∫ (a + b sinh2(e + fx))3∕2 tanh(e + fx) dx

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((a+b*sinh(f*x+e)^2)^(1/2)/(a-b)^(1/2))/f+1/3*(a+b*sinh(f*x+e)^2)^(3/2)/f+(a-b)*(a+b*sinh(

f*x+e)^2)^(1/2)/f

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Rubi [A] time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A] time = 0.15, 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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fricas [B] time = 1.11, size = 1052, normalized size = 11.69 \[ \text {result too large to display} \]

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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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn

ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular val

ue [0] was discarded and replaced randomly by 0=[91]Warning, need to choose a branch for the root of a polynom

ial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[74]Wa

rning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular v

alue [0] was discarded and replaced randomly by 0=[-62]Warning, need to choose a branch for the root of a poly

nomial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[-4

4]Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regul

ar value [0] was discarded and replaced randomly by 0=[-3]Warning, need to choose a branch for the root of a p

olynomial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=

[-6]Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non reg

ular value [0] was discarded and replaced randomly by 0=[-77]Warning, need to choose a branch for the root of

a polynomial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by

0=[-10]Evaluation time: 0.81index.cc index_m operator + Error: Bad Argument Value

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maple [C] time = 0.20, size = 69, normalized size = 0.77 \[ \frac {\mathit {`\,int/indef0`\,}\left (\frac {\sinh \left (f x +e \right ) \left (b^{2} \left (\sinh ^{4}\left (f x +e \right )\right )+2 a b \left (\sinh ^{2}\left (f x +e \right )\right )+a^{2}\right )}{\cosh \left (f x +e \right )^{2} \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tanh \left (f x + e\right )\,{d x} \]

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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {tanh}\left (e+f\,x\right )\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tanh {\left (e + f x \right )}\, dx \]

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]

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

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