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

Optimal. Leaf size=296 \[ \frac{2 d (c+d x) \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}\right )}{f^2 \sqrt{a^2+b^2}}-\frac{2 d (c+d x) \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )}{f^2 \sqrt{a^2+b^2}}-\frac{2 d^2 \text{PolyLog}\left (3,-\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}\right )}{f^3 \sqrt{a^2+b^2}}+\frac{2 d^2 \text{PolyLog}\left (3,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )}{f^3 \sqrt{a^2+b^2}}+\frac{(c+d x)^2 \log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}+1\right )}{f \sqrt{a^2+b^2}}-\frac{(c+d x)^2 \log \left (\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}+1\right )}{f \sqrt{a^2+b^2}} \]

[Out]

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

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Rubi [A]  time = 0.668979, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3322, 2264, 2190, 2531, 2282, 6589} \[ \frac{2 d (c+d x) \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}\right )}{f^2 \sqrt{a^2+b^2}}-\frac{2 d (c+d x) \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )}{f^2 \sqrt{a^2+b^2}}-\frac{2 d^2 \text{PolyLog}\left (3,-\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}\right )}{f^3 \sqrt{a^2+b^2}}+\frac{2 d^2 \text{PolyLog}\left (3,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )}{f^3 \sqrt{a^2+b^2}}+\frac{(c+d x)^2 \log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}+1\right )}{f \sqrt{a^2+b^2}}-\frac{(c+d x)^2 \log \left (\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}+1\right )}{f \sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 2531

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 2282

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 6589

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

Mathematica [A]  time = 0.155571, size = 233, normalized size = 0.79 \[ \frac{\frac{2 d \left (f (c+d x) \text{PolyLog}\left (2,\frac{b e^{e+f x}}{\sqrt{a^2+b^2}-a}\right )-d \text{PolyLog}\left (3,\frac{b e^{e+f x}}{\sqrt{a^2+b^2}-a}\right )\right )}{f^2}-\frac{2 d \left (f (c+d x) \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )-d \text{PolyLog}\left (3,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )\right )}{f^2}+(c+d x)^2 \log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}+1\right )-(c+d x)^2 \log \left (\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}+1\right )}{f \sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.12, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{2}}{a+b\sinh \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.5205, size = 1733, normalized size = 5.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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