3.222 \(\int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx\)

Optimal. Leaf size=327 \[ \frac {6 \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^4}+\frac {6 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^4}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^4}{4 b} \]

[Out]

-1/4*x^4/b+x^3*ln(1+b*exp(d*x+c)/(a-(a^2-b^2)^(1/2)))/b/d+x^3*ln(1+b*exp(d*x+c)/(a+(a^2-b^2)^(1/2)))/b/d+3*x^2
*polylog(2,-b*exp(d*x+c)/(a-(a^2-b^2)^(1/2)))/b/d^2+3*x^2*polylog(2,-b*exp(d*x+c)/(a+(a^2-b^2)^(1/2)))/b/d^2-6
*x*polylog(3,-b*exp(d*x+c)/(a-(a^2-b^2)^(1/2)))/b/d^3-6*x*polylog(3,-b*exp(d*x+c)/(a+(a^2-b^2)^(1/2)))/b/d^3+6
*polylog(4,-b*exp(d*x+c)/(a-(a^2-b^2)^(1/2)))/b/d^4+6*polylog(4,-b*exp(d*x+c)/(a+(a^2-b^2)^(1/2)))/b/d^4

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Rubi [A]  time = 0.48, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5562, 2190, 2531, 6609, 2282, 6589} \[ \frac {3 x^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}\right )}{b d^2}-\frac {6 x \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}-\frac {6 x \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}\right )}{b d^3}+\frac {6 \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^4}+\frac {6 \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}\right )}{b d^4}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Sinh[c + d*x])/(a + b*Cosh[c + d*x]),x]

[Out]

-x^4/(4*b) + (x^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 - b^2])])/(b*d) + (x^3*Log[1 + (b*E^(c + d*x))/(a + Sq
rt[a^2 - b^2])])/(b*d) + (3*x^2*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 - b^2]))])/(b*d^2) + (3*x^2*PolyLog
[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 - b^2]))])/(b*d^2) - (6*x*PolyLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 - b^2
]))])/(b*d^3) - (6*x*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 - b^2]))])/(b*d^3) + (6*PolyLog[4, -((b*E^(c +
d*x))/(a - Sqrt[a^2 - b^2]))])/(b*d^4) + (6*PolyLog[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 - b^2]))])/(b*d^4)

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 5562

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)])/(Cosh[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), 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]

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx &=-\frac {x^4}{4 b}+\int \frac {e^{c+d x} x^3}{a-\sqrt {a^2-b^2}+b e^{c+d x}} \, dx+\int \frac {e^{c+d x} x^3}{a+\sqrt {a^2-b^2}+b e^{c+d x}} \, dx\\ &=-\frac {x^4}{4 b}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 \int x^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b d}-\frac {3 \int x^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b d}\\ &=-\frac {x^4}{4 b}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {6 \int x \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b d^2}-\frac {6 \int x \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b d^2}\\ &=-\frac {x^4}{4 b}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {6 \int \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b d^3}+\frac {6 \int \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b d^3}\\ &=-\frac {x^4}{4 b}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {b x}{-a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b d^4}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b d^4}\\ &=-\frac {x^4}{4 b}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {6 \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^4}+\frac {6 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 326, normalized size = 1.00 \[ \frac {6 \text {Li}_4\left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}-a}\right )}{b d^4}+\frac {6 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^4}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}-\frac {6 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Sinh[c + d*x])/(a + b*Cosh[c + d*x]),x]

[Out]

-1/4*x^4/b + (x^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 - b^2])])/(b*d) + (x^3*Log[1 + (b*E^(c + d*x))/(a + Sq
rt[a^2 - b^2])])/(b*d) + (3*x^2*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 - b^2]))])/(b*d^2) + (3*x^2*PolyLog
[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 - b^2]))])/(b*d^2) - (6*x*PolyLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 - b^2
]))])/(b*d^3) - (6*x*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 - b^2]))])/(b*d^3) + (6*PolyLog[4, (b*E^(c + d
*x))/(-a + Sqrt[a^2 - b^2])])/(b*d^4) + (6*PolyLog[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 - b^2]))])/(b*d^4)

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fricas [C]  time = 0.57, size = 624, normalized size = 1.91 \[ -\frac {d^{4} x^{4} - 12 \, d^{2} x^{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 ) - 12 \, d^{2} x^{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 ) + 4 \, c^{3} \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 ) + 4 \, c^{3} \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 ) + 24 \, d x {\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 ) + 24 \, d x {\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 ) - 4 \, {\left (d^{3} x^{3} + c^{3}\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 ) - 4 \, {\left (d^{3} x^{3} + c^{3}\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 ) - 24 \, {\rm polylog}\left (4, -\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 ) - 24 \, {\rm polylog}\left (4, -\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 )}{4 \, b d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(d*x+c)/(a+b*cosh(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(d^4*x^4 - 12*d^2*x^2*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) - 12*d^2*x^2*dilog(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*s
inh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) + 4*c^3*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((
a^2 - b^2)/b^2) + 2*a) + 4*c^3*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) +
24*d*x*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) + 24*d*x*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) - 4*(d^3*x^3 + c^3)*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) - 4*(d^3*x^3 + c^3)*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) - 24*polylog(4, -(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) - 24*polylog(4, -(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*d^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sinh \left (d x + c\right )}{b \cosh \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(d*x+c)/(a+b*cosh(d*x+c)),x, algorithm="giac")

[Out]

integrate(x^3*sinh(d*x + c)/(b*cosh(d*x + c) + a), x)

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maple [F]  time = 0.56, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sinh \left (d x +c \right )}{a +b \cosh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinh(d*x+c)/(a+b*cosh(d*x+c)),x)

[Out]

int(x^3*sinh(d*x+c)/(a+b*cosh(d*x+c)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x^{4}}{4 \, b} - \frac {1}{2} \, \int \frac {4 \, {\left (a x^{3} e^{\left (d x + c\right )} + b x^{3}\right )}}{b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} + b^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(d*x+c)/(a+b*cosh(d*x+c)),x, algorithm="maxima")

[Out]

1/4*x^4/b - 1/2*integrate(4*(a*x^3*e^(d*x + c) + b*x^3)/(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 \[ \int \frac {x^3\,\mathrm {sinh}\left (c+d\,x\right )}{a+b\,\mathrm {cosh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*sinh(c + d*x))/(a + b*cosh(c + d*x)),x)

[Out]

int((x^3*sinh(c + d*x))/(a + b*cosh(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sinh {\left (c + d x \right )}}{a + b \cosh {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*sinh(d*x+c)/(a+b*cosh(d*x+c)),x)

[Out]

Integral(x**3*sinh(c + d*x)/(a + b*cosh(c + d*x)), x)

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