3.5.75 \(\int (a+b \sinh ^2(e+f x))^{3/2} \tanh ^2(e+f x) \, dx\) [475]

Optimal. Leaf size=260 \[ \frac {4 b \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}-\frac {(7 a-8 b) E\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(3 a-4 b) F\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(7 a-8 b) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}-\frac {\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh (e+f x)}{f} \]

[Out]

4/3*b*cosh(f*x+e)*sinh(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/f-1/3*(7*a-8*b)*(1/(1+sinh(f*x+e)^2))^(1/2)*(1+sinh(f*
x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)/(1+sinh(f*x+e)^2)^(1/2),(1-b/a)^(1/2))*sech(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/
2)/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)+1/3*(3*a-4*b)*(1/(1+sinh(f*x+e)^2))^(1/2)*(1+sinh(f*x+e)^2)^(
1/2)*EllipticF(sinh(f*x+e)/(1+sinh(f*x+e)^2)^(1/2),(1-b/a)^(1/2))*sech(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/f/(sec
h(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)-(a+b*sinh(f*x+e)^2)^(3/2)*tanh(f*x+e)/f+1/3*(7*a-8*b)*(a+b*sinh(f*x+e)
^2)^(1/2)*tanh(f*x+e)/f

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Rubi [A]
time = 0.17, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3275, 478, 542, 545, 429, 506, 422} \begin {gather*} \frac {(3 a-4 b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right )}{3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {(7 a-8 b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right )}{3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {\tanh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{f}+\frac {(7 a-8 b) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}+\frac {4 b \sinh (e+f x) \cosh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*Cosh[e + f*x]*Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*f) - ((7*a - 8*b)*EllipticE[ArcTan[Sinh[e + f
*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/
a]) + ((3*a - 4*b)*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*f*S
qrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((7*a - 8*b)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3
*f) - ((a + b*Sinh[e + f*x]^2)^(3/2)*Tanh[e + f*x])/f

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sq
rt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 429

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*
Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*n*(p + 1))), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[c + d*x^2])), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 542

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
f*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(n*(p + q + 1) + 1))), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

Rule 545

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 3275

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff^(m + 1)*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), Subst[Int[x^m*((a + b*ff^2*
x^2)^p/(1 - ff^2*x^2)^((m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^2(e+f x) \, dx &=\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^{3/2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac {\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh (e+f x)}{f}+\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2} \left (a+4 b x^2\right )}{\sqrt {1+x^2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac {4 b \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}-\frac {\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh (e+f x)}{f}+\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {a (3 a-4 b)+(7 a-8 b) b x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 f}\\ &=\frac {4 b \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}-\frac {\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh (e+f x)}{f}+\frac {\left (a (3 a-4 b) \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 f}+\frac {\left ((7 a-8 b) b \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 f}\\ &=\frac {4 b \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}+\frac {(3 a-4 b) F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(7 a-8 b) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}-\frac {\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh (e+f x)}{f}-\frac {\left ((7 a-8 b) \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 f}\\ &=\frac {4 b \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}-\frac {(7 a-8 b) E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(3 a-4 b) F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(7 a-8 b) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}-\frac {\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh (e+f x)}{f}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.26, size = 188, normalized size = 0.72 \begin {gather*} \frac {-8 i a (7 a-8 b) \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )+32 i a (a-b) \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )+\sqrt {2} \left (-24 a^2+40 a b-13 b^2-4 (2 a-3 b) b \cosh (2 (e+f x))+b^2 \cosh (4 (e+f x))\right ) \tanh (e+f x)}{24 f \sqrt {2 a-b+b \cosh (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*a*(7*a - 8*b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] + (32*I)*a*(a - b)*S
qrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] + Sqrt[2]*(-24*a^2 + 40*a*b - 13*b^2 - 4*(2
*a - 3*b)*b*Cosh[2*(e + f*x)] + b^2*Cosh[4*(e + f*x)])*Tanh[e + f*x])/(24*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]
])

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Maple [A]
time = 1.57, size = 413, normalized size = 1.59

method	result	size
default	\(\frac {\sqrt {-\frac {b}{a}}\, b^{2} \left (\sinh ^{5}\left (f x +e \right )\right )-2 \sqrt {-\frac {b}{a}}\, a b \left (\sinh ^{3}\left (f x +e \right )\right )+4 \sqrt {-\frac {b}{a}}\, b^{2} \left (\sinh ^{3}\left (f x +e \right )\right )+3 a^{2} \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-11 a b \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )+8 b^{2} \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )+7 \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a b -8 \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b^{2}-3 \sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}\, a^{2}+4 \sqrt {-\frac {b}{a}}\, a b \sinh \left (f x +e \right )}{3 \sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f}\)	\(413\)

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)^2,x,method=_RETURNVERBOSE)

[Out]

1/3*((-1/a*b)^(1/2)*b^2*sinh(f*x+e)^5-2*(-1/a*b)^(1/2)*a*b*sinh(f*x+e)^3+4*(-1/a*b)^(1/2)*b^2*sinh(f*x+e)^3+3*
a^2*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))-11*a
*b*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))+8*b^2
*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))+7*((a+b
*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a*b-8*((a+b*s
inh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b^2-3*sinh(f*x+
e)*(-1/a*b)^(1/2)*a^2+4*(-1/a*b)^(1/2)*a*b*sinh(f*x+e))/(-1/a*b)^(1/2)/cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.11, size = 25, normalized size = 0.10 \begin {gather*} {\rm integral}\left ({\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tanh \left (f x + e\right )^{2}, x\right ) \end {gather*}

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)^2,x, algorithm="fricas")

[Out]

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

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

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)**2,x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.58Error: Bad Argument Type

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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