3.399 \(\int x^5 (d+e x)^3 (a+b x^2)^p \, dx\)

Optimal. Leaf size=247 \[ \frac{a^2 d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^4 (p+1)}-\frac{a d \left (2 b d^2-9 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^4 (p+2)}+\frac{d \left (b d^2-9 a e^2\right ) \left (a+b x^2\right )^{p+3}}{2 b^4 (p+3)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+4}}{2 b^4 (p+4)}-\frac{e x^7 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (7 a e^2-3 b d^2 (2 p+9)\right ) \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )}{7 b (2 p+9)}+\frac{e^3 x^7 \left (a+b x^2\right )^{p+1}}{b (2 p+9)} \]

[Out]

(a^2*d*(b*d^2 - 3*a*e^2)*(a + b*x^2)^(1 + p))/(2*b^4*(1 + p)) + (e^3*x^7*(a + b*x^2)^(1 + p))/(b*(9 + 2*p)) -
(a*d*(2*b*d^2 - 9*a*e^2)*(a + b*x^2)^(2 + p))/(2*b^4*(2 + p)) + (d*(b*d^2 - 9*a*e^2)*(a + b*x^2)^(3 + p))/(2*b
^4*(3 + p)) + (3*d*e^2*(a + b*x^2)^(4 + p))/(2*b^4*(4 + p)) - (e*(7*a*e^2 - 3*b*d^2*(9 + 2*p))*x^7*(a + b*x^2)
^p*Hypergeometric2F1[7/2, -p, 9/2, -((b*x^2)/a)])/(7*b*(9 + 2*p)*(1 + (b*x^2)/a)^p)

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Rubi [A]  time = 0.246802, antiderivative size = 241, normalized size of antiderivative = 0.98, 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 = {1652, 446, 77, 459, 365, 364} \[ \frac{a^2 d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^4 (p+1)}-\frac{a d \left (2 b d^2-9 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^4 (p+2)}+\frac{d \left (b d^2-9 a e^2\right ) \left (a+b x^2\right )^{p+3}}{2 b^4 (p+3)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+4}}{2 b^4 (p+4)}+\frac{1}{7} e x^7 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac{7 a e^2}{2 b p+9 b}\right ) \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )+\frac{e^3 x^7 \left (a+b x^2\right )^{p+1}}{b (2 p+9)} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(d + e*x)^3*(a + b*x^2)^p,x]

[Out]

(a^2*d*(b*d^2 - 3*a*e^2)*(a + b*x^2)^(1 + p))/(2*b^4*(1 + p)) + (e^3*x^7*(a + b*x^2)^(1 + p))/(b*(9 + 2*p)) -
(a*d*(2*b*d^2 - 9*a*e^2)*(a + b*x^2)^(2 + p))/(2*b^4*(2 + p)) + (d*(b*d^2 - 9*a*e^2)*(a + b*x^2)^(3 + p))/(2*b
^4*(3 + p)) + (3*d*e^2*(a + b*x^2)^(4 + p))/(2*b^4*(4 + p)) + (e*(3*d^2 - (7*a*e^2)/(9*b + 2*b*p))*x^7*(a + b*
x^2)^p*Hypergeometric2F1[7/2, -p, 9/2, -((b*x^2)/a)])/(7*(1 + (b*x^2)/a)^p)

Rule 1652

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[x^m*Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2] && IGtQ[m, -2] &&  !
IntegerQ[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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 459

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

Rubi steps

\begin{align*} \int x^5 (d+e x)^3 \left (a+b x^2\right )^p \, dx &=\int x^5 \left (a+b x^2\right )^p \left (d^3+3 d e^2 x^2\right ) \, dx+\int x^6 \left (a+b x^2\right )^p \left (3 d^2 e+e^3 x^2\right ) \, dx\\ &=\frac{e^3 x^7 \left (a+b x^2\right )^{1+p}}{b (9+2 p)}+\frac{1}{2} \operatorname{Subst}\left (\int x^2 (a+b x)^p \left (d^3+3 d e^2 x\right ) \, dx,x,x^2\right )+\left (e \left (3 d^2-\frac{7 a e^2}{9 b+2 b p}\right )\right ) \int x^6 \left (a+b x^2\right )^p \, dx\\ &=\frac{e^3 x^7 \left (a+b x^2\right )^{1+p}}{b (9+2 p)}+\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{a^2 d \left (-b d^2+3 a e^2\right ) (a+b x)^p}{b^3}+\frac{a d \left (-2 b d^2+9 a e^2\right ) (a+b x)^{1+p}}{b^3}+\frac{\left (b d^3-9 a d e^2\right ) (a+b x)^{2+p}}{b^3}+\frac{3 d e^2 (a+b x)^{3+p}}{b^3}\right ) \, dx,x,x^2\right )+\left (e \left (3 d^2-\frac{7 a e^2}{9 b+2 b p}\right ) \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int x^6 \left (1+\frac{b x^2}{a}\right )^p \, dx\\ &=\frac{a^2 d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^4 (1+p)}+\frac{e^3 x^7 \left (a+b x^2\right )^{1+p}}{b (9+2 p)}-\frac{a d \left (2 b d^2-9 a e^2\right ) \left (a+b x^2\right )^{2+p}}{2 b^4 (2+p)}+\frac{d \left (b d^2-9 a e^2\right ) \left (a+b x^2\right )^{3+p}}{2 b^4 (3+p)}+\frac{3 d e^2 \left (a+b x^2\right )^{4+p}}{2 b^4 (4+p)}+\frac{1}{7} e \left (3 d^2-\frac{7 a e^2}{9 b+2 b p}\right ) x^7 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )\\ \end{align*}

Mathematica [A]  time = 0.233041, size = 249, normalized size = 1.01 \[ \frac{1}{126} \left (a+b x^2\right )^p \left (\frac{63 d^3 \left (a+b x^2\right ) \left (2 a^2-2 a b (p+1) x^2+b^2 \left (p^2+3 p+2\right ) x^4\right )}{b^3 (p+1) (p+2) (p+3)}+\frac{189 d e^2 \left (a+b x^2\right ) \left (6 a^2 b (p+1) x^2-6 a^3-3 a b^2 \left (p^2+3 p+2\right ) x^4+b^3 \left (p^3+6 p^2+11 p+6\right ) x^6\right )}{b^4 (p+1) (p+2) (p+3) (p+4)}+54 d^2 e x^7 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )+14 e^3 x^9 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{9}{2},-p;\frac{11}{2};-\frac{b x^2}{a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(d + e*x)^3*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^p*((63*d^3*(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)) + (189*d*e^2*(a + b*x^2)*(-6*a^3 + 6*a^2*b*(1 + p)*x^2 - 3*a*b^2*(2 + 3*p + p^2)*x^4 + b^3*(6 + 11*
p + 6*p^2 + p^3)*x^6))/(b^4*(1 + p)*(2 + p)*(3 + p)*(4 + p)) + (54*d^2*e*x^7*Hypergeometric2F1[7/2, -p, 9/2, -
((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (14*e^3*x^9*Hypergeometric2F1[9/2, -p, 11/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^
p))/126

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Maple [F]  time = 0.539, size = 0, normalized size = 0. \begin{align*} \int{x}^{5} \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(e*x+d)^3*(b*x^2+a)^p,x)

[Out]

int(x^5*(e*x+d)^3*(b*x^2+a)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{6} +{\left (p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + 2 \, a^{3}\right )}{\left (b x^{2} + a\right )}^{p} d^{3}}{2 \,{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} + \int{\left (e^{3} x^{8} + 3 \, d e^{2} x^{7} + 3 \, d^{2} e x^{6}\right )}{\left (b x^{2} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)^3*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

1/2*((p^2 + 3*p + 2)*b^3*x^6 + (p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + 2*a^3)*(b*x^2 + a)^p*d^3/((p^3 + 6*p^2 +
11*p + 6)*b^3) + integrate((e^3*x^8 + 3*d*e^2*x^7 + 3*d^2*e*x^6)*(b*x^2 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{3} x^{8} + 3 \, d e^{2} x^{7} + 3 \, d^{2} e x^{6} + d^{3} x^{5}\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)^3*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

integral((e^3*x^8 + 3*d*e^2*x^7 + 3*d^2*e*x^6 + d^3*x^5)*(b*x^2 + a)^p, 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**5*(e*x+d)**3*(b*x**2+a)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{3}{\left (b x^{2} + a\right )}^{p} x^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)^3*(b*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((e*x + d)^3*(b*x^2 + a)^p*x^5, x)