\(\int \frac {(a^2+2 a b x+b^2 x^2)^p}{(d+e x)^{5/2}} \, dx\) [1755]

Optimal result

Integrand size = 28, antiderivative size = 83 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (-\frac {e (a+b x)}{b d-a e}\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-2 p,-\frac {1}{2},\frac {b (d+e x)}{b d-a e}\right )}{3 e (d+e x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {660, 72, 71} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (a^2+2 a b x+b^2 x^2\right )^p \left (-\frac {e (a+b x)}{b d-a e}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-2 p,-\frac {1}{2},\frac {b (d+e x)}{b d-a e}\right )}{3 e (d+e x)^{3/2}} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^p/(d + e*x)^(5/2),x]

[Out]

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

Rule 71

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

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \left (\left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \frac {\left (a b+b^2 x\right )^{2 p}}{(d+e x)^{5/2}} \, dx \\ & = \left (\left (\frac {e \left (a b+b^2 x\right )}{-b^2 d+a b e}\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \frac {\left (-\frac {a e}{b d-a e}-\frac {b e x}{b d-a e}\right )^{2 p}}{(d+e x)^{5/2}} \, dx \\ & = -\frac {2 \left (-\frac {e (a+b x)}{b d-a e}\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (-\frac {3}{2},-2 p;-\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )}{3 e (d+e x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (\frac {e (a+b x)}{-b d+a e}\right )^{-2 p} \left ((a+b x)^2\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-2 p,-\frac {1}{2},\frac {b (d+e x)}{b d-a e}\right )}{3 e (d+e x)^{3/2}} \]

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^p/(d + e*x)^(5/2),x]

[Out]

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

Maple [F]

\[\int \frac {\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{\left (e x +d \right )^{\frac {5}{2}}}d x\]

[In]

int((b^2*x^2+2*a*b*x+a^2)^p/(e*x+d)^(5/2),x)

[Out]

int((b^2*x^2+2*a*b*x+a^2)^p/(e*x+d)^(5/2),x)

Fricas [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, dx=\int { \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{{\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^p/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{p}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**p/(e*x+d)**(5/2),x)

[Out]

Integral(((a + b*x)**2)**p/(d + e*x)**(5/2), x)

Maxima [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, dx=\int { \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{{\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^p/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2)^p/(e*x + d)^(5/2), x)

Giac [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, dx=\int { \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{{\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^p/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2)^p/(e*x + d)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^p}{(d+e x)^{5/2}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^{5/2}} \,d x \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^p/(d + e*x)^(5/2),x)

[Out]

int((a^2 + b^2*x^2 + 2*a*b*x)^p/(d + e*x)^(5/2), x)