3.32 \(\int \frac{\text{csch}^4(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx\)
Optimal. Leaf size=75 \[ -\frac{a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{d (a+b)^{5/2}}-\frac{\coth ^3(c+d x)}{3 d (a+b)}+\frac{a \coth (c+d x)}{d (a+b)^2} \]
[Out]
-((a*Sqrt[b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/((a + b)^(5/2)*d)) + (a*Coth[c + d*x])/((a + b)^2*d
) - Coth[c + d*x]^3/(3*(a + b)*d)
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Rubi [A] time = 0.0989715, antiderivative size = 75, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used =
{4132, 453, 325, 208} \[ -\frac{a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{d (a+b)^{5/2}}-\frac{\coth ^3(c+d x)}{3 d (a+b)}+\frac{a \coth (c+d x)}{d (a+b)^2} \]
Antiderivative was successfully verified.
[In]
Int[Csch[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]
[Out]
-((a*Sqrt[b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/((a + b)^(5/2)*d)) + (a*Coth[c + d*x])/((a + b)^2*d
) - Coth[c + d*x]^3/(3*(a + b)*d)
Rule 4132
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]
Rule 453
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Rule 325
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]
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 \text{sech}^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{x^4 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{\coth ^3(c+d x)}{3 (a+b) d}-\frac{a \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{(a+b) d}\\ &=\frac{a \coth (c+d x)}{(a+b)^2 d}-\frac{\coth ^3(c+d x)}{3 (a+b) d}-\frac{(a b) \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b)^2 d}\\ &=-\frac{a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{(a+b)^{5/2} d}+\frac{a \coth (c+d x)}{(a+b)^2 d}-\frac{\coth ^3(c+d x)}{3 (a+b) d}\\ \end{align*}
Mathematica [B] time = 2.0267, size = 216, normalized size = 2.88 \[ \frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\frac{1}{4} \sqrt{a+b} \text{csch}(c) \sqrt{b (\cosh (c)-\sinh (c))^4} \text{csch}^3(c+d x) ((b-2 a) \sinh (2 c+3 d x)+6 a \sinh (d x)-3 b \sinh (2 c+d x))+3 a b (\sinh (2 c)-\cosh (2 c)) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )\right )}{6 d (a+b)^{5/2} \sqrt{b (\cosh (c)-\sinh (c))^4} \left (a+b \text{sech}^2(c+d x)\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[Csch[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]
[Out]
((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(3*a*b*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*
Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(-Cosh[2*c] + Sinh[2*c]) + (Sqr
t[a + b]*Csch[c]*Csch[c + d*x]^3*Sqrt[b*(Cosh[c] - Sinh[c])^4]*(6*a*Sinh[d*x] - 3*b*Sinh[2*c + d*x] + (-2*a +
b)*Sinh[2*c + 3*d*x]))/4))/(6*(a + b)^(5/2)*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^4])
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Maple [B] time = 0.076, size = 258, normalized size = 3.4 \begin{align*} -{\frac{a}{24\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{b}{24\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{3\,a}{8\,d \left ( a+b \right ) ^{2}}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{b}{8\,d \left ( a+b \right ) ^{2}}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{a}{2\,d}\sqrt{b}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}}+{\frac{a}{2\,d}\sqrt{b}\ln \left ( -\sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}-\sqrt{a+b} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}}-{\frac{1}{24\,d \left ( a+b \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{3\,a}{8\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{8\,d \left ( a+b \right ) ^{2}} \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*sech(d*x+c)^2),x)
[Out]
-1/24/d/(a+b)^2*a*tanh(1/2*d*x+1/2*c)^3-1/24/d/(a+b)^2*b*tanh(1/2*d*x+1/2*c)^3+3/8/d/(a+b)^2*a*tanh(1/2*d*x+1/
2*c)-1/8/d/(a+b)^2*tanh(1/2*d*x+1/2*c)*b-1/2/d*a*b^(1/2)/(a+b)^(5/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*ta
nh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/2/d*a*b^(1/2)/(a+b)^(5/2)*ln(-(a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tan
h(1/2*d*x+1/2*c)*b^(1/2)-(a+b)^(1/2))-1/24/d/(a+b)/tanh(1/2*d*x+1/2*c)^3+3/8/d/(a+b)^2/tanh(1/2*d*x+1/2*c)*a-1
/8/d/(a+b)^2/tanh(1/2*d*x+1/2*c)*b
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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(csch(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 3.00318, size = 4602, normalized size = 61.36 \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*sech(d*x+c)^2),x, algorithm="fricas")
[Out]
[-1/6*(12*b*cosh(d*x + c)^4 + 48*b*cosh(d*x + c)*sinh(d*x + c)^3 + 12*b*sinh(d*x + c)^4 + 24*a*cosh(d*x + c)^2
+ 24*(3*b*cosh(d*x + c)^2 + a)*sinh(d*x + c)^2 - 3*(a*cosh(d*x + c)^6 + 6*a*cosh(d*x + c)*sinh(d*x + c)^5 + a
*sinh(d*x + c)^6 - 3*a*cosh(d*x + c)^4 + 3*(5*a*cosh(d*x + c)^2 - a)*sinh(d*x + c)^4 + 4*(5*a*cosh(d*x + c)^3
- 3*a*cosh(d*x + c))*sinh(d*x + c)^3 + 3*a*cosh(d*x + c)^2 + 3*(5*a*cosh(d*x + c)^4 - 6*a*cosh(d*x + c)^2 + a)
*sinh(d*x + c)^2 + 6*(a*cosh(d*x + c)^5 - 2*a*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - a)*sqrt(b/(a
+ b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*c
osh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d
*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*((a^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d
*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c)^4
+ 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^
2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + 48*(b*cos
h(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - 8*a + 4*b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^6 + 6*(a^2 + 2
*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b + b^2)*d*sinh(d*x + c)^6 - 3*(a^2 + 2*a*b + b^2)*d*
cosh(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d)*sinh(d*x + c)^4 + 3*(a^2
+ 2*a*b + b^2)*d*cosh(d*x + c)^2 + 4*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 - 3*(a^2 + 2*a*b + b^2)*d*cosh(
d*x + c))*sinh(d*x + c)^3 + 3*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^4 - 6*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)
^2 + (a^2 + 2*a*b + b^2)*d)*sinh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d + 6*((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^5
- 2*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*d*cosh(d*x + c))*sinh(d*x + c)), -1/3*(6*b*co
sh(d*x + c)^4 + 24*b*cosh(d*x + c)*sinh(d*x + c)^3 + 6*b*sinh(d*x + c)^4 + 12*a*cosh(d*x + c)^2 + 12*(3*b*cosh
(d*x + c)^2 + a)*sinh(d*x + c)^2 + 3*(a*cosh(d*x + c)^6 + 6*a*cosh(d*x + c)*sinh(d*x + c)^5 + a*sinh(d*x + c)^
6 - 3*a*cosh(d*x + c)^4 + 3*(5*a*cosh(d*x + c)^2 - a)*sinh(d*x + c)^4 + 4*(5*a*cosh(d*x + c)^3 - 3*a*cosh(d*x
+ c))*sinh(d*x + c)^3 + 3*a*cosh(d*x + c)^2 + 3*(5*a*cosh(d*x + c)^4 - 6*a*cosh(d*x + c)^2 + a)*sinh(d*x + c)^
2 + 6*(a*cosh(d*x + c)^5 - 2*a*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - a)*sqrt(-b/(a + b))*arctan(1
/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b) + 2
4*(b*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - 4*a + 2*b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^6 + 6*
(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b + b^2)*d*sinh(d*x + c)^6 - 3*(a^2 + 2*a*b +
b^2)*d*cosh(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d)*sinh(d*x + c)^4
+ 3*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^2 + 4*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 - 3*(a^2 + 2*a*b + b^2)
*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^4 - 6*(a^2 + 2*a*b + b^2)*d*cosh(
d*x + c)^2 + (a^2 + 2*a*b + b^2)*d)*sinh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d + 6*((a^2 + 2*a*b + b^2)*d*cosh(d*
x + c)^5 - 2*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*d*cosh(d*x + c))*sinh(d*x + c))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{4}{\left (c + d x \right )}}{a + b \operatorname{sech}^{2}{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)**4/(a+b*sech(d*x+c)**2),x)
[Out]
Integral(csch(c + d*x)**4/(a + b*sech(c + d*x)**2), x)
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Giac [A] time = 1.28063, size = 173, normalized size = 2.31 \begin{align*} -\frac{a b \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{{\left (a^{2} d + 2 \, a b d + b^{2} d\right )} \sqrt{-a b - b^{2}}} - \frac{2 \,{\left (3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a + b\right )}}{3 \,{\left (a^{2} d + 2 \, a b d + b^{2} d\right )}{\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*sech(d*x+c)^2),x, algorithm="giac")
[Out]
-a*b*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/((a^2*d + 2*a*b*d + b^2*d)*sqrt(-a*b - b^2)) -
2/3*(3*b*e^(4*d*x + 4*c) + 6*a*e^(2*d*x + 2*c) - 2*a + b)/((a^2*d + 2*a*b*d + b^2*d)*(e^(2*d*x + 2*c) - 1)^3)