3.469 \(\int (a+b \sinh ^2(e+f x))^{3/2} \tanh ^3(e+f x) \, dx\)
Optimal. Leaf size=156 \[ \frac{(2 a-5 b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{6 f (a-b)}+\frac{(2 a-5 b) \sqrt{a+b \sinh ^2(e+f x)}}{2 f}-\frac{(2 a-5 b) \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{2 f}+\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{2 f (a-b)} \]
[Out]
-((2*a - 5*b)*Sqrt[a - b]*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/(2*f) + ((2*a - 5*b)*Sqrt[a + b*Si
nh[e + f*x]^2])/(2*f) + ((2*a - 5*b)*(a + b*Sinh[e + f*x]^2)^(3/2))/(6*(a - b)*f) + (Sech[e + f*x]^2*(a + b*Si
nh[e + f*x]^2)^(5/2))/(2*(a - b)*f)
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Rubi [A] time = 0.158917, antiderivative size = 156, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3194, 78, 50, 63, 208} \[ \frac{(2 a-5 b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{6 f (a-b)}+\frac{(2 a-5 b) \sqrt{a+b \sinh ^2(e+f x)}}{2 f}-\frac{(2 a-5 b) \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{2 f}+\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{2 f (a-b)} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sinh[e + f*x]^2)^(3/2)*Tanh[e + f*x]^3,x]
[Out]
-((2*a - 5*b)*Sqrt[a - b]*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/(2*f) + ((2*a - 5*b)*Sqrt[a + b*Si
nh[e + f*x]^2])/(2*f) + ((2*a - 5*b)*(a + b*Sinh[e + f*x]^2)^(3/2))/(6*(a - b)*f) + (Sech[e + f*x]^2*(a + b*Si
nh[e + f*x]^2)^(5/2))/(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 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
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 \left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^3(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x (a+b x)^{3/2}}{(1+x)^2} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{2 (a-b) f}+\frac{(2 a-5 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{1+x} \, dx,x,\sinh ^2(e+f x)\right )}{4 (a-b) f}\\ &=\frac{(2 a-5 b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{6 (a-b) f}+\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{2 (a-b) f}+\frac{(2 a-5 b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1+x} \, dx,x,\sinh ^2(e+f x)\right )}{4 f}\\ &=\frac{(2 a-5 b) \sqrt{a+b \sinh ^2(e+f x)}}{2 f}+\frac{(2 a-5 b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{6 (a-b) f}+\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{2 (a-b) f}+\frac{((2 a-5 b) (a-b)) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{4 f}\\ &=\frac{(2 a-5 b) \sqrt{a+b \sinh ^2(e+f x)}}{2 f}+\frac{(2 a-5 b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{6 (a-b) f}+\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{2 (a-b) f}+\frac{((2 a-5 b) (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 b f}\\ &=-\frac{(2 a-5 b) \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{2 f}+\frac{(2 a-5 b) \sqrt{a+b \sinh ^2(e+f x)}}{2 f}+\frac{(2 a-5 b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{6 (a-b) f}+\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{2 (a-b) f}\\ \end{align*}
Mathematica [A] time = 0.54235, size = 122, normalized size = 0.78 \[ \frac{(2 a-5 b) \left (\sqrt{a+b \sinh ^2(e+f x)} \left (4 a+b \sinh ^2(e+f x)-3 b\right )-3 (a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )\right )+3 \text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{6 f (a-b)} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Sinh[e + f*x]^2)^(3/2)*Tanh[e + f*x]^3,x]
[Out]
(3*Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2)^(5/2) + (2*a - 5*b)*(-3*(a - b)^(3/2)*ArcTanh[Sqrt[a + b*Sinh[e + f
*x]^2]/Sqrt[a - b]] + Sqrt[a + b*Sinh[e + f*x]^2]*(4*a - 3*b + b*Sinh[e + f*x]^2)))/(6*(a - b)*f)
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Maple [C] time = 0.125, size = 71, normalized size = 0.5 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{ \left ( \sinh \left ( fx+e \right ) \right ) ^{3} \left ({b}^{2} \left ( \sinh \left ( fx+e \right ) \right ) ^{4}+2\,ab \left ( \sinh \left ( fx+e \right ) \right ) ^{2}+{a}^{2} \right ) }{ \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((a+b*sinh(f*x+e)^2)^(3/2)*tanh(f*x+e)^3,x)
[Out]
`int/indef0`(sinh(f*x+e)^3*(b^2*sinh(f*x+e)^4+2*a*b*sinh(f*x+e)^2+a^2)/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{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tanh \left (f x + e\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sinh(f*x+e)^2)^(3/2)*tanh(f*x+e)^3,x, algorithm="maxima")
[Out]
integrate((b*sinh(f*x + e)^2 + a)^(3/2)*tanh(f*x + e)^3, x)
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Fricas [B] time = 7.38749, size = 6527, normalized size = 41.84 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sinh(f*x+e)^2)^(3/2)*tanh(f*x+e)^3,x, algorithm="fricas")
[Out]
[-1/24*(6*((2*a - 5*b)*cosh(f*x + e)^7 + 7*(2*a - 5*b)*cosh(f*x + e)*sinh(f*x + e)^6 + (2*a - 5*b)*sinh(f*x +
e)^7 + 2*(2*a - 5*b)*cosh(f*x + e)^5 + (21*(2*a - 5*b)*cosh(f*x + e)^2 + 4*a - 10*b)*sinh(f*x + e)^5 + 5*(7*(2
*a - 5*b)*cosh(f*x + e)^3 + 2*(2*a - 5*b)*cosh(f*x + e))*sinh(f*x + e)^4 + (2*a - 5*b)*cosh(f*x + e)^3 + (35*(
2*a - 5*b)*cosh(f*x + e)^4 + 20*(2*a - 5*b)*cosh(f*x + e)^2 + 2*a - 5*b)*sinh(f*x + e)^3 + (21*(2*a - 5*b)*cos
h(f*x + e)^5 + 20*(2*a - 5*b)*cosh(f*x + e)^3 + 3*(2*a - 5*b)*cosh(f*x + e))*sinh(f*x + e)^2 + (7*(2*a - 5*b)*
cosh(f*x + e)^6 + 10*(2*a - 5*b)*cosh(f*x + e)^4 + 3*(2*a - 5*b)*cosh(f*x + e)^2)*sinh(f*x + e))*sqrt(a - b)*l
og((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)*sinh(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) + s
inh(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))*sinh(f*x + e) + 1)) - sqrt(2)*(b*cosh(f*x + e)^8 + 8*b*cosh(f*x + e)*sinh(
f*x + e)^7 + b*sinh(f*x + e)^8 + 8*(2*a - 3*b)*cosh(f*x + e)^6 + 4*(7*b*cosh(f*x + e)^2 + 4*a - 6*b)*sinh(f*x
+ e)^6 + 8*(7*b*cosh(f*x + e)^3 + 6*(2*a - 3*b)*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(28*a - 37*b)*cosh(f*x + e)
^4 + 2*(35*b*cosh(f*x + e)^4 + 60*(2*a - 3*b)*cosh(f*x + e)^2 + 28*a - 37*b)*sinh(f*x + e)^4 + 8*(7*b*cosh(f*x
+ e)^5 + 20*(2*a - 3*b)*cosh(f*x + e)^3 + (28*a - 37*b)*cosh(f*x + e))*sinh(f*x + e)^3 + 8*(2*a - 3*b)*cosh(f
*x + e)^2 + 4*(7*b*cosh(f*x + e)^6 + 30*(2*a - 3*b)*cosh(f*x + e)^4 + 3*(28*a - 37*b)*cosh(f*x + e)^2 + 4*a -
6*b)*sinh(f*x + e)^2 + 8*(b*cosh(f*x + e)^7 + 6*(2*a - 3*b)*cosh(f*x + e)^5 + (28*a - 37*b)*cosh(f*x + e)^3 +
2*(2*a - 3*b)*cosh(f*x + e))*sinh(f*x + e) + 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)))/(f*cosh(f*x + e)^7 + 7*f*cosh(f*x + e)*sinh(f*x
+ e)^6 + f*sinh(f*x + e)^7 + 2*f*cosh(f*x + e)^5 + (21*f*cosh(f*x + e)^2 + 2*f)*sinh(f*x + e)^5 + 5*(7*f*cosh
(f*x + e)^3 + 2*f*cosh(f*x + e))*sinh(f*x + e)^4 + f*cosh(f*x + e)^3 + (35*f*cosh(f*x + e)^4 + 20*f*cosh(f*x +
e)^2 + f)*sinh(f*x + e)^3 + (21*f*cosh(f*x + e)^5 + 20*f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e))*sinh(f*x + e)^2
+ (7*f*cosh(f*x + e)^6 + 10*f*cosh(f*x + e)^4 + 3*f*cosh(f*x + e)^2)*sinh(f*x + e)), -1/24*(12*((2*a - 5*b)*c
osh(f*x + e)^7 + 7*(2*a - 5*b)*cosh(f*x + e)*sinh(f*x + e)^6 + (2*a - 5*b)*sinh(f*x + e)^7 + 2*(2*a - 5*b)*cos
h(f*x + e)^5 + (21*(2*a - 5*b)*cosh(f*x + e)^2 + 4*a - 10*b)*sinh(f*x + e)^5 + 5*(7*(2*a - 5*b)*cosh(f*x + e)^
3 + 2*(2*a - 5*b)*cosh(f*x + e))*sinh(f*x + e)^4 + (2*a - 5*b)*cosh(f*x + e)^3 + (35*(2*a - 5*b)*cosh(f*x + e)
^4 + 20*(2*a - 5*b)*cosh(f*x + e)^2 + 2*a - 5*b)*sinh(f*x + e)^3 + (21*(2*a - 5*b)*cosh(f*x + e)^5 + 20*(2*a -
5*b)*cosh(f*x + e)^3 + 3*(2*a - 5*b)*cosh(f*x + e))*sinh(f*x + e)^2 + (7*(2*a - 5*b)*cosh(f*x + e)^6 + 10*(2*
a - 5*b)*cosh(f*x + e)^4 + 3*(2*a - 5*b)*cosh(f*x + e)^2)*sinh(f*x + e))*sqrt(-a + b)*arctan(-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*cosh(f*x + e)*sinh(f*x +
e) + sinh(f*x + e)^2))/((a - b)*cosh(f*x + e) + (a - b)*sinh(f*x + e))) - sqrt(2)*(b*cosh(f*x + e)^8 + 8*b*cos
h(f*x + e)*sinh(f*x + e)^7 + b*sinh(f*x + e)^8 + 8*(2*a - 3*b)*cosh(f*x + e)^6 + 4*(7*b*cosh(f*x + e)^2 + 4*a
- 6*b)*sinh(f*x + e)^6 + 8*(7*b*cosh(f*x + e)^3 + 6*(2*a - 3*b)*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(28*a - 37*
b)*cosh(f*x + e)^4 + 2*(35*b*cosh(f*x + e)^4 + 60*(2*a - 3*b)*cosh(f*x + e)^2 + 28*a - 37*b)*sinh(f*x + e)^4 +
8*(7*b*cosh(f*x + e)^5 + 20*(2*a - 3*b)*cosh(f*x + e)^3 + (28*a - 37*b)*cosh(f*x + e))*sinh(f*x + e)^3 + 8*(2
*a - 3*b)*cosh(f*x + e)^2 + 4*(7*b*cosh(f*x + e)^6 + 30*(2*a - 3*b)*cosh(f*x + e)^4 + 3*(28*a - 37*b)*cosh(f*x
+ e)^2 + 4*a - 6*b)*sinh(f*x + e)^2 + 8*(b*cosh(f*x + e)^7 + 6*(2*a - 3*b)*cosh(f*x + e)^5 + (28*a - 37*b)*co
sh(f*x + e)^3 + 2*(2*a - 3*b)*cosh(f*x + e))*sinh(f*x + e) + 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)))/(f*cosh(f*x + e)^7 + 7*f*cosh(f
*x + e)*sinh(f*x + e)^6 + f*sinh(f*x + e)^7 + 2*f*cosh(f*x + e)^5 + (21*f*cosh(f*x + e)^2 + 2*f)*sinh(f*x + e)
^5 + 5*(7*f*cosh(f*x + e)^3 + 2*f*cosh(f*x + e))*sinh(f*x + e)^4 + f*cosh(f*x + e)^3 + (35*f*cosh(f*x + e)^4 +
20*f*cosh(f*x + e)^2 + f)*sinh(f*x + e)^3 + (21*f*cosh(f*x + e)^5 + 20*f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e))
*sinh(f*x + e)^2 + (7*f*cosh(f*x + e)^6 + 10*f*cosh(f*x + e)^4 + 3*f*cosh(f*x + e)^2)*sinh(f*x + e))]
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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((a+b*sinh(f*x+e)**2)**(3/2)*tanh(f*x+e)**3,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tanh \left (f x + e\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sinh(f*x+e)^2)^(3/2)*tanh(f*x+e)^3,x, algorithm="giac")
[Out]
integrate((b*sinh(f*x + e)^2 + a)^(3/2)*tanh(f*x + e)^3, x)