\(\int \frac {(d+e x) (d^2-e^2 x^2)^p}{x} \, dx\) [246]

Optimal result

Integrand size = 23, antiderivative size = 104 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^p}{x} \, dx=e x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d (1+p)} \]

[Out]

e*x*(-e^2*x^2+d^2)^p*hypergeom([1/2, -p],[3/2],e^2*x^2/d^2)/((1-e^2*x^2/d^2)^p)-1/2*(-e^2*x^2+d^2)^(p+1)*hyper
geom([1, p+1],[2+p],1-e^2*x^2/d^2)/d/(p+1)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {778, 272, 67, 252, 251} \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^p}{x} \, dx=e x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,1-\frac {e^2 x^2}{d^2}\right )}{2 d (p+1)} \]

[In]

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

[Out]

(e*x*(d^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p - ((d^2 - e^2*x^2
)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (e^2*x^2)/d^2])/(2*d*(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 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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 778

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

Rubi steps \begin{align*} \text {integral}& = d \int \frac {\left (d^2-e^2 x^2\right )^p}{x} \, dx+e \int \left (d^2-e^2 x^2\right )^p \, dx \\ & = \frac {1}{2} d \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^p}{x} \, dx,x,x^2\right )+\left (e \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 \\ & = e 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 )-\frac {\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 d (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^p}{x} \, dx=e x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d (1+p)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.75 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^p}{x} \, dx=- \frac {d e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + d^{2 p} e x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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