3.2288 \(\int \frac{(d+e x)^m (f+g x)}{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\)

Optimal. Leaf size=205 \[ \frac{(d+e x)^m (-b e+c d-c e x) \left (\frac{c (d+e x)}{2 c d-b e}\right )^{\frac{1}{2}-m} (b e g (2 m+1)-2 c (d g m+e f (m+1))) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{c d-b e-c e x}{2 c d-b e}\right )}{c^2 e^2 (m+1) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac{g (d+e x)^m \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.312948, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 44, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {794, 679, 677, 70, 69} \[ \frac{(d+e x)^m (-b e+c d-c e x) \left (\frac{c (d+e x)}{2 c d-b e}\right )^{\frac{1}{2}-m} (b e g (2 m+1)-2 c (d g m+e f (m+1))) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{c d-b e-c e x}{2 c d-b e}\right )}{c^2 e^2 (m+1) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac{g (d+e x)^m \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 679

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPart[m]*(d + e*
x)^FracPart[m])/(1 + (e*x)/d)^FracPart[m], Int[(1 + (e*x)/d)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c,
d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] || GtQ
[d, 0])

Rule 677

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^m*(a + b*x + c*x^2
)^FracPart[p])/((1 + (e*x)/d)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)
/e)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !Int
egerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (IntegerQ[3*p] || IntegerQ[4*p]))

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

Mathematica [A]  time = 0.34664, size = 155, normalized size = 0.76 \[ -\frac{2 (d+e x)^m \sqrt{(d+e x) (c (d-e x)-b e)} \left (\frac{e \left (\frac{c (d+e x)}{2 c d-b e}\right )^{-m-\frac{1}{2}} (b e g (2 m+1)-2 c (d g m+e f (m+1))) \, _2F_1\left (\frac{1}{2},-m-\frac{1}{2};\frac{3}{2};\frac{-c d+b e+c e x}{b e-2 c d}\right )}{c}+e (e f-d g)\right )}{e^3 (2 m+1) (b e-2 c d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.073, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{m} \left ( gx+f \right ){\frac{1}{\sqrt{-c{e}^{2}{x}^{2}-b{e}^{2}x-bde+c{d}^{2}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{m} \left (f + g x\right )}{\sqrt{- \left (d + e x\right ) \left (b e - c d + c e x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**m*(f + g*x)/sqrt(-(d + e*x)*(b*e - c*d + c*e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x + f\right )}{\left (e x + d\right )}^{m}}{\sqrt{-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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