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

Optimal. Leaf size=171 \[ -\frac {d^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a (p+1)}-\frac {e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2-3 b d^2 (2 p+3)\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b (2 p+3)}+\frac {3 d e^2 \left (a+b x^2\right )^{p+1}}{2 b (p+1)}+\frac {e^3 x \left (a+b x^2\right )^{p+1}}{b (2 p+3)} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 165, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1652, 446, 80, 65, 388, 246, 245} \[ e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac {a e^2}{2 b p+3 b}\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )-\frac {d^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a (p+1)}+\frac {3 d e^2 \left (a+b x^2\right )^{p+1}}{2 b (p+1)}+\frac {e^3 x \left (a+b x^2\right )^{p+1}}{b (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(a + b*x^2)^p)/x,x]

[Out]

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

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 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 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])

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 388

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 170, normalized size = 0.99 \[ \frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (-3 b d^3 \left (a+b x^2\right ) \left (\frac {b x^2}{a}+1\right )^p \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )+18 a b d^2 e (p+1) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+a e^2 \left (9 d \left (a+b x^2\right ) \left (\frac {b x^2}{a}+1\right )^p+2 b e (p+1) x^3 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )\right )\right )}{6 a b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^p*(18*a*b*d^2*e*(1 + p)*x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] - 3*b*d^3*(a + b*x^2)*(1
+ (b*x^2)/a)^p*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a] + a*e^2*(9*d*(a + b*x^2)*(1 + (b*x^2)/a)^p +
2*b*e*(1 + p)*x^3*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)])))/(6*a*b*(1 + p)*(1 + (b*x^2)/a)^p)

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} {\left (b x^{2} + a\right )}^{p}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{3} \left (b \,x^{2}+a \right )^{p}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^p*(d + e*x)^3)/x,x)

[Out]

int(((a + b*x^2)^p*(d + e*x)^3)/x, x)

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sympy [A]  time = 17.62, size = 144, normalized size = 0.84 \[ 3 a^{p} d^{2} e x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {a^{p} e^{3} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} - \frac {b^{p} d^{3} x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + 3 d e^{2} \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a**p*d**2*e*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*e**3*x**3*hyper((3/2, -p), (5/2,), b
*x**2*exp_polar(I*pi)/a)/3 - b**p*d**3*x**(2*p)*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*x**2)
)/(2*gamma(1 - p)) + 3*d*e**2*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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