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

Optimal. Leaf size=264 \[ \frac {a^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^2 \sqrt {a^2+b^2}}-\frac {a^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^2 \sqrt {a^2+b^2}}+\frac {a^2 (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b^2 d \sqrt {a^2+b^2}}-\frac {a^2 (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b^2 d \sqrt {a^2+b^2}}-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}-\frac {f \sinh (c+d x)}{b d^2}+\frac {(e+f x) \cosh (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5557, 3296, 2637, 3322, 2264, 2190, 2279, 2391} \[ \frac {a^2 f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^2 \sqrt {a^2+b^2}}-\frac {a^2 f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b^2 d^2 \sqrt {a^2+b^2}}+\frac {a^2 (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b^2 d \sqrt {a^2+b^2}}-\frac {a^2 (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b^2 d \sqrt {a^2+b^2}}-\frac {a e x}{b^2}-\frac {a f x^2}{2 b^2}-\frac {f \sinh (c+d x)}{b d^2}+\frac {(e+f x) \cosh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 5557

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

Rubi steps

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

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Mathematica [A]  time = 2.13, size = 299, normalized size = 1.13 \[ \frac {\frac {2 a^2 \left (-2 d e \tanh ^{-1}\left (\frac {a+b \sinh (c+d x)+b \cosh (c+d x)}{\sqrt {a^2+b^2}}\right )+f \text {Li}_2\left (\frac {b (\cosh (c+d x)+\sinh (c+d x))}{\sqrt {a^2+b^2}-a}\right )-f \text {Li}_2\left (-\frac {b (\cosh (c+d x)+\sinh (c+d x))}{a+\sqrt {a^2+b^2}}\right )+f (c+d x) \log \left (\frac {b (\sinh (c+d x)+\cosh (c+d x))}{a-\sqrt {a^2+b^2}}+1\right )-f (c+d x) \log \left (\frac {b (\sinh (c+d x)+\cosh (c+d x))}{\sqrt {a^2+b^2}+a}+1\right )+2 c f \tanh ^{-1}\left (\frac {a+b \sinh (c+d x)+b \cosh (c+d x)}{\sqrt {a^2+b^2}}\right )\right )}{\sqrt {a^2+b^2}}+a (c+d x) (c f-d (2 e+f x))+2 b d (e+f x) \cosh (c+d x)-2 b f \sinh (c+d x)}{2 b^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(c + d*x)*(c*f - d*(2*e + f*x)) + 2*b*d*(e + f*x)*Cosh[c + d*x] + (2*a^2*(-2*d*e*ArcTanh[(a + b*Cosh[c + d*
x] + b*Sinh[c + d*x])/Sqrt[a^2 + b^2]] + 2*c*f*ArcTanh[(a + b*Cosh[c + d*x] + b*Sinh[c + d*x])/Sqrt[a^2 + b^2]
] + f*(c + d*x)*Log[1 + (b*(Cosh[c + d*x] + Sinh[c + d*x]))/(a - Sqrt[a^2 + b^2])] - f*(c + d*x)*Log[1 + (b*(C
osh[c + d*x] + Sinh[c + d*x]))/(a + Sqrt[a^2 + b^2])] + f*PolyLog[2, (b*(Cosh[c + d*x] + Sinh[c + d*x]))/(-a +
Sqrt[a^2 + b^2])] - f*PolyLog[2, -((b*(Cosh[c + d*x] + Sinh[c + d*x]))/(a + Sqrt[a^2 + b^2]))]))/Sqrt[a^2 + b
^2] - 2*b*f*Sinh[c + d*x])/(2*b^2*d^2)

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fricas [B]  time = 0.54, size = 946, normalized size = 3.58 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*b + b^3)*d*f*x + (a^2*b + b^3)*d*e + ((a^2*b + b^3)*d*f*x + (a^2*b + b^3)*d*e - (a^2*b + b^3)*f)*cos
h(d*x + c)^2 + ((a^2*b + b^3)*d*f*x + (a^2*b + b^3)*d*e - (a^2*b + b^3)*f)*sinh(d*x + c)^2 + 2*(a^2*b*f*cosh(d
*x + c) + a^2*b*f*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) - 2*(a^2*b*f*cosh(d*x + c) + a^2*b*f*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) - 2*((a^2*b*d*e - a^2*b*c*f)*cosh(d*x + c) + (a^2*b*d*e - a^2*b*c*f)*sinh(d*x + c))*s
qrt((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) + 2*((a^2*b*
d*e - a^2*b*c*f)*cosh(d*x + c) + (a^2*b*d*e - a^2*b*c*f)*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) + 2*((a^2*b*d*f*x + a^2*b*c*f)*cosh(d*x + c) + (a
^2*b*d*f*x + a^2*b*c*f)*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) - 2*((a^2*b*d*f*x + a^2*b*c*f)*cosh(d*x + c) + (a^2
*b*d*f*x + a^2*b*c*f)*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 + b^3)*f - ((a^3 + a*b^2)*d^2*f*x^2 + 2*(a^3
+ a*b^2)*d^2*e*x)*cosh(d*x + c) - ((a^3 + a*b^2)*d^2*f*x^2 + 2*(a^3 + a*b^2)*d^2*e*x - 2*((a^2*b + b^3)*d*f*x
+ (a^2*b + b^3)*d*e - (a^2*b + b^3)*f)*cosh(d*x + c))*sinh(d*x + c))/((a^2*b^2 + b^4)*d^2*cosh(d*x + c) + (a^
2*b^2 + b^4)*d^2*sinh(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.14, size = 510, normalized size = 1.93 \[ -\frac {a f \,x^{2}}{2 b^{2}}-\frac {a e x}{b^{2}}+\frac {\left (d f x +d e -f \right ) {\mathrm e}^{d x +c}}{2 d^{2} b}+\frac {\left (d f x +d e +f \right ) {\mathrm e}^{-d x -c}}{2 d^{2} b}-\frac {2 a^{2} e \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{2} \sqrt {a^{2}+b^{2}}}+\frac {a^{2} f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \,b^{2} \sqrt {a^{2}+b^{2}}}+\frac {a^{2} f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} b^{2} \sqrt {a^{2}+b^{2}}}-\frac {a^{2} f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \,b^{2} \sqrt {a^{2}+b^{2}}}-\frac {a^{2} f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} b^{2} \sqrt {a^{2}+b^{2}}}+\frac {a^{2} f \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} b^{2} \sqrt {a^{2}+b^{2}}}-\frac {a^{2} f \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} b^{2} \sqrt {a^{2}+b^{2}}}+\frac {2 a^{2} f c \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{2} b^{2} \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*f*x^2/b^2-a*e*x/b^2+1/2*(d*f*x+d*e-f)/d^2/b*exp(d*x+c)+1/2*(d*f*x+d*e+f)/d^2/b*exp(-d*x-c)-2/d*a^2/b^2*
e/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))+1/d*a^2/b^2*f/(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+1/d^2*a^2/b^2*f/(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-1/d*a^2/b^2*f/(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-1/d^2*a^2/b^2*f/(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+1/d^2*a^
2/b^2*f/(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)))-1/d^2*a^2/b^2*f/(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^2*a^2/b^2*f*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 \[ \frac {1}{2} \, {\left (4 \, a^{2} \int \frac {x e^{\left (d x + c\right )}}{b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b^{2} e^{\left (d x + c\right )} - b^{3}}\,{d x} - \frac {{\left (a d^{2} x^{2} e^{c} - {\left (b d x e^{\left (2 \, c\right )} - b e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} - {\left (b d x + b\right )} e^{\left (-d x\right )}\right )} e^{\left (-c\right )}}{b^{2} d^{2}}\right )} f + \frac {1}{2} \, e {\left (\frac {2 \, a^{2} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{2} d} - \frac {2 \, {\left (d x + c\right )} a}{b^{2} d} + \frac {e^{\left (d x + c\right )}}{b d} + \frac {e^{\left (-d x - c\right )}}{b d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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