3.5.5 \(\int \frac {(d+e x)^3 (a+b x^2)^p}{x} \, dx\) [405]

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

[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 = 0.09, 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 = {1666, 457, 81, 67, 396, 252, 251} \begin {gather*} -\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)}+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 {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)} \end {gather*}

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 67

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

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 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 396

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 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 1666

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} \text {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 \text {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 = 0.26, size = 170, normalized size = 0.99 \begin {gather*} \frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (18 a b d^2 e (1+p) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )-3 b d^3 \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p \, _2F_1\left (1,1+p;2+p;1+\frac {b x^2}{a}\right )+a e^2 \left (9 d \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p+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 )}{6 a b (1+p)} \end {gather*}

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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Maple [F]
time = 0.01, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

[Out]

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

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Sympy [A]
time = 7.37, size = 144, normalized size = 0.84 \begin {gather*} 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 ) \end {gather*}

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

[Out]

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

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

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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