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

Optimal. Leaf size=107 \[ \frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b^3 d \sqrt{a^2+b^2}}+\frac{x \left (2 a^2-b^2\right )}{2 b^3}-\frac{a \cosh (c+d x)}{b^2 d}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 b d} \]

[Out]

((2*a^2 - b^2)*x)/(2*b^3) + (2*a^3*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b^3*Sqrt[a^2 + b^2]*d)
- (a*Cosh[c + d*x])/(b^2*d) + (Cosh[c + d*x]*Sinh[c + d*x])/(2*b*d)

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Rubi [A]  time = 0.220769, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2793, 3023, 2735, 2660, 618, 204} \[ \frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b^3 d \sqrt{a^2+b^2}}+\frac{x \left (2 a^2-b^2\right )}{2 b^3}-\frac{a \cosh (c+d x)}{b^2 d}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 - b^2)*x)/(2*b^3) + (2*a^3*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b^3*Sqrt[a^2 + b^2]*d)
- (a*Cosh[c + d*x])/(b^2*d) + (Cosh[c + d*x]*Sinh[c + d*x])/(2*b*d)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}-\frac{\int \frac{a+b \sinh (c+d x)+2 a \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx}{2 b}\\ &=-\frac{a \cosh (c+d x)}{b^2 d}+\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}-\frac{i \int \frac{-i a b+i \left (2 a^2-b^2\right ) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx}{2 b^2}\\ &=\frac{\left (2 a^2-b^2\right ) x}{2 b^3}-\frac{a \cosh (c+d x)}{b^2 d}+\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}-\frac{a^3 \int \frac{1}{a+b \sinh (c+d x)} \, dx}{b^3}\\ &=\frac{\left (2 a^2-b^2\right ) x}{2 b^3}-\frac{a \cosh (c+d x)}{b^2 d}+\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}+\frac{\left (2 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-2 i b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{b^3 d}\\ &=\frac{\left (2 a^2-b^2\right ) x}{2 b^3}-\frac{a \cosh (c+d x)}{b^2 d}+\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}-\frac{\left (4 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 i b+2 a \tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{b^3 d}\\ &=\frac{\left (2 a^2-b^2\right ) x}{2 b^3}+\frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b^3 \sqrt{a^2+b^2} d}-\frac{a \cosh (c+d x)}{b^2 d}+\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 0.305172, size = 101, normalized size = 0.94 \[ \frac{-2 \left (b^2-2 a^2\right ) (c+d x)-\frac{8 a^3 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-4 a b \cosh (c+d x)+b^2 \sinh (2 (c+d x))}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-2*a^2 + b^2)*(c + d*x) - (8*a^3*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 4
*a*b*Cosh[c + d*x] + b^2*Sinh[2*(c + d*x)])/(4*b^3*d)

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Maple [B]  time = 0.03, size = 262, normalized size = 2.5 \begin{align*} -{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{a}^{2}}{d{b}^{3}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{2\,bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{{a}^{3}}{d{b}^{3}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{{a}^{2}}{d{b}^{3}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{2\,bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)^2+1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)-1/d/b^2/(tanh(1/2*d*x+1/2*c)+1)*a+1/d/b^3*l
n(tanh(1/2*d*x+1/2*c)+1)*a^2-1/2/d/b*ln(tanh(1/2*d*x+1/2*c)+1)-2/d*a^3/b^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*ta
nh(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))+1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)^2+1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)+1/d/
b^2/(tanh(1/2*d*x+1/2*c)-1)*a-1/d/b^3*ln(tanh(1/2*d*x+1/2*c)-1)*a^2+1/2/d/b*ln(tanh(1/2*d*x+1/2*c)-1)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.42845, size = 1434, normalized size = 13.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(2*a^4 + a^2*b^2 - b^4)*d*x*cosh(d*x + c)^2 + (a^2*b^2 + b^4)*cosh(d*x + c)^4 + (a^2*b^2 + b^4)*sinh(d*
x + c)^4 - a^2*b^2 - b^4 - 4*(a^3*b + a*b^3)*cosh(d*x + c)^3 - 4*(a^3*b + a*b^3 - (a^2*b^2 + b^4)*cosh(d*x + c
))*sinh(d*x + c)^3 + 2*(2*(2*a^4 + a^2*b^2 - b^4)*d*x + 3*(a^2*b^2 + b^4)*cosh(d*x + c)^2 - 6*(a^3*b + a*b^3)*
cosh(d*x + c))*sinh(d*x + c)^2 + 8*(a^3*cosh(d*x + c)^2 + 2*a^3*cosh(d*x + c)*sinh(d*x + c) + a^3*sinh(d*x + c
)^2)*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b
^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x
+ c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) - 4*(a^3*b + a*b
^3)*cosh(d*x + c) - 4*(a^3*b + a*b^3 - 2*(2*a^4 + a^2*b^2 - b^4)*d*x*cosh(d*x + c) - (a^2*b^2 + b^4)*cosh(d*x
+ c)^3 + 3*(a^3*b + a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/((a^2*b^3 + b^5)*d*cosh(d*x + c)^2 + 2*(a^2*b^3 + b
^5)*d*cosh(d*x + c)*sinh(d*x + c) + (a^2*b^3 + b^5)*d*sinh(d*x + c)^2)

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

[Out]

Timed out

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Giac [A]  time = 1.20252, size = 217, normalized size = 2.03 \begin{align*} -\frac{a^{3} \log \left (\frac{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} b^{3} d} + \frac{{\left (2 \, a^{2} - b^{2}\right )}{\left (d x + c\right )}}{2 \, b^{3} d} - \frac{{\left (4 \, a b e^{\left (d x + c\right )} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b^{3} d} + \frac{b d e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a d e^{\left (d x + c\right )}}{8 \, b^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*log(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + c) + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(
a^2 + b^2)*b^3*d) + 1/2*(2*a^2 - b^2)*(d*x + c)/(b^3*d) - 1/8*(4*a*b*e^(d*x + c) + b^2)*e^(-2*d*x - 2*c)/(b^3*
d) + 1/8*(b*d*e^(2*d*x + 2*c) - 4*a*d*e^(d*x + c))/(b^2*d^2)