3.3.24 \(\int \frac {(e+f x)^2 \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [224]

Optimal. Leaf size=337 \[ \frac {(e+f x)^3}{3 b f}-\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}+\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}-\frac {2 a f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {2 a f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {2 a f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {2 a f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3} \]

[Out]

1/3*(f*x+e)^3/b/f-a*(f*x+e)^2*ln(1+b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b/d/(a^2+b^2)^(1/2)+a*(f*x+e)^2*ln(1+b*ex
p(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/d/(a^2+b^2)^(1/2)-2*a*f*(f*x+e)*polylog(2,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b
/d^2/(a^2+b^2)^(1/2)+2*a*f*(f*x+e)*polylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/d^2/(a^2+b^2)^(1/2)+2*a*f^2*
polylog(3,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b/d^3/(a^2+b^2)^(1/2)-2*a*f^2*polylog(3,-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 = 0.49, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5676, 32, 3403, 2296, 2221, 2611, 2320, 6724} \begin {gather*} \frac {2 a f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {2 a f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {2 a f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^2 \sqrt {a^2+b^2}}+\frac {2 a f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {a (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d \sqrt {a^2+b^2}}+\frac {a (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d \sqrt {a^2+b^2}}+\frac {(e+f x)^3}{3 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3403

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 5676

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]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac {\int (e+f x)^2 \, dx}{b}-\frac {a \int \frac {(e+f x)^2}{a+b \sinh (c+d x)} \, dx}{b}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {(2 a) \int \frac {e^{c+d x} (e+f x)^2}{-b+2 a e^{c+d x}+b e^{2 (c+d x)}} \, dx}{b}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {(2 a) \int \frac {e^{c+d x} (e+f x)^2}{2 a-2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{\sqrt {a^2+b^2}}+\frac {(2 a) \int \frac {e^{c+d x} (e+f x)^2}{2 a+2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{\sqrt {a^2+b^2}}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}+\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}+\frac {(2 a f) \int (e+f x) \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}-\frac {(2 a f) \int (e+f x) \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}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}+\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}-\frac {2 a f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {2 a f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {\left (2 a f^2\right ) \int \text {Li}_2\left (-\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 a f^2\right ) \int \text {Li}_2\left (-\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 {(e+f x)^3}{3 b f}-\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}+\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}-\frac {2 a f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {2 a f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {\left (2 a f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {b x}{-a+\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 a f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b \sqrt {a^2+b^2} d^3}\\ &=\frac {(e+f x)^3}{3 b f}-\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}+\frac {a (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}-\frac {2 a f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {2 a f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {2 a f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {2 a f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}\\ \end {align*}

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Mathematica [A]
time = 2.04, size = 610, normalized size = 1.81 \begin {gather*} \frac {x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 b}+\frac {a \left (2 d^2 e^2 \sqrt {\left (a^2+b^2\right ) e^{2 c}} \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-2 \sqrt {a^2+b^2} d^2 e e^c f x \log \left (1+\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-\sqrt {a^2+b^2} d^2 e^c f^2 x^2 \log \left (1+\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+2 \sqrt {a^2+b^2} d^2 e e^c f x \log \left (1+\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+\sqrt {a^2+b^2} d^2 e^c f^2 x^2 \log \left (1+\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-2 \sqrt {a^2+b^2} d e^c f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+2 \sqrt {a^2+b^2} d e^c f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+2 \sqrt {a^2+b^2} e^c f^2 \text {PolyLog}\left (3,-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-2 \sqrt {a^2+b^2} e^c f^2 \text {PolyLog}\left (3,-\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )\right )}{b \sqrt {a^2+b^2} d^3 \sqrt {\left (a^2+b^2\right ) e^{2 c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.47, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{2} \sinh \left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 956 vs. \(2 (308) = 616\).
time = 0.36, size = 956, normalized size = 2.84 \begin {gather*} \frac {{\left (a^{2} + b^{2}\right )} d^{3} f^{2} x^{3} + 3 \, {\left (a^{2} + b^{2}\right )} d^{3} f x^{2} \cosh \left (1\right ) + 3 \, {\left (a^{2} + b^{2}\right )} d^{3} x \cosh \left (1\right )^{2} + 3 \, {\left (a^{2} + b^{2}\right )} d^{3} x \sinh \left (1\right )^{2} + 6 \, a b f^{2} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm polylog}\left (3, \frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}}}{b}\right ) - 6 \, a b f^{2} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm polylog}\left (3, \frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}}}{b}\right ) - 6 \, {\left (a b d f^{2} x + a b d f \cosh \left (1\right ) + a b d f \sinh \left (1\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + 6 \, {\left (a b d f^{2} x + a b d f \cosh \left (1\right ) + a b d f \sinh \left (1\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + 3 \, {\left (a b c^{2} f^{2} - 2 \, a b c d f \cosh \left (1\right ) + a b d^{2} \cosh \left (1\right )^{2} + a b d^{2} \sinh \left (1\right )^{2} - 2 \, {\left (a b c d f - a b d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - 3 \, {\left (a b c^{2} f^{2} - 2 \, a b c d f \cosh \left (1\right ) + a b d^{2} \cosh \left (1\right )^{2} + a b d^{2} \sinh \left (1\right )^{2} - 2 \, {\left (a b c d f - a b d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) - 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - 3 \, {\left (a b d^{2} f^{2} x^{2} - a b c^{2} f^{2} + 2 \, {\left (a b d^{2} f x + a b c d f\right )} \cosh \left (1\right ) + 2 \, {\left (a b d^{2} f x + a b c d f\right )} \sinh \left (1\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) + 3 \, {\left (a b d^{2} f^{2} x^{2} - a b c^{2} f^{2} + 2 \, {\left (a b d^{2} f x + a b c d f\right )} \cosh \left (1\right ) + 2 \, {\left (a b d^{2} f x + a b c d f\right )} \sinh \left (1\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) + 3 \, {\left ({\left (a^{2} + b^{2}\right )} d^{3} f x^{2} + 2 \, {\left (a^{2} + b^{2}\right )} d^{3} x \cosh \left (1\right )\right )} \sinh \left (1\right )}{3 \, {\left (a^{2} b + b^{3}\right )} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((a^2 + b^2)*d^3*f^2*x^3 + 3*(a^2 + b^2)*d^3*f*x^2*cosh(1) + 3*(a^2 + b^2)*d^3*x*cosh(1)^2 + 3*(a^2 + b^2)
*d^3*x*sinh(1)^2 + 6*a*b*f^2*sqrt((a^2 + b^2)/b^2)*polylog(3, (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) - 6*a*b*f^2*sqrt((a^2 + b^2)/b^2)*polylog(3, (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) - 6*(a*b*d*f^2*x + a*b*d*
f*cosh(1) + a*b*d*f*sinh(1))*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) + 6*(a*b*d*f^2*x + a*b*d*f*cosh(1) + a*b*d*f*sinh(1))*sq
rt((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) + 3*(a*b*c^2*f^2 - 2*a*b*c*d*f*cosh(1) + a*b*d^2*cosh(1)^2 + a*b*d^2*sinh(1)^2 - 2*(a*b
*c*d*f - a*b*d^2*cosh(1))*sinh(1))*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) - 3*(a*b*c^2*f^2 - 2*a*b*c*d*f*cosh(1) + a*b*d^2*cosh(1)^2 + a*b*d^2*sinh(1)^2 - 2*(a*
b*c*d*f - a*b*d^2*cosh(1))*sinh(1))*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) - 3*(a*b*d^2*f^2*x^2 - a*b*c^2*f^2 + 2*(a*b*d^2*f*x + a*b*c*d*f)*cosh(1) + 2*(a*b*d^2
*f*x + a*b*c*d*f)*sinh(1))*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) + 3*(a*b*d^2*f^2*x^2 - a*b*c^2*f^2 + 2*(a*b*d^2*f*x + a*b*c*d*f
)*cosh(1) + 2*(a*b*d^2*f*x + a*b*c*d*f)*sinh(1))*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) + 3*((a^2 + b^2)*d^3*f*x^2 + 2*(a^2 + b^2
)*d^3*x*cosh(1))*sinh(1))/((a^2*b + b^3)*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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