3.240 \(\int \frac{\text{csch}^4(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\)
Optimal. Leaf size=148 \[ \frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{7/4} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{7/4} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\coth ^3(c+d x)}{3 a d}+\frac{\coth (c+d x)}{a d} \]
[Out]
(b*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*a^(7/4)*Sqrt[Sqrt[a] - Sqrt[b]]*d) + (b*ArcTan
h[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*a^(7/4)*Sqrt[Sqrt[a] + Sqrt[b]]*d) + Coth[c + d*x]/(a*d
) - Coth[c + d*x]^3/(3*a*d)
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Rubi [A] time = 0.211268, antiderivative size = 148, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used =
{3217, 1287, 1166, 208} \[ \frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{7/4} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{7/4} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\coth ^3(c+d x)}{3 a d}+\frac{\coth (c+d x)}{a d} \]
Antiderivative was successfully verified.
[In]
Int[Csch[c + d*x]^4/(a - b*Sinh[c + d*x]^4),x]
[Out]
(b*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*a^(7/4)*Sqrt[Sqrt[a] - Sqrt[b]]*d) + (b*ArcTan
h[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*a^(7/4)*Sqrt[Sqrt[a] + Sqrt[b]]*d) + Coth[c + d*x]/(a*d
) - Coth[c + d*x]^3/(3*a*d)
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 1166
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]
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{\text{csch}^4(c+d x)}{a-b \sinh ^4(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^4 \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}{a x^4}-\frac{1}{a x^2}+\frac{b-b x^2}{a \left (a-2 a x^2+(a-b) x^4\right )}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\coth (c+d x)}{a d}-\frac{\coth ^3(c+d x)}{3 a d}+\frac{\operatorname{Subst}\left (\int \frac{b-b x^2}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{a d}\\ &=\frac{\coth (c+d x)}{a d}-\frac{\coth ^3(c+d x)}{3 a d}-\frac{\left (\left (\sqrt{a}+\sqrt{b}\right ) b\right ) \operatorname{Subst}\left (\int \frac{1}{-a-\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{2 a^{3/2} d}-\frac{\left (\left (1-\frac{\sqrt{b}}{\sqrt{a}}\right ) b\right ) \operatorname{Subst}\left (\int \frac{1}{-a+\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{2 a d}\\ &=\frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{7/4} \sqrt{\sqrt{a}-\sqrt{b}} d}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{7/4} \sqrt{\sqrt{a}+\sqrt{b}} d}+\frac{\coth (c+d x)}{a d}-\frac{\coth ^3(c+d x)}{3 a d}\\ \end{align*}
Mathematica [A] time = 2.19296, size = 165, normalized size = 1.11 \[ \frac{\frac{3 b \tanh ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}+a}}-\frac{3 b \tan ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}-a}}+4 \sqrt{a} \coth (c+d x)-2 \sqrt{a} \coth (c+d x) \text{csch}^2(c+d x)}{6 a^{3/2} d} \]
Antiderivative was successfully verified.
[In]
Integrate[Csch[c + d*x]^4/(a - b*Sinh[c + d*x]^4),x]
[Out]
((-3*b*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/Sqrt[-a + Sqrt[a]*Sqrt[b]] + (3
*b*ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/Sqrt[a + Sqrt[a]*Sqrt[b]] + 4*Sqrt[
a]*Coth[c + d*x] - 2*Sqrt[a]*Coth[c + d*x]*Csch[c + d*x]^2)/(6*a^(3/2)*d)
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Maple [C] time = 0.075, size = 179, normalized size = 1.2 \begin{align*} -{\frac{1}{24\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{3}{8\,da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{b}{4\,da}\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}}^{6}-3\,{{\it \_R}}^{4}+3\,{{\it \_R}}^{2}-1}{{{\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}{24\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{3}{8\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csch(d*x+c)^4/(a-b*sinh(d*x+c)^4),x)
[Out]
-1/24/d/a*tanh(1/2*d*x+1/2*c)^3+3/8/d/a*tanh(1/2*d*x+1/2*c)-1/4/d/a*b*sum((_R^6-3*_R^4+3*_R^2-1)/(_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))-
1/24/d/a/tanh(1/2*d*x+1/2*c)^3+3/8/d/a/tanh(1/2*d*x+1/2*c)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -16 \, b \int \frac{e^{\left (4 \, d x + 4 \, c\right )}}{a b e^{\left (8 \, d x + 8 \, c\right )} - 4 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 4 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a b - 2 \,{\left (8 \, a^{2} e^{\left (4 \, c\right )} - 3 \, a b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )}}\,{d x} - \frac{4 \,{\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}{3 \,{\left (a d e^{\left (6 \, d x + 6 \, c\right )} - 3 \, a d e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a d e^{\left (2 \, d x + 2 \, c\right )} - a d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^4/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")
[Out]
-16*b*integrate(e^(4*d*x + 4*c)/(a*b*e^(8*d*x + 8*c) - 4*a*b*e^(6*d*x + 6*c) - 4*a*b*e^(2*d*x + 2*c) + a*b - 2
*(8*a^2*e^(4*c) - 3*a*b*e^(4*c))*e^(4*d*x)), x) - 4/3*(3*e^(2*d*x + 2*c) - 1)/(a*d*e^(6*d*x + 6*c) - 3*a*d*e^(
4*d*x + 4*c) + 3*a*d*e^(2*d*x + 2*c) - a*d)
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Fricas [B] time = 2.21904, size = 5195, normalized size = 35.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^4/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")
[Out]
1/12*(3*(a*d*cosh(d*x + c)^6 + 6*a*d*cosh(d*x + c)*sinh(d*x + c)^5 + a*d*sinh(d*x + c)^6 - 3*a*d*cosh(d*x + c)
^4 + 3*(5*a*d*cosh(d*x + c)^2 - a*d)*sinh(d*x + c)^4 + 3*a*d*cosh(d*x + c)^2 + 4*(5*a*d*cosh(d*x + c)^3 - 3*a*
d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a*d*cosh(d*x + c)^4 - 6*a*d*cosh(d*x + c)^2 + a*d)*sinh(d*x + c)^2 - a
*d + 6*(a*d*cosh(d*x + c)^5 - 2*a*d*cosh(d*x + c)^3 + a*d*cosh(d*x + c))*sinh(d*x + c))*sqrt(((a^4 - a^3*b)*d^
2*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)) + b^2)/((a^4 - a^3*b)*d^2))*log(b^4*cosh(d*x + c)^2 + 2*b^4*cosh(d
*x + c)*sinh(d*x + c) + b^4*sinh(d*x + c)^2 - b^4 + 2*(a^5*b - a^4*b^2)*d^2*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2
)*d^4)) + 2*(a^2*b^3*d - (a^7 - a^6*b)*d^3*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)))*sqrt(((a^4 - a^3*b)*d^2*
sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)) + b^2)/((a^4 - a^3*b)*d^2))) - 3*(a*d*cosh(d*x + c)^6 + 6*a*d*cosh(d
*x + c)*sinh(d*x + c)^5 + a*d*sinh(d*x + c)^6 - 3*a*d*cosh(d*x + c)^4 + 3*(5*a*d*cosh(d*x + c)^2 - a*d)*sinh(d
*x + c)^4 + 3*a*d*cosh(d*x + c)^2 + 4*(5*a*d*cosh(d*x + c)^3 - 3*a*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a*d
*cosh(d*x + c)^4 - 6*a*d*cosh(d*x + c)^2 + a*d)*sinh(d*x + c)^2 - a*d + 6*(a*d*cosh(d*x + c)^5 - 2*a*d*cosh(d*
x + c)^3 + a*d*cosh(d*x + c))*sinh(d*x + c))*sqrt(((a^4 - a^3*b)*d^2*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4))
+ b^2)/((a^4 - a^3*b)*d^2))*log(b^4*cosh(d*x + c)^2 + 2*b^4*cosh(d*x + c)*sinh(d*x + c) + b^4*sinh(d*x + c)^2
- b^4 + 2*(a^5*b - a^4*b^2)*d^2*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)) - 2*(a^2*b^3*d - (a^7 - a^6*b)*d^3*
sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)))*sqrt(((a^4 - a^3*b)*d^2*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)) +
b^2)/((a^4 - a^3*b)*d^2))) + 3*(a*d*cosh(d*x + c)^6 + 6*a*d*cosh(d*x + c)*sinh(d*x + c)^5 + a*d*sinh(d*x + c)
^6 - 3*a*d*cosh(d*x + c)^4 + 3*(5*a*d*cosh(d*x + c)^2 - a*d)*sinh(d*x + c)^4 + 3*a*d*cosh(d*x + c)^2 + 4*(5*a*
d*cosh(d*x + c)^3 - 3*a*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a*d*cosh(d*x + c)^4 - 6*a*d*cosh(d*x + c)^2 +
a*d)*sinh(d*x + c)^2 - a*d + 6*(a*d*cosh(d*x + c)^5 - 2*a*d*cosh(d*x + c)^3 + a*d*cosh(d*x + c))*sinh(d*x + c)
)*sqrt(-((a^4 - a^3*b)*d^2*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)) - b^2)/((a^4 - a^3*b)*d^2))*log(b^4*cosh(
d*x + c)^2 + 2*b^4*cosh(d*x + c)*sinh(d*x + c) + b^4*sinh(d*x + c)^2 - b^4 - 2*(a^5*b - a^4*b^2)*d^2*sqrt(b^5/
((a^9 - 2*a^8*b + a^7*b^2)*d^4)) + 2*(a^2*b^3*d + (a^7 - a^6*b)*d^3*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)))
*sqrt(-((a^4 - a^3*b)*d^2*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)) - b^2)/((a^4 - a^3*b)*d^2))) - 3*(a*d*cosh
(d*x + c)^6 + 6*a*d*cosh(d*x + c)*sinh(d*x + c)^5 + a*d*sinh(d*x + c)^6 - 3*a*d*cosh(d*x + c)^4 + 3*(5*a*d*cos
h(d*x + c)^2 - a*d)*sinh(d*x + c)^4 + 3*a*d*cosh(d*x + c)^2 + 4*(5*a*d*cosh(d*x + c)^3 - 3*a*d*cosh(d*x + c))*
sinh(d*x + c)^3 + 3*(5*a*d*cosh(d*x + c)^4 - 6*a*d*cosh(d*x + c)^2 + a*d)*sinh(d*x + c)^2 - a*d + 6*(a*d*cosh(
d*x + c)^5 - 2*a*d*cosh(d*x + c)^3 + a*d*cosh(d*x + c))*sinh(d*x + c))*sqrt(-((a^4 - a^3*b)*d^2*sqrt(b^5/((a^9
- 2*a^8*b + a^7*b^2)*d^4)) - b^2)/((a^4 - a^3*b)*d^2))*log(b^4*cosh(d*x + c)^2 + 2*b^4*cosh(d*x + c)*sinh(d*x
+ c) + b^4*sinh(d*x + c)^2 - b^4 - 2*(a^5*b - a^4*b^2)*d^2*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)) - 2*(a^2
*b^3*d + (a^7 - a^6*b)*d^3*sqrt(b^5/((a^9 - 2*a^8*b + a^7*b^2)*d^4)))*sqrt(-((a^4 - a^3*b)*d^2*sqrt(b^5/((a^9
- 2*a^8*b + a^7*b^2)*d^4)) - b^2)/((a^4 - a^3*b)*d^2))) - 48*cosh(d*x + c)^2 - 96*cosh(d*x + c)*sinh(d*x + c)
- 48*sinh(d*x + c)^2 + 16)/(a*d*cosh(d*x + c)^6 + 6*a*d*cosh(d*x + c)*sinh(d*x + c)^5 + a*d*sinh(d*x + c)^6 -
3*a*d*cosh(d*x + c)^4 + 3*(5*a*d*cosh(d*x + c)^2 - a*d)*sinh(d*x + c)^4 + 3*a*d*cosh(d*x + c)^2 + 4*(5*a*d*cos
h(d*x + c)^3 - 3*a*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a*d*cosh(d*x + c)^4 - 6*a*d*cosh(d*x + c)^2 + a*d)*
sinh(d*x + c)^2 - a*d + 6*(a*d*cosh(d*x + c)^5 - 2*a*d*cosh(d*x + c)^3 + a*d*cosh(d*x + c))*sinh(d*x + c))
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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(csch(d*x+c)**4/(a-b*sinh(d*x+c)**4),x)
[Out]
Timed out
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Giac [A] time = 2.85349, size = 46, normalized size = 0.31 \begin{align*} -\frac{4 \,{\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}{3 \, a d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^4/(a-b*sinh(d*x+c)^4),x, algorithm="giac")
[Out]
-4/3*(3*e^(2*d*x + 2*c) - 1)/(a*d*(e^(2*d*x + 2*c) - 1)^3)