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

Optimal. Leaf size=264 \[ -\frac {2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3}-\frac {2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^3}+\frac {2 f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^2}+\frac {2 f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^2}+\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d}+\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d}-\frac {(e+f x)^3}{3 b f} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5561, 2190, 2531, 2282, 6589} \[ \frac {2 f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^2}+\frac {2 f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b d^2}-\frac {2 f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3}-\frac {2 f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b d^3}+\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d}+\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d}-\frac {(e+f x)^3}{3 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 0]

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

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Mathematica [A]  time = 0.13, size = 244, normalized size = 0.92 \[ \frac {\frac {6 f \left (d (e+f x) \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-f \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )\right )}{d^3}+\frac {6 f \left (d (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-f \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{d^3}+\frac {3 (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d}+\frac {3 (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d}-\frac {(e+f x)^3}{f}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((e + f*x)^3/f) + (3*(e + f*x)^2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/d + (3*(e + f*x)^2*Log[1 +
(b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/d + (6*f*(d*(e + f*x)*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2]
)] - f*PolyLog[3, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])]))/d^3 + (6*f*(d*(e + f*x)*PolyLog[2, -((b*E^(c + d*x
))/(a + Sqrt[a^2 + b^2]))] - f*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))]))/d^3)/(3*b)

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fricas [C]  time = 0.54, size = 609, normalized size = 2.31 \[ -\frac {d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x + 6 \, f^{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 \, f^{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 (d f^{2} x + d e f\right )} {\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 (d f^{2} x + d e f\right )} {\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 (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \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 (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \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 (d^{2} f^{2} x^{2} + 2 \, d^{2} e f x + 2 \, c d e f - c^{2} f^{2}\right )} \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 (d^{2} f^{2} x^{2} + 2 \, d^{2} e f x + 2 \, c d e f - c^{2} f^{2}\right )} \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 \, b d^{3}} \]

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)),x, algorithm="fricas")

[Out]

-1/3*(d^3*f^2*x^3 + 3*d^3*e*f*x^2 + 3*d^3*e^2*x + 6*f^2*polylog(3, (a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cos
h(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2))/b) + 6*f^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*(d*f^2*x + d*e*f)*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*(d*f^2*x + d*e*f)
*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*(d^2*e^2 - 2*c*d*e*f + c^2*f^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*(d^2*e^2 - 2*c*d*e*f + c^2*f^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*(d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) + (b*c
osh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) - 3*(d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2
*f^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*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \cosh \left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x+e)^2*cosh(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 \[ \frac {e^{2} \log \left (b \sinh \left (d x + c\right ) + a\right )}{b d} + \frac {f^{2} x^{3} + 3 \, e f x^{2}}{3 \, b} - \int -\frac {2 \, {\left (b f^{2} x^{2} + 2 \, b e f x - {\left (a f^{2} x^{2} e^{c} + 2 \, a e f x e^{c}\right )} e^{\left (d x\right )}\right )}}{b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} - b^{2}}\,{d x} \]

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)),x, algorithm="maxima")

[Out]

e^2*log(b*sinh(d*x + c) + a)/(b*d) + 1/3*(f^2*x^3 + 3*e*f*x^2)/b - integrate(-2*(b*f^2*x^2 + 2*b*e*f*x - (a*f^
2*x^2*e^c + 2*a*e*f*x*e^c)*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \]

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)),x)

[Out]

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

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

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)),x)

[Out]

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

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