3.776 \(\int (d+e x)^m \sqrt {f+g x} (a d e+(c d^2+a e^2) x+c d e x^2)^{-m} \, dx\)

Optimal. Leaf size=105 \[ \frac {2 (f+g x)^{3/2} (d+e x)^m \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \left (-\frac {g (a e+c d x)}{c d f-a e g}\right )^m \, _2F_1\left (\frac {3}{2},m;\frac {5}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{3 g} \]

[Out]

2/3*(-g*(c*d*x+a*e)/(-a*e*g+c*d*f))^m*(e*x+d)^m*(g*x+f)^(3/2)*hypergeom([3/2, m],[5/2],c*d*(g*x+f)/(-a*e*g+c*d
*f))/g/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m)

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Rubi [A]  time = 0.08, antiderivative size = 105, normalized size of antiderivative = 1.00, 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} \[ \frac {2 (f+g x)^{3/2} (d+e x)^m \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \left (-\frac {g (a e+c d x)}{c d f-a e g}\right )^m \, _2F_1\left (\frac {3}{2},m;\frac {5}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{3 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-((g*(a*e + c*d*x))/(c*d*f - a*e*g)))^m*(d + e*x)^m*(f + g*x)^(3/2)*Hypergeometric2F1[3/2, m, 5/2, (c*d*(f
+ g*x))/(c*d*f - a*e*g)])/(3*g*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^m)

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]))

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 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]

Rubi steps

\begin {align*} \int (d+e x)^m \sqrt {f+g x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx &=\left ((a e+c d x)^m (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}\right ) \int (a e+c d x)^{-m} \sqrt {f+g x} \, dx\\ &=\left (\left (\frac {g (a e+c d x)}{-c d f+a e g}\right )^m (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}\right ) \int \sqrt {f+g x} \left (-\frac {a e g}{c d f-a e g}-\frac {c d g x}{c d f-a e g}\right )^{-m} \, dx\\ &=\frac {2 \left (-\frac {g (a e+c d x)}{c d f-a e g}\right )^m (d+e x)^m (f+g x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, _2F_1\left (\frac {3}{2},m;\frac {5}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{3 g}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 93, normalized size = 0.89 \[ \frac {2 (f+g x)^{3/2} (d+e x)^m ((d+e x) (a e+c d x))^{-m} \left (\frac {g (a e+c d x)}{a e g-c d f}\right )^m \, _2F_1\left (\frac {3}{2},m;\frac {5}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{3 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((g*(a*e + c*d*x))/(-(c*d*f) + a*e*g))^m*(d + e*x)^m*(f + g*x)^(3/2)*Hypergeometric2F1[3/2, m, 5/2, (c*d*(f
+ g*x))/(c*d*f - a*e*g)])/(3*g*((a*e + c*d*x)*(d + e*x))^m)

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {g x + f} {\left (e x + d\right )}^{m}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^(1/2)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {g x + f} {\left (e x + d\right )}^{m}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^(1/2)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \sqrt {g x +f}\, \left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{-m} \left (e x +d \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m*(g*x+f)^(1/2)/((c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^m),x)

[Out]

int((e*x+d)^m*(g*x+f)^(1/2)/((c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^m),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {g x + f} {\left (e x + d\right )}^{m}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^(1/2)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {f+g\,x}\,{\left (d+e\,x\right )}^m}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^m} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^(1/2)*(d + e*x)^m)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^m,x)

[Out]

int(((f + g*x)^(1/2)*(d + e*x)^m)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^m, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*(g*x+f)**(1/2)/((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**m),x)

[Out]

Timed out

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