3.238 \(\int \frac{1}{a-b \sinh ^4(c+d x)} \, dx\)

Optimal. Leaf size=115 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt{\sqrt{a}+\sqrt{b}}} \]

[Out]

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

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Rubi [A]  time = 0.0976941, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3209, 1166, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt{\sqrt{a}+\sqrt{b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*Sinh[c + d*x]^4)^(-1),x]

[Out]

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

Rule 3209

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

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

Mathematica [A]  time = 0.2393, size = 128, normalized size = 1.11 \[ \frac{\frac{\tanh ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}+a}}-\frac{\tan ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}-a}}}{2 \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*Sinh[c + d*x]^4)^(-1),x]

[Out]

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

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Maple [C]  time = 0.033, size = 102, normalized size = 0.9 \begin{align*}{\frac{1}{4\,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}}^{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 ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-b*sinh(d*x+c)^4),x)

[Out]

1/4/d*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=RootO
f(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{1}{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(1/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")

[Out]

-integrate(1/(b*sinh(d*x + c)^4 - a), x)

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Fricas [B]  time = 2.16271, size = 2130, normalized size = 18.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{1}{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(1/(a-b*sinh(d*x+c)^4),x, algorithm="giac")

[Out]

integrate(-1/(b*sinh(d*x + c)^4 - a), x)