2} e^{\left (d x + c\right )}}{b^{2} d e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b d e^{\left (d x + c\right )} - b^{2} d} - 2 \, \int \frac {x e^{\left (d x + c\right )}}{b^{2} d e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b d e^{\left (d x + c\right )} - b^{2} d}\,{d x}\right )} f^{2} - 2 \, e f {\left (\frac {2 \, x e^{\left (d x + c\right )}}{b^{2} d e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b d e^{\left (d x + c\right )} - b^{2} d} - \frac {\log \left (\frac {b e^{\left (d x + c\right )} + a - \sqrt {a^{2} + b^{2}}}{b e^{\left (d x + c\right )} + a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b d^{2}}\right )} - \frac {2 \, e^{2} e^{\left (-d x - c\right )}}{{\left (2 \, a b e^{\left (-d x - c\right )} - b^{2} e^{\left (-2 \, d x - 2 \, c\right )} + b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(x^2*e^(d*x + c)/(b^2*d*e^(2*d*x + 2*c) + 2*a*b*d*e^(d*x + c) - b^2*d) - 2*integrate(x*e^(d*x + c)/(b^2*d*e
^(2*d*x + 2*c) + 2*a*b*d*e^(d*x + c) - b^2*d), x))*f^2 - 2*e*f*(2*x*e^(d*x + c)/(b^2*d*e^(2*d*x + 2*c) + 2*a*b
*d*e^(d*x + c) - b^2*d) - log((b*e^(d*x + c) + a - sqrt(a^2 + b^2))/(b*e^(d*x + c) + a + sqrt(a^2 + b^2)))/(sq
rt(a^2 + b^2)*b*d^2)) - 2*e^2*e^(-d*x - c)/((2*a*b*e^(-d*x - c) - b^2*e^(-2*d*x - 2*c) + b^2)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{{\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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