3.408 \(\int \frac{x^4 (a+b x^2)^p}{d+e x} \, dx\)

Optimal. Leaf size=199 \[ \frac{x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{5}{2};-p,1;\frac{7}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{5 d}+\frac{\left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 e^3 (p+1)}+\frac{\left (a+b x^2\right )^{p+2}}{2 b^2 e (p+2)}-\frac{d^4 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^3 (p+1) \left (a e^2+b d^2\right )} \]

[Out]

((b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^2*e^3*(1 + p)) + (a + b*x^2)^(2 + p)/(2*b^2*e*(2 + p)) + (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*(1 + (b*x^2)/a)^p) - (d^4*(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^3*(b*d^2 + a*e^2)*(1 + p))

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Rubi [A]  time = 0.226079, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {959, 511, 510, 446, 88, 68} \[ \frac{x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{5}{2};-p,1;\frac{7}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{5 d}+\frac{\left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 e^3 (p+1)}+\frac{\left (a+b x^2\right )^{p+2}}{2 b^2 e (p+2)}-\frac{d^4 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^3 (p+1) \left (a e^2+b d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^2)^p)/(d + e*x),x]

[Out]

((b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^2*e^3*(1 + p)) + (a + b*x^2)^(2 + p)/(2*b^2*e*(2 + p)) + (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*(1 + (b*x^2)/a)^p) - (d^4*(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^3*(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 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])

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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]

Rubi steps

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

Mathematica [F]  time = 0.673851, size = 0, normalized size = 0. \[ \int \frac{x^4 \left (a+b x^2\right )^p}{d+e x} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(x^4*(a + b*x^2)^p)/(d + e*x),x]

[Out]

Integrate[(x^4*(a + b*x^2)^p)/(d + e*x), x]

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Maple [F]  time = 0.658, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{4} \left ( b{x}^{2}+a \right ) ^{p}}{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)^p/(e*x+d),x)

[Out]

int(x^4*(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^{4}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^p/(e*x+d),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^p*x^4/(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^{4}}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^p/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^p*x^4/(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**4*(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^{4}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^p/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p*x^4/(e*x + d), x)