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

Optimal. Leaf size=159 \[ \frac{d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{a}-\frac{3 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}-\frac{d^3 \left (a+b x^2\right )^{p+1}}{a x}+\frac{e^3 \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.189275, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1807, 1652, 446, 80, 65, 12, 246, 245} \[ \frac{d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{a}-\frac{3 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}-\frac{d^3 \left (a+b x^2\right )^{p+1}}{a x}+\frac{e^3 \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

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 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 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

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

Rubi steps

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

Mathematica [A]  time = 0.123825, size = 154, normalized size = 0.97 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (e x \left (\left (a+b x^2\right ) \left (\frac{b x^2}{a}+1\right )^p \left (a e^2-3 b d^2 \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )\right )+6 a b d e (p+1) x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )\right )-2 a b d^3 (p+1) \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )\right )}{2 a b (p+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^p*(-2*a*b*d^3*(1 + p)*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)] + e*x*(6*a*b*d*e*(1 + p)*x*H
ypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + (a + b*x^2)*(1 + (b*x^2)/a)^p*(a*e^2 - 3*b*d^2*Hypergeometric2F
1[1, 1 + p, 2 + p, 1 + (b*x^2)/a]))))/(2*a*b*(1 + p)*x*(1 + (b*x^2)/a)^p)

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Maple [F]  time = 0.515, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 29.6593, size = 143, normalized size = 0.9 \begin{align*} - \frac{a^{p} d^{3}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - p \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{x} + 3 a^{p} d e^{2} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} - \frac{3 b^{p} d^{2} e x^{2 p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ 1 - p \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + e^{3} \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\begin{cases} \frac{\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + b x^{2} \right )} & \text{otherwise} \end{cases}}{2 b} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**p*d**3*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x + 3*a**p*d*e**2*x*hyper((1/2, -p), (3/2,), b*
x**2*exp_polar(I*pi)/a) - 3*b**p*d**2*e*x**(2*p)*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*x**2
))/(2*gamma(1 - p)) + e**3*Piecewise((a**p*x**2/2, Eq(b, 0)), (Piecewise(((a + b*x**2)**(p + 1)/(p + 1), Ne(p,
-1)), (log(a + b*x**2), True))/(2*b), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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