3.382 \(\int x^4 (d+e x) (a+b x^2)^p \, dx\)

Optimal. Leaf size=125 \[ \frac {a^2 e \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}-\frac {a e \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {e \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac {1}{5} d x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {764, 365, 364, 266, 43} \[ \frac {a^2 e \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}-\frac {a e \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {e \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac {1}{5} d x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4*(d + e*x)*(a + b*x^2)^p,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 764

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 112, normalized size = 0.90 \[ \frac {1}{10} \left (a+b x^2\right )^p \left (\frac {5 e \left (a+b x^2\right ) \left (2 a^2-2 a b (p+1) x^2+b^2 \left (p^2+3 p+2\right ) x^4\right )}{b^3 (p+1) (p+2) (p+3)}+2 d x^5 \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^p*((5*e*(a + b*x^2)*(2*a^2 - 2*a*b*(1 + p)*x^2 + b^2*(2 + 3*p + p^2)*x^4))/(b^3*(1 + p)*(2 + p)*(
3 + p)) + (2*d*x^5*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/10

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x^{5} + d x^{4}\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x^5 + d*x^4)*(b*x^2 + a)^p, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right ) x^{4} \left (b \,x^{2}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,{\left (b\,x^2+a\right )}^p\,\left (d+e\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*x^2)^p*(d + e*x),x)

[Out]

int(x^4*(a + b*x^2)^p*(d + e*x), x)

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sympy [C]  time = 21.96, size = 1012, normalized size = 8.10 \[ \frac {a^{p} d x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + e \left (\begin {cases} \frac {a^{p} x^{6}}{6} & \text {for}\: b = 0 \\\frac {2 a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {3 a^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} & \text {for}\: p = -3 \\- \frac {2 a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {b^{2} x^{4}}{2 a b^{3} + 2 b^{4} x^{2}} & \text {for}\: p = -2 \\\frac {a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{3}} + \frac {a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{3}} - \frac {a x^{2}}{2 b^{2}} + \frac {x^{4}}{4 b} & \text {for}\: p = -1 \\\frac {2 a^{3} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} - \frac {2 a^{2} b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {a b^{2} p^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {a b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {b^{3} p^{2} x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {3 b^{3} p x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {2 b^{3} x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*d*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + e*Piecewise((a**p*x**6/6, Eq(b, 0)), (2*a**
2*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*a**2*log(I*sqrt(a)*sqrt(1/b) +
x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 3*a**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x
**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(I*sqrt(a)*sqrt(
1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4)
+ 2*b**2*x**4*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(I*s
qrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4), Eq(p, -3)), (-2*a**2*log(-I*sqrt(a)*sqrt(1/
b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a**2/(2*a*
b**3 + 2*b**4*x**2) - 2*a*b*x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log(I*sqr
t(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) + b**2*x**4/(2*a*b**3 + 2*b**4*x**2), Eq(p, -2)), (a**2*log(-I*sq
rt(a)*sqrt(1/b) + x)/(2*b**3) + a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(2*b**3) - a*x**2/(2*b**2) + x**4/(4*b), Eq(
p, -1)), (2*a**3*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) - 2*a**2*b*p*x**2*(a + b*x
**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p**2*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 1
2*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 1
2*b**3) + b**3*p**2*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + 3*b**3*p*x**6*(a
+ b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + 2*b**3*x**6*(a + b*x**2)**p/(2*b**3*p**3 +
12*b**3*p**2 + 22*b**3*p + 12*b**3), True))

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