3.409 \(\int \frac{x^3 (a+b x^2)^p}{d+e x} \, dx\)

Optimal. Leaf size=163 \[ -\frac{e x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{5}{2};-p,1;\frac{7}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{5 d^2}+\frac{d^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^2 (p+1) \left (a e^2+b d^2\right )}-\frac{d \left (a+b x^2\right )^{p+1}}{2 b e^2 (p+1)} \]

[Out]

-(d*(a + b*x^2)^(1 + p))/(2*b*e^2*(1 + p)) - (e*x^5*(a + b*x^2)^p*AppellF1[5/2, -p, 1, 7/2, -((b*x^2)/a), (e^2
*x^2)/d^2])/(5*d^2*(1 + (b*x^2)/a)^p) + (d^3*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a +
b*x^2))/(b*d^2 + a*e^2)])/(2*e^2*(b*d^2 + a*e^2)*(1 + p))

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Rubi [A]  time = 0.147575, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {959, 446, 80, 68, 511, 510} \[ -\frac{e x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{5}{2};-p,1;\frac{7}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{5 d^2}+\frac{d^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^2 (p+1) \left (a e^2+b d^2\right )}-\frac{d \left (a+b x^2\right )^{p+1}}{2 b e^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*x^2)^p)/(d + e*x),x]

[Out]

-(d*(a + b*x^2)^(1 + p))/(2*b*e^2*(1 + p)) - (e*x^5*(a + b*x^2)^p*AppellF1[5/2, -p, 1, 7/2, -((b*x^2)/a), (e^2
*x^2)/d^2])/(5*d^2*(1 + (b*x^2)/a)^p) + (d^3*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a +
b*x^2))/(b*d^2 + a*e^2)])/(2*e^2*(b*d^2 + a*e^2)*(1 + p))

Rule 959

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[(d*(g*x)^n)/x^n, In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[(e*(g*x)^n)/x^n, Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

Rule 446

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

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

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 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])

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b x^2\right )^p}{d+e x} \, dx &=d \int \frac{x^3 \left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx-e \int \frac{x^4 \left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx\\ &=\frac{1}{2} d \operatorname{Subst}\left (\int \frac{x (a+b x)^p}{d^2-e^2 x} \, dx,x,x^2\right )-\left (e \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \frac{x^4 \left (1+\frac{b x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx\\ &=-\frac{d \left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}-\frac{e x^5 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{5}{2};-p,1;\frac{7}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{5 d^2}+\frac{d^3 \operatorname{Subst}\left (\int \frac{(a+b x)^p}{d^2-e^2 x} \, dx,x,x^2\right )}{2 e^2}\\ &=-\frac{d \left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}-\frac{e x^5 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{5}{2};-p,1;\frac{7}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{5 d^2}+\frac{d^3 \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac{e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e^2 \left (b d^2+a e^2\right ) (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.355292, size = 260, normalized size = 1.6 \[ \frac{\left (a+b x^2\right )^p \left (\frac{e \left (\frac{b x^2}{a}+1\right )^{-p} \left (6 b d^2 (p+1) x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+e \left (2 b e (p+1) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )-3 d \left (b x^2 \left (\frac{b x^2}{a}+1\right )^p+a \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )\right )\right )\right )}{b (p+1)}-\frac{3 d^3 \left (\frac{e \left (x-\sqrt{-\frac{a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac{d-\sqrt{-\frac{a}{b}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{b}} e}{d+e x}\right )}{p}\right )}{6 e^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*(a + b*x^2)^p)/(d + e*x),x]

[Out]

((a + b*x^2)^p*((-3*d^3*AppellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(
d + e*x)])/(p*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p) + (e*(6*b*d^2*(1 + p
)*x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + e*(-3*d*(b*x^2*(1 + (b*x^2)/a)^p + a*(-1 + (1 + (b*x^2)/a)
^p)) + 2*b*e*(1 + p)*x^3*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)])))/(b*(1 + p)*(1 + (b*x^2)/a)^p)))/(6*e
^4)

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Maple [F]  time = 0.648, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p}{x}^{3}}{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^p/(e*x+d),x)

[Out]

int(x^3*(b*x^2+a)^p/(e*x+d),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{p} x^{3}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^p/(e*x+d),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^p*x^3/(e*x + d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{p} x^{3}}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^p/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^p*x^3/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)**p/(e*x+d),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{p} x^{3}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^p/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p*x^3/(e*x + d), x)