3.6.13 \(\int (a+b \sinh ^2(c+d x))^p \tanh ^3(c+d x) \, dx\) [513]

Optimal. Leaf size=110 \[ -\frac {(a-b (1+p)) \, _2F_1\left (1,1+p;2+p;\frac {a+b \sinh ^2(c+d x)}{a-b}\right ) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 (a-b)^2 d (1+p)}+\frac {\text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 (a-b) d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 110, 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, 70} \begin {gather*} \frac {\text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^{p+1}}{2 d (a-b)}-\frac {(a-b (p+1)) \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}{a-b}\right )}{2 d (p+1) (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 70

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

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

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Mathematica [A]
time = 0.20, size = 90, normalized size = 0.82 \begin {gather*} \frac {\left ((-a+b+b p) \, _2F_1\left (1,1+p;2+p;\frac {a+b \sinh ^2(c+d x)}{a-b}\right )+(a-b) (1+p) \text {sech}^2(c+d x)\right ) \left (a+b \sinh ^2(c+d x)\right )^{1+p}}{2 (a-b)^2 d (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 2.10, size = 0, normalized size = 0.00 \[\int \left (a +b \left (\sinh ^{2}\left (d x +c \right )\right )\right )^{p} \left (\tanh ^{3}\left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F]
time = 0.44, size = 25, normalized size = 0.23 \begin {gather*} {\rm integral}\left ({\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{p} \tanh \left (d x + c\right )^{3}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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