\(\int \frac {(d+e x)^3 (d^2-e^2 x^2)^p}{x} \, dx\) [264]

Optimal result

Integrand size = 25, antiderivative size = 171 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x} \, dx=-\frac {3 d \left (d^2-e^2 x^2\right )^{1+p}}{2 (1+p)}-\frac {e x \left (d^2-e^2 x^2\right )^{1+p}}{3+2 p}+\frac {2 d^2 e (5+3 p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{3+2 p}-\frac {d \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )}{2 (1+p)} \]

[Out]

-3/2*d*(-e^2*x^2+d^2)^(p+1)/(p+1)-e*x*(-e^2*x^2+d^2)^(p+1)/(3+2*p)+2*d^2*e*(5+3*p)*x*(-e^2*x^2+d^2)^p*hypergeo
m([1/2, -p],[3/2],e^2*x^2/d^2)/(3+2*p)/((1-e^2*x^2/d^2)^p)-1/2*d*(-e^2*x^2+d^2)^(p+1)*hypergeom([1, p+1],[2+p]
,1-e^2*x^2/d^2)/(p+1)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1666, 457, 81, 67, 396, 252, 251} \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x} \, dx=\frac {2 d^2 e (3 p+5) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{2 p+3}-\frac {d \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,1-\frac {e^2 x^2}{d^2}\right )}{2 (p+1)}-\frac {e x \left (d^2-e^2 x^2\right )^{p+1}}{2 p+3}-\frac {3 d \left (d^2-e^2 x^2\right )^{p+1}}{2 (p+1)} \]

[In]

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

[Out]

(-3*d*(d^2 - e^2*x^2)^(1 + p))/(2*(1 + p)) - (e*x*(d^2 - e^2*x^2)^(1 + p))/(3 + 2*p) + (2*d^2*e*(5 + 3*p)*x*(d
^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^2])/((3 + 2*p)*(1 - (e^2*x^2)/d^2)^p) - (d*(d^2 -
e^2*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (e^2*x^2)/d^2])/(2*(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*} \text {integral}& = \int \frac {\left (d^2-e^2 x^2\right )^p \left (d^3+3 d e^2 x^2\right )}{x} \, dx+\int \left (d^2-e^2 x^2\right )^p \left (3 d^2 e+e^3 x^2\right ) \, dx \\ & = -\frac {e x \left (d^2-e^2 x^2\right )^{1+p}}{3+2 p}+\frac {1}{2} \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^p \left (d^3+3 d e^2 x\right )}{x} \, dx,x,x^2\right )+\frac {\left (2 d^2 e (5+3 p)\right ) \int \left (d^2-e^2 x^2\right )^p \, dx}{3+2 p} \\ & = -\frac {3 d \left (d^2-e^2 x^2\right )^{1+p}}{2 (1+p)}-\frac {e x \left (d^2-e^2 x^2\right )^{1+p}}{3+2 p}+\frac {1}{2} d^3 \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^p}{x} \, dx,x,x^2\right )+\frac {\left (2 d^2 e (5+3 p) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx}{3+2 p} \\ & = -\frac {3 d \left (d^2-e^2 x^2\right )^{1+p}}{2 (1+p)}-\frac {e x \left (d^2-e^2 x^2\right )^{1+p}}{3+2 p}+\frac {2 d^2 e (5+3 p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{3+2 p}-\frac {d \left (d^2-e^2 x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1-\frac {e^2 x^2}{d^2}\right )}{2 (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x} \, dx=\frac {1}{6} \left (d^2-e^2 x^2\right )^p \left (-\frac {9 d \left (d^2-e^2 x^2\right )}{1+p}+18 d^2 e x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\frac {3 d \left (d^2-e^2 x^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )}{1+p}+2 e^3 x^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )\right ) \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (e x +d \right )^{3} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{x}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x} \,d x } \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.81 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x} \, dx=- \frac {d^{3} e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + 3 d^{2} d^{2 p} e x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + 3 d e^{2} \left (\begin {cases} \frac {x^{2} \left (d^{2}\right )^{p}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\begin {cases} \frac {\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (d^{2} - e^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2 p} e^{3} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} \]

[In]

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

[Out]

-d**3*e**(2*p)*x**(2*p)*exp(I*pi*p)*gamma(-p)*hyper((-p, -p), (1 - p,), d**2/(e**2*x**2))/(2*gamma(1 - p)) + 3
*d**2*d**(2*p)*e*x*hyper((1/2, -p), (3/2,), e**2*x**2*exp_polar(2*I*pi)/d**2) + 3*d*e**2*Piecewise((x**2*(d**2
)**p/2, Eq(e**2, 0)), (-Piecewise(((d**2 - e**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(d**2 - e**2*x**2), Tr
ue))/(2*e**2), True)) + d**(2*p)*e**3*x**3*hyper((3/2, -p), (5/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/3

Maxima [F]

\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^3}{x} \,d x \]

[In]

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

[Out]

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