3.518 \(\int (a+b \sinh ^2(c+d x))^p \tanh ^2(c+d x) \, dx\)

Optimal. Leaf size=103 \[ \frac {\sinh ^2(c+d x) \sqrt {\cosh ^2(c+d x)} \tanh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \left (\frac {b \sinh ^2(c+d x)}{a}+1\right )^{-p} F_1\left (\frac {3}{2};\frac {3}{2},-p;\frac {5}{2};-\sinh ^2(c+d x),-\frac {b \sinh ^2(c+d x)}{a}\right )}{3 d} \]

[Out]

1/3*AppellF1(3/2,3/2,-p,5/2,-sinh(d*x+c)^2,-b*sinh(d*x+c)^2/a)*sinh(d*x+c)^2*(a+b*sinh(d*x+c)^2)^p*(cosh(d*x+c
)^2)^(1/2)*tanh(d*x+c)/d/((1+b*sinh(d*x+c)^2/a)^p)

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Rubi [A]  time = 0.11, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3196, 511, 510} \[ \frac {\sinh ^2(c+d x) \sqrt {\cosh ^2(c+d x)} \tanh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \left (\frac {b \sinh ^2(c+d x)}{a}+1\right )^{-p} F_1\left (\frac {3}{2};\frac {3}{2},-p;\frac {5}{2};-\sinh ^2(c+d x),-\frac {b \sinh ^2(c+d x)}{a}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x]^2)^p*Tanh[c + d*x]^2,x]

[Out]

(AppellF1[3/2, 3/2, -p, 5/2, -Sinh[c + d*x]^2, -((b*Sinh[c + d*x]^2)/a)]*Sqrt[Cosh[c + d*x]^2]*Sinh[c + d*x]^2
*(a + b*Sinh[c + d*x]^2)^p*Tanh[c + d*x])/(3*d*(1 + (b*Sinh[c + d*x]^2)/a)^p)

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

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

Rule 3196

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(c+d x)\right )^p \tanh ^2(c+d x) \, dx &=\frac {\left (\sqrt {\cosh ^2(c+d x)} \text {sech}(c+d x)\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^p}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {\left (\sqrt {\cosh ^2(c+d x)} \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \left (1+\frac {b \sinh ^2(c+d x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (1+\frac {b x^2}{a}\right )^p}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {F_1\left (\frac {3}{2};\frac {3}{2},-p;\frac {5}{2};-\sinh ^2(c+d x),-\frac {b \sinh ^2(c+d x)}{a}\right ) \sqrt {\cosh ^2(c+d x)} \sinh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \left (1+\frac {b \sinh ^2(c+d x)}{a}\right )^{-p} \tanh (c+d x)}{3 d}\\ \end {align*}

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Mathematica [F]  time = 5.65, size = 0, normalized size = 0.00 \[ \int \left (a+b \sinh ^2(c+d x)\right )^p \tanh ^2(c+d x) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Sinh[c + d*x]^2)^p*Tanh[c + d*x]^2,x]

[Out]

Integrate[(a + b*Sinh[c + d*x]^2)^p*Tanh[c + d*x]^2, x]

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{p} \tanh \left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(d*x+c)^2)^p*tanh(d*x+c)^2,x, algorithm="fricas")

[Out]

integral((b*sinh(d*x + c)^2 + a)^p*tanh(d*x + c)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{p} \tanh \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(d*x+c)^2)^p*tanh(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((b*sinh(d*x + c)^2 + a)^p*tanh(d*x + c)^2, x)

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maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \left (a +b \left (\sinh ^{2}\left (d x +c \right )\right )\right )^{p} \left (\tanh ^{2}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sinh(d*x+c)^2)^p*tanh(d*x+c)^2,x)

[Out]

int((a+b*sinh(d*x+c)^2)^p*tanh(d*x+c)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{p} \tanh \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(d*x+c)^2)^p*tanh(d*x+c)^2,x, algorithm="maxima")

[Out]

integrate((b*sinh(d*x + c)^2 + a)^p*tanh(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(c + d*x)^2*(a + b*sinh(c + d*x)^2)^p,x)

[Out]

int(tanh(c + d*x)^2*(a + b*sinh(c + d*x)^2)^p, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(d*x+c)**2)**p*tanh(d*x+c)**2,x)

[Out]

Timed out

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