3.470 \(\int (a+b \sinh ^2(e+f x))^{3/2} \tanh (e+f x) \, dx\)

Optimal. Leaf size=90 \[ \frac{(a-b) \sqrt{a+b \sinh ^2(e+f x)}}{f}+\frac{\left (a+b \sinh ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]

[Out]

-(((a - b)^(3/2)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/f) + ((a - b)*Sqrt[a + b*Sinh[e + f*x]^2])/
f + (a + b*Sinh[e + f*x]^2)^(3/2)/(3*f)

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Rubi [A]  time = 0.090232, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3194, 50, 63, 208} \[ \frac{(a-b) \sqrt{a+b \sinh ^2(e+f x)}}{f}+\frac{\left (a+b \sinh ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[e + f*x]^2)^(3/2)*Tanh[e + f*x],x]

[Out]

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

Mathematica [A]  time = 0.146198, size = 86, normalized size = 0.96 \[ \frac{\left (4 a+b \cosh ^2(e+f x)-4 b\right ) \sqrt{a+b \cosh ^2(e+f x)-b}-3 (a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \cosh ^2(e+f x)-b}}{\sqrt{a-b}}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[e + f*x]^2)^(3/2)*Tanh[e + f*x],x]

[Out]

(-3*(a - b)^(3/2)*ArcTanh[Sqrt[a - b + b*Cosh[e + f*x]^2]/Sqrt[a - b]] + (4*a - 4*b + b*Cosh[e + f*x]^2)*Sqrt[
a - b + b*Cosh[e + f*x]^2])/(3*f)

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Maple [C]  time = 0.296, size = 69, normalized size = 0.8 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{\sinh \left ( fx+e \right ) \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 ) ^{2}}{\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),x)

[Out]

`int/indef0`(sinh(f*x+e)*(b^2*sinh(f*x+e)^4+2*a*b*sinh(f*x+e)^2+a^2)/cosh(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(1/2),s
inh(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 )\,{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),x, algorithm="maxima")

[Out]

integrate((b*sinh(f*x + e)^2 + a)^(3/2)*tanh(f*x + e), x)

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Fricas [B]  time = 6.38133, size = 2827, normalized size = 31.41 \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),x, algorithm="fricas")

[Out]

[-1/24*(12*((a - b)*cosh(f*x + e)^3 + 3*(a - b)*cosh(f*x + e)^2*sinh(f*x + e) + 3*(a - b)*cosh(f*x + e)*sinh(f
*x + e)^2 + (a - b)*sinh(f*x + e)^3)*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)*sinh(f*x + e)^2 + 4*sq
rt(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)*si
nh(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))*sinh(f*x + e) + 1)) - sq
rt(2)*(b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(8*a - 7*b)*cosh(f*x + e)
^2 + 2*(3*b*cosh(f*x + e)^2 + 8*a - 7*b)*sinh(f*x + e)^2 + 4*(b*cosh(f*x + e)^3 + (8*a - 7*b)*cosh(f*x + e))*s
inh(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)*si
nh(f*x + e) + sinh(f*x + e)^2)))/(f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e)^2*sinh(f*x + e) + 3*f*cosh(f*x + e)*si
nh(f*x + e)^2 + f*sinh(f*x + e)^3), -1/24*(24*((a - b)*cosh(f*x + e)^3 + 3*(a - b)*cosh(f*x + e)^2*sinh(f*x +
e) + 3*(a - b)*cosh(f*x + e)*sinh(f*x + e)^2 + (a - b)*sinh(f*x + e)^3)*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)^4 + 4*b*cosh
(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(8*a - 7*b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 8*a -
 7*b)*sinh(f*x + e)^2 + 4*(b*cosh(f*x + e)^3 + (8*a - 7*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)^3 + 3*f*cosh(f*x + e)^2*sinh(f*x + e) + 3*f*cosh(f*x + e)*sinh(f*x + e)^2 + f*sinh(f*x + e)^3)]

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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),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 )\,{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),x, algorithm="giac")

[Out]

integrate((b*sinh(f*x + e)^2 + a)^(3/2)*tanh(f*x + e), x)