3.480 \(\int \frac{\tanh ^3(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\)
Optimal. Leaf size=89 \[ \frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{2 f (a-b)}-\frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{2 f (a-b)^{3/2}} \]
[Out]
-((2*a - b)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/(2*(a - b)^(3/2)*f) + (Sech[e + f*x]^2*Sqrt[a +
b*Sinh[e + f*x]^2])/(2*(a - b)*f)
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Rubi [A] time = 0.112257, antiderivative size = 89, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used =
{3194, 78, 63, 208} \[ \frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{2 f (a-b)}-\frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{2 f (a-b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[e + f*x]^3/Sqrt[a + b*Sinh[e + f*x]^2],x]
[Out]
-((2*a - b)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/(2*(a - b)^(3/2)*f) + (Sech[e + f*x]^2*Sqrt[a +
b*Sinh[e + f*x]^2])/(2*(a - b)*f)
Rule 3194
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rule 78
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ
[p, n]))))
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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{\tanh ^3(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{(1+x)^2 \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=\frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{2 (a-b) f}+\frac{(2 a-b) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{4 (a-b) f}\\ &=\frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{2 (a-b) f}+\frac{(2 a-b) \operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^2(e+f x)}\right )}{2 (a-b) b f}\\ &=-\frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{2 (a-b)^{3/2} f}+\frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{2 (a-b) f}\\ \end{align*}
Mathematica [A] time = 0.116503, size = 85, normalized size = 0.96 \[ -\frac{\frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}}-\frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{a-b}}{2 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Tanh[e + f*x]^3/Sqrt[a + b*Sinh[e + f*x]^2],x]
[Out]
-(((2*a - b)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/(a - b)^(3/2) - (Sech[e + f*x]^2*Sqrt[a + b*Sin
h[e + f*x]^2])/(a - b))/(2*f)
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Maple [C] time = 0.129, size = 43, normalized size = 0.5 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{ \left ( \sinh \left ( fx+e \right ) \right ) ^{3}}{ \left ( \cosh \left ( fx+e \right ) \right ) ^{4}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}},\sinh \left ( fx+e \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x)
[Out]
`int/indef0`(sinh(f*x+e)^3/cosh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (f x + e\right )^{3}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(tanh(f*x + e)^3/sqrt(b*sinh(f*x + e)^2 + a), x)
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Fricas [B] time = 2.74653, size = 3425, normalized size = 38.48 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/4*(((2*a - b)*cosh(f*x + e)^4 + 4*(2*a - b)*cosh(f*x + e)*sinh(f*x + e)^3 + (2*a - b)*sinh(f*x + e)^4 + 2*(
2*a - b)*cosh(f*x + e)^2 + 2*(3*(2*a - b)*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^2 + 4*((2*a - b)*cosh(f*x +
e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f*x + e) + 2*a - b)*sqrt(a - b)*log((b*cosh(f*x + e)^4 + 4*b*cosh(f*x +
e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(4*a - 3*b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - 3*b)*s
inh(f*x + e)^2 - 4*sqrt(2)*sqrt(a - b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2
- 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))*(cosh(f*x + e) + sinh(f*x + e)) + 4*(b*cosh(f*x + e)^3 +
(4*a - 3*b)*cosh(f*x + e))*sinh(f*x + e) + b)/(cosh(f*x + e)^4 + 4*cosh(f*x + e)*sinh(f*x + e)^3 + sinh(f*x +
e)^4 + 2*(3*cosh(f*x + e)^2 + 1)*sinh(f*x + e)^2 + 2*cosh(f*x + e)^2 + 4*(cosh(f*x + e)^3 + cosh(f*x + e))*sin
h(f*x + e) + 1)) + 4*sqrt(2)*((a - b)*cosh(f*x + e) + (a - b)*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(
f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/((a^2 - 2*a*b + b^
2)*f*cosh(f*x + e)^4 + 4*(a^2 - 2*a*b + b^2)*f*cosh(f*x + e)*sinh(f*x + e)^3 + (a^2 - 2*a*b + b^2)*f*sinh(f*x
+ e)^4 + 2*(a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^2 + 2*(3*(a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^2 + (a^2 - 2*a*b +
b^2)*f)*sinh(f*x + e)^2 + (a^2 - 2*a*b + b^2)*f + 4*((a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^3 + (a^2 - 2*a*b + b
^2)*f*cosh(f*x + e))*sinh(f*x + e)), -1/2*(((2*a - b)*cosh(f*x + e)^4 + 4*(2*a - b)*cosh(f*x + e)*sinh(f*x + e
)^3 + (2*a - b)*sinh(f*x + e)^4 + 2*(2*a - b)*cosh(f*x + e)^2 + 2*(3*(2*a - b)*cosh(f*x + e)^2 + 2*a - b)*sinh
(f*x + e)^2 + 4*((2*a - b)*cosh(f*x + e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f*x + e) + 2*a - b)*sqrt(-a + b)*ar
ctan(-1/2*sqrt(2)*sqrt(-a + b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cos
h(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/((a - b)*cosh(f*x + e) + (a - b)*sinh(f*x + e))) - 2*sqrt(2)*((a
- b)*cosh(f*x + e) + (a - b)*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x +
e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/((a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^4 + 4*(a^2 - 2
*a*b + b^2)*f*cosh(f*x + e)*sinh(f*x + e)^3 + (a^2 - 2*a*b + b^2)*f*sinh(f*x + e)^4 + 2*(a^2 - 2*a*b + b^2)*f*
cosh(f*x + e)^2 + 2*(3*(a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^2 + (a^2 - 2*a*b + b^2)*f)*sinh(f*x + e)^2 + (a^2 -
2*a*b + b^2)*f + 4*((a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^3 + (a^2 - 2*a*b + b^2)*f*cosh(f*x + e))*sinh(f*x + e
))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{3}{\left (e + f x \right )}}{\sqrt{a + b \sinh ^{2}{\left (e + f x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)**3/(a+b*sinh(f*x+e)**2)**(1/2),x)
[Out]
Integral(tanh(e + f*x)**3/sqrt(a + b*sinh(e + f*x)**2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (f x + e\right )^{3}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
integrate(tanh(f*x + e)^3/sqrt(b*sinh(f*x + e)^2 + a), x)