3.762 \(\int \frac{(d+e x)^{3/2} (f+g x)^n}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{(d+e x)^{3/2} (f+g x)^{n+1} (a e+c d x) \, _2F_1\left (1,n+\frac{1}{2};n+2;\frac{c d (f+g x)}{c d f-a e g}\right )}{(n+1) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)} \]

[Out]

-(((a*e + c*d*x)*(d + e*x)^(3/2)*(f + g*x)^(1 + n)*Hypergeometric2F1[1, 1/2 + n, 2 + n, (c*d*(f + g*x))/(c*d*f
- a*e*g)])/((c*d*f - a*e*g)*(1 + n)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)))

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Rubi [A]  time = 0.110115, antiderivative size = 110, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {891, 70, 69} \[ -\frac{2 \sqrt{d+e x} (f+g x)^n \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (-\frac{1}{2},-n;\frac{1}{2};-\frac{g (a e+c d x)}{c d f-a e g}\right )}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^(3/2)*(f + g*x)^n)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

(-2*Sqrt[d + e*x]*(f + g*x)^n*Hypergeometric2F1[-1/2, -n, 1/2, -((g*(a*e + c*d*x))/(c*d*f - a*e*g))])/(c*d*((c
*d*(f + g*x))/(c*d*f - a*e*g))^n*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 891

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

Rule 70

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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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]))

Rubi steps

\begin{align*} \int \frac{(d+e x)^{3/2} (f+g x)^n}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac{\left (\sqrt{a e+c d x} \sqrt{d+e x}\right ) \int \frac{(f+g x)^n}{(a e+c d x)^{3/2}} \, dx}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac{\left (\sqrt{a e+c d x} \sqrt{d+e x} (f+g x)^n \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{-n}\right ) \int \frac{\left (\frac{c d f}{c d f-a e g}+\frac{c d g x}{c d f-a e g}\right )^n}{(a e+c d x)^{3/2}} \, dx}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac{2 \sqrt{d+e x} (f+g x)^n \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (-\frac{1}{2},-n;\frac{1}{2};-\frac{g (a e+c d x)}{c d f-a e g}\right )}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0467142, size = 98, normalized size = 0.94 \[ -\frac{2 \sqrt{d+e x} (f+g x)^n \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (-\frac{1}{2},-n;\frac{1}{2};\frac{g (a e+c d x)}{a e g-c d f}\right )}{c d \sqrt{(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^(3/2)*(f + g*x)^n)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

(-2*Sqrt[d + e*x]*(f + g*x)^n*Hypergeometric2F1[-1/2, -n, 1/2, (g*(a*e + c*d*x))/(-(c*d*f) + a*e*g)])/(c*d*Sqr
t[(a*e + c*d*x)*(d + e*x)]*((c*d*(f + g*x))/(c*d*f - a*e*g))^n)

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Maple [F]  time = 1.721, size = 0, normalized size = 0. \begin{align*} \int{ \left ( gx+f \right ) ^{n} \left ( ex+d \right ) ^{{\frac{3}{2}}} \left ( ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(g*x+f)^n/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

int((e*x+d)^(3/2)*(g*x+f)^n/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{3}{2}}{\left (g x + f\right )}^{n}}{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(g*x+f)^n/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(3/2)*(g*x + f)^n/(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}{\left (g x + f\right )}^{n}}{c^{2} d^{2} e x^{3} + a^{2} d e^{2} +{\left (c^{2} d^{3} + 2 \, a c d e^{2}\right )} x^{2} +{\left (2 \, a c d^{2} e + a^{2} e^{3}\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(g*x+f)^n/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*(g*x + f)^n/(c^2*d^2*e*x^3 + a^2*d*e^2 + (c
^2*d^3 + 2*a*c*d*e^2)*x^2 + (2*a*c*d^2*e + a^2*e^3)*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(3/2)*(g*x+f)**n/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(g*x+f)^n/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x