3.3.90 \(\int \frac {(e+f x)^2 \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [290]

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

[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 = 0.31, 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 = {5680, 2221, 2611, 2320, 6724} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(e + f*x)^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*Po
lyLog[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 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 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 5680

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 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 \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 ) \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 d^3}-\frac {\left (2 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 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 = 0.11, size = 244, normalized size = 0.92 \begin {gather*} \frac {-\frac {(e+f x)^3}{f}+\frac {3 (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d}+\frac {3 (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d}+\frac {6 f \left (d (e+f x) \text {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )-f \text {PolyLog}\left (3,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )\right )}{d^3}+\frac {6 f \left (d (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-f \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{d^3}}{3 b} \end {gather*}

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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Maple [F]
time = 0.77, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

1/3*(f^2*x^3 + 3*f*x^2*e)/b + e^2*log(b*sinh(d*x + c) + a)/(b*d) - integrate(-2*(b*f^2*x^2 + 2*b*f*x*e - (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. 738 vs. \(2 (247) = 494\).
time = 0.36, size = 738, normalized size = 2.80 \begin {gather*} -\frac {d^{3} f^{2} x^{3} + 3 \, d^{3} f x^{2} \cosh \left (1\right ) + 3 \, d^{3} x \cosh \left (1\right )^{2} + 3 \, d^{3} x \sinh \left (1\right )^{2} + 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 f \cosh \left (1\right ) + d f \sinh \left (1\right )\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 f \cosh \left (1\right ) + d f \sinh \left (1\right )\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 (c^{2} f^{2} - 2 \, c d f \cosh \left (1\right ) + d^{2} \cosh \left (1\right )^{2} + d^{2} \sinh \left (1\right )^{2} - 2 \, {\left (c d f - d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\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 (c^{2} f^{2} - 2 \, c d f \cosh \left (1\right ) + d^{2} \cosh \left (1\right )^{2} + d^{2} \sinh \left (1\right )^{2} - 2 \, {\left (c d f - d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\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} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} \cosh \left (1\right ) + 2 \, {\left (d^{2} f x + c d f\right )} \sinh \left (1\right )\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} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} \cosh \left (1\right ) + 2 \, {\left (d^{2} f x + c d f\right )} \sinh \left (1\right )\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^{3} f x^{2} + 2 \, d^{3} x \cosh \left (1\right )\right )} \sinh \left (1\right )}{3 \, b d^{3}} \end {gather*}

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*f*x^2*cosh(1) + 3*d^3*x*cosh(1)^2 + 3*d^3*x*sinh(1)^2 + 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*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*f*cosh(1) + d*f*sinh(1))*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqr
t((a^2 + b^2)/b^2) - b)/b + 1) - 6*(d*f^2*x + d*f*cosh(1) + d*f*sinh(1))*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*(c^2*f^2 - 2*c*d*f*cosh(1) + d
^2*cosh(1)^2 + d^2*sinh(1)^2 - 2*(c*d*f - d^2*cosh(1))*sinh(1))*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*(c^2*f^2 - 2*c*d*f*cosh(1) + d^2*cosh(1)^2 + d^2*sinh(1)^2 - 2*(c*d*f - d^2
*cosh(1))*sinh(1))*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 - c^2*f^2 + 2*(d^2*f*x + c*d*f)*cosh(1) + 2*(d^2*f*x + c*d*f)*sinh(1))*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*(d^2*f^2*x^2 - c^2*f^2 + 2*(d^2*f*
x + c*d*f)*cosh(1) + 2*(d^2*f*x + c*d*f)*sinh(1))*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*(d^3*f*x^2 + 2*d^3*x*cosh(1))*sinh(1))/(b*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} \cosh {\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*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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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*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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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