3.5.9 \(\int \frac {x^3 (a+b x^2)^p}{d+e x} \, dx\) [409]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {973, 457, 81, 70, 525, 524} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*x^2)^p)/(d + e*x),x]

[Out]

-1/2*(d*(a + b*x^2)^(1 + p))/(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 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 81

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 457

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 973

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]

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 \text {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 \text {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*}

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Mathematica [A]
time = 0.34, size = 260, normalized size = 1.60 \begin {gather*} \frac {\left (a+b x^2\right )^p \left (-\frac {3 d^3 \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\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}+\frac {e \left (1+\frac {b x^2}{a}\right )^{-p} \left (6 b d^2 (1+p) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+e \left (-3 d \left (b x^2 \left (1+\frac {b x^2}{a}\right )^p+a \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )\right )+2 b e (1+p) x^3 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )\right )\right )}{b (1+p)}\right )}{6 e^4} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (b \,x^{2}+a \right )^{p}}{e x +d}\, dx\]

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 = 0.00, 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(x^3*(b*x^2+a)^p/(e*x+d),x, algorithm="maxima")

[Out]

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

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*x^2)^p)/(d + e*x),x)

[Out]

int((x^3*(a + b*x^2)^p)/(d + e*x), x)

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