3.235 \(\int \frac{\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\)

Optimal. Leaf size=175 \[ -\frac{a^{3/4} \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{a^{3/4} \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{1}{4 b d (1-\tanh (c+d x))}+\frac{1}{4 b d (\tanh (c+d x)+1)}+\frac{x}{2 b} \]

[Out]

x/(2*b) - (a^(3/4)*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(3/2
)*d) + (a^(3/4)*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(3/2)*d
) - 1/(4*b*d*(1 - Tanh[c + d*x])) + 1/(4*b*d*(1 + Tanh[c + d*x]))

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Rubi [A]  time = 0.258026, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3217, 1287, 207, 1130, 208} \[ -\frac{a^{3/4} \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{a^{3/4} \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{1}{4 b d (1-\tanh (c+d x))}+\frac{1}{4 b d (\tanh (c+d x)+1)}+\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^6/(a - b*Sinh[c + d*x]^4),x]

[Out]

x/(2*b) - (a^(3/4)*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(3/2
)*d) + (a^(3/4)*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(3/2)*d
) - 1/(4*b*d*(1 - Tanh[c + d*x])) + 1/(4*b*d*(1 + Tanh[c + d*x]))

Rule 3217

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p)/(1 + ff^2
*x^2)^(m/2 + 2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 1287

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Ex
pandIntegrand[((f*x)^m*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^
2 - 4*a*c, 0] && IntegerQ[q] && IntegerQ[m]

Rule 207

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

Rule 1130

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2*(b/q + 1))/2, Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2*(b/q - 1))/2, Int[(d*x)^(m - 2)/(b
/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sinh ^6(c+d x)}{a-b \sinh ^4(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2 \left (a-2 a x^2+(a-b) x^4\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{4 b (-1+x)^2}-\frac{1}{4 b (1+x)^2}-\frac{1}{2 b \left (-1+x^2\right )}+\frac{a x^2}{b \left (a-2 a x^2+(a-b) x^4\right )}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{1}{4 b d (1-\tanh (c+d x))}+\frac{1}{4 b d (1+\tanh (c+d x))}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{2 b d}+\frac{a \operatorname{Subst}\left (\int \frac{x^2}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{b d}\\ &=\frac{x}{2 b}-\frac{1}{4 b d (1-\tanh (c+d x))}+\frac{1}{4 b d (1+\tanh (c+d x))}+\frac{\left (a \left (\sqrt{a}+\sqrt{b}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{2 b^{3/2} d}+\frac{\left (a \left (1-\frac{\sqrt{a}}{\sqrt{b}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a+\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{2 b d}\\ &=\frac{x}{2 b}-\frac{a^{3/4} \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{\sqrt{a}-\sqrt{b}} b^{3/2} d}+\frac{a^{3/4} \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{\sqrt{a}+\sqrt{b}} b^{3/2} d}-\frac{1}{4 b d (1-\tanh (c+d x))}+\frac{1}{4 b d (1+\tanh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.867199, size = 158, normalized size = 0.9 \[ \frac{\frac{2 a \tanh ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}+a}}+\frac{2 a \tan ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}-a}}+2 \sqrt{b} (c+d x)-\sqrt{b} \sinh (2 (c+d x))}{4 b^{3/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^6/(a - b*Sinh[c + d*x]^4),x]

[Out]

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

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Maple [C]  time = 0.051, size = 223, normalized size = 1.3 \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{1}{2\,bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{a}{bd}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{8}-4\,a{{\it \_Z}}^{6}+ \left ( 6\,a-16\,b \right ){{\it \_Z}}^{4}-4\,a{{\it \_Z}}^{2}+a \right ) }{\frac{{{\it \_R}}^{4}-{{\it \_R}}^{2}}{{{\it \_R}}^{7}a-3\,{{\it \_R}}^{5}a+3\,{{\it \_R}}^{3}a-8\,{{\it \_R}}^{3}b-{\it \_R}\,a}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \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{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)^6/(a-b*sinh(d*x+c)^4),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)+1/2/d/b*ln(tanh(1/2*d*x+1/2*c)+1)-1/d*a/b*su
m((_R^4-_R^2)/(_R^7*a-3*_R^5*a+3*_R^3*a-8*_R^3*b-_R*a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(a*_Z^8-4*a*_Z^6+(6
*a-16*b)*_Z^4-4*a*_Z^2+a))-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)-1/2/d/b*ln(tanh(1
/2*d*x+1/2*c)-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (4 \, d x e^{\left (2 \, d x + 2 \, c\right )} - e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b d} - \frac{1}{64} \, \int \frac{256 \,{\left (a e^{\left (6 \, d x + 6 \, c\right )} - 2 \, a e^{\left (4 \, d x + 4 \, c\right )} + a e^{\left (2 \, d x + 2 \, c\right )}\right )}}{b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 4 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 4 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2} - 2 \,{\left (8 \, a b e^{\left (4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^6/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")

[Out]

1/8*(4*d*x*e^(2*d*x + 2*c) - e^(4*d*x + 4*c) + 1)*e^(-2*d*x - 2*c)/(b*d) - 1/64*integrate(256*(a*e^(6*d*x + 6*
c) - 2*a*e^(4*d*x + 4*c) + a*e^(2*d*x + 2*c))/(b^2*e^(8*d*x + 8*c) - 4*b^2*e^(6*d*x + 6*c) - 4*b^2*e^(2*d*x +
2*c) + b^2 - 2*(8*a*b*e^(4*c) - 3*b^2*e^(4*c))*e^(4*d*x)), x)

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Fricas [B]  time = 2.39824, size = 3123, normalized size = 17.85 \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)^6/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")

[Out]

1/8*(4*d*x*cosh(d*x + c)^2 - cosh(d*x + c)^4 - 4*cosh(d*x + c)*sinh(d*x + c)^3 - sinh(d*x + c)^4 + 2*(2*d*x -
3*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 2*(b*d*cosh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + c) + b*d*sinh(d*x
+ c)^2)*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2)/((a*b^3 - b^4)*d^2))*log(a^2
*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 + 2*(a^2*b^2 - a*b^3)*d^2*sqrt(a^3/
((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2 + 2*((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2*
b*d)*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2)/((a*b^3 - b^4)*d^2))) + 2*(b*d*c
osh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + c) + b*d*sinh(d*x + c)^2)*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^3/((a
^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2)/((a*b^3 - b^4)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d
*x + c) + a^2*sinh(d*x + c)^2 + 2*(a^2*b^2 - a*b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2 - 2*((
a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2*b*d)*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^3/((a^2*b
^5 - 2*a*b^6 + b^7)*d^4)) + a^2)/((a*b^3 - b^4)*d^2))) + 2*(b*d*cosh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x
+ c) + b*d*sinh(d*x + c)^2)*sqrt(-((a*b^3 - b^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2)/((a*b^3
- b^4)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 - 2*(a^2*b^2 -
a*b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2 + 2*((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^
6 + b^7)*d^4)) + a^2*b*d)*sqrt(-((a*b^3 - b^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2)/((a*b^3 -
b^4)*d^2))) - 2*(b*d*cosh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + c) + b*d*sinh(d*x + c)^2)*sqrt(-((a*b^3
- b^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2)/((a*b^3 - b^4)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a
^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 - 2*(a^2*b^2 - a*b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 +
b^7)*d^4)) - a^2 - 2*((a*b^4 - b^5)*d^3*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a^2*b*d)*sqrt(-((a*b^3 - b
^4)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a^2)/((a*b^3 - b^4)*d^2))) + 4*(2*d*x*cosh(d*x + c) - cosh
(d*x + c)^3)*sinh(d*x + c) + 1)/(b*d*cosh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + c) + b*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)**6/(a-b*sinh(d*x+c)**4),x)

[Out]

Timed out

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Giac [A]  time = 1.30782, size = 82, normalized size = 0.47 \begin{align*} -\frac{{\left (2 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b d} + \frac{d x + c}{2 \, b d} - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{8 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^6/(a-b*sinh(d*x+c)^4),x, algorithm="giac")

[Out]

-1/8*(2*e^(2*d*x + 2*c) - 1)*e^(-2*d*x - 2*c)/(b*d) + 1/2*(d*x + c)/(b*d) - 1/8*e^(2*d*x + 2*c)/(b*d)