3.23.89 \(\int \frac {(d+e x)^m (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}} \, dx\) [2289]

Optimal. Leaf size=210 \[ \frac {g (d+e x)^m}{c e^2 (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(b e g (1-2 m)-2 c (e f (1-m)-d g m)) (d+e x)^m \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {c d-b e-c e x}{2 c d-b e}\right )}{c e^2 (2 c d-b e) (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 210, normalized size of antiderivative = 1.00, 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 = {808, 693, 691, 72, 71} \begin {gather*} \frac {g (d+e x)^m}{c e^2 (1-m) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(d+e x)^m \left (\frac {c (d+e x)}{2 c d-b e}\right )^{\frac {1}{2}-m} (b e g (1-2 m)-2 c (e f (1-m)-d g m)) \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {c d-b e-c e x}{2 c d-b e}\right )}{c e^2 (1-m) (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(g*(d + e*x)^m)/(c*e^2*(1 - m)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]) - ((b*e*g*(1 - 2*m) - 2*c*(e*f*(1 -
m) - d*g*m))*(d + e*x)^m*((c*(d + e*x))/(2*c*d - b*e))^(1/2 - m)*Hypergeometric2F1[-1/2, 3/2 - m, 1/2, (c*d -
b*e - c*e*x)/(2*c*d - b*e)])/(c*e^2*(2*c*d - b*e)*(1 - m)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^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 691

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/e)
*x)^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 693

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 808

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])

Rubi steps

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

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Mathematica [A]
time = 0.60, size = 176, normalized size = 0.84 \begin {gather*} \frac {2 (d+e x)^m \left (e (2 c d-b e) (c e f+c d g-b e g)-e (b e g (1-2 m)+2 c (e f (-1+m)+d g 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 {3}{2}-m;\frac {3}{2};\frac {-c d+b e+c e x}{-2 c d+b e}\right )\right )}{c e^3 (-2 c d+b e)^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m} \left (g x +f \right )}{\left (-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}\right )^{\frac {3}{2}}}\, dx\]

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)^(3/2),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)^(3/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

integrate((g*x + f)*(x*e + d)^m/(-c*x^2*e^2 + c*d^2 - b*x*e^2 - b*d*e)^(3/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*d^2 - b*d*e - (c*x^2 + b*x)*e^2)*(g*x + f)*(x*e + d)^m/(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 +
(c^2*x^4 + 2*b*c*x^3 + b^2*x^2)*e^4 - 2*(c^2*d^2*x^2 + b*c*d^2*x - (b*c*d*x^2 + b^2*d*x)*e)*e^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{m} \left (f + g x\right )}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)**(3/2),x)

[Out]

Integral((d + e*x)**m*(f + g*x)/(-(d + e*x)*(b*e - c*d + c*e*x))**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

integrate((g*x + f)*(x*e + d)^m/(-c*x^2*e^2 + c*d^2 - b*x*e^2 - b*d*e)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (f+g\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(d + e*x)^m)/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2),x)

[Out]

int(((f + g*x)*(d + e*x)^m)/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2), x)

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