3.6.16 \(\int \coth ^3(c+d x) (a+b \sinh ^2(c+d x))^p \, dx\) [516]

Optimal. Leaf size=94 \[ -\frac {\text {csch}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 a d}-\frac {(a+b p) \, _2F_1\left (1,1+p;2+p;1+\frac {b \sinh ^2(c+d x)}{a}\right ) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 a^2 d (1+p)} \]

[Out]

-1/2*csch(d*x+c)^2*(a+b*sinh(d*x+c)^2)^(1+p)/a/d-1/2*(b*p+a)*hypergeom([1, 1+p],[2+p],1+b*sinh(d*x+c)^2/a)*(a+
b*sinh(d*x+c)^2)^(1+p)/a^2/d/(1+p)

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Rubi [A]
time = 0.06, antiderivative size = 94, 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 = {3273, 79, 67} \begin {gather*} -\frac {(a+b p) \left (a+b \sinh ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sinh ^2(c+d x)}{a}+1\right )}{2 a^2 d (p+1)}-\frac {\text {csch}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^{p+1}}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^3*(a + b*Sinh[c + d*x]^2)^p,x]

[Out]

-1/2*(Csch[c + d*x]^2*(a + b*Sinh[c + d*x]^2)^(1 + p))/(a*d) - ((a + b*p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1
+ (b*Sinh[c + d*x]^2)/a]*(a + b*Sinh[c + d*x]^2)^(1 + p))/(2*a^2*d*(1 + p))

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 79

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] || L
tQ[p, n]))))

Rule 3273

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 \coth ^3(c+d x) \left (a+b \sinh ^2(c+d x)\right )^p \, dx &=\frac {\text {Subst}\left (\int \frac {(1+x) (a+b x)^p}{x^2} \, dx,x,\sinh ^2(c+d x)\right )}{2 d}\\ &=-\frac {\text {csch}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 a d}+\frac {(a+b p) \text {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,\sinh ^2(c+d x)\right )}{2 a d}\\ &=-\frac {\text {csch}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 a d}-\frac {(a+b p) \, _2F_1\left (1,1+p;2+p;1+\frac {b \sinh ^2(c+d x)}{a}\right ) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 a^2 d (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 71, normalized size = 0.76 \begin {gather*} -\frac {\left (a \text {csch}^2(c+d x)+\frac {(a+b p) \, _2F_1\left (1,1+p;2+p;1+\frac {b \sinh ^2(c+d x)}{a}\right )}{1+p}\right ) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^3*(a + b*Sinh[c + d*x]^2)^p,x]

[Out]

-1/2*((a*Csch[c + d*x]^2 + ((a + b*p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sinh[c + d*x]^2)/a])/(1 + p))*
(a + b*Sinh[c + d*x]^2)^(1 + p))/(a^2*d)

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Maple [F]
time = 1.54, size = 0, normalized size = 0.00 \[\int \left (\coth ^{3}\left (d x +c \right )\right ) \left (a +b \left (\sinh ^{2}\left (d x +c \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x)

[Out]

int(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x)

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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(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sinh(d*x + c)^2 + a)^p*coth(d*x + c)^3, x)

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Fricas [F]
time = 0.50, size = 25, normalized size = 0.27 \begin {gather*} {\rm integral}\left ({\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{p} \coth \left (d x + c\right )^{3}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x, algorithm="fricas")

[Out]

integral((b*sinh(d*x + c)^2 + a)^p*coth(d*x + c)^3, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**3*(a+b*sinh(d*x+c)**2)**p,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3*(a+b*sinh(d*x+c)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sinh(d*x + c)^2 + a)^p*coth(d*x + c)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {coth}\left (c+d\,x\right )}^3\,{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^3*(a + b*sinh(c + d*x)^2)^p,x)

[Out]

int(coth(c + d*x)^3*(a + b*sinh(c + d*x)^2)^p, x)

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