3.237 \(\int \frac{\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\)

Optimal. Leaf size=125 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt{b} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt{b} d \sqrt{\sqrt{a}-\sqrt{b}}} \]

[Out]

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

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Rubi [A]  time = 0.120626, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3217, 1130, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt{b} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt{b} d \sqrt{\sqrt{a}-\sqrt{b}}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^2/(a - b*Sinh[c + d*x]^4),x]

[Out]

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

Mathematica [A]  time = 0.349201, size = 127, normalized size = 1.02 \[ \frac{\frac{\tanh ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}+a}}+\frac{\tan ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}-a}}}{2 \sqrt{b} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^2/(a - b*Sinh[c + d*x]^4),x]

[Out]

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

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Maple [C]  time = 0.031, size = 94, normalized size = 0.8 \begin{align*} -{\frac{1}{d}\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 ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4),x)

[Out]

-1/d*sum((_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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")

[Out]

-integrate(sinh(d*x + c)^2/(b*sinh(d*x + c)^4 - a), x)

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Fricas [B]  time = 2.21265, size = 2130, normalized size = 17.04 \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)^2/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")

[Out]

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

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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)**2/(a-b*sinh(d*x+c)**4),x)

[Out]

Timed out

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Giac [A]  time = 1.616, size = 1, normalized size = 0.01 \begin{align*} 0 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4),x, algorithm="giac")

[Out]

0