3.2263 \(\int \frac{(d+e x)^{3/2} (f+g x)}{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\)

Optimal. Leaf size=193 \[ -\frac{2 \sqrt{d+e x} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+3 c d g+5 c e f)}{15 c^2 e^2}-\frac{4 (2 c d-b e) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+3 c d g+5 c e f)}{15 c^3 e^2 \sqrt{d+e x}}-\frac{2 g (d+e x)^{3/2} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2} \]

[Out]

(-4*(2*c*d - b*e)*(5*c*e*f + 3*c*d*g - 4*b*e*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(15*c^3*e^2*Sqrt[d
+ e*x]) - (2*(5*c*e*f + 3*c*d*g - 4*b*e*g)*Sqrt[d + e*x]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(15*c^2*e^
2) - (2*g*(d + e*x)^(3/2)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(5*c*e^2)

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Rubi [A]  time = 0.324414, antiderivative size = 193, normalized size of antiderivative = 1., 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 = {794, 656, 648} \[ -\frac{2 \sqrt{d+e x} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+3 c d g+5 c e f)}{15 c^2 e^2}-\frac{4 (2 c d-b e) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+3 c d g+5 c e f)}{15 c^3 e^2 \sqrt{d+e x}}-\frac{2 g (d+e x)^{3/2} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(2*c*d - b*e)*(5*c*e*f + 3*c*d*g - 4*b*e*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(15*c^3*e^2*Sqrt[d
+ e*x]) - (2*(5*c*e*f + 3*c*d*g - 4*b*e*g)*Sqrt[d + e*x]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(15*c^2*e^
2) - (2*g*(d + e*x)^(3/2)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(5*c*e^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 656

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 648

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^{3/2} (f+g x)}{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx &=-\frac{2 g (d+e x)^{3/2} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2}-\frac{\left (2 \left (\frac{1}{2} e \left (-2 c e^2 f+b e^2 g\right )+\frac{3}{2} \left (-c e^3 f+\left (-c d e^2+b e^3\right ) g\right )\right )\right ) \int \frac{(d+e x)^{3/2}}{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{5 c e^3}\\ &=-\frac{2 (5 c e f+3 c d g-4 b e g) \sqrt{d+e x} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^2 e^2}-\frac{2 g (d+e x)^{3/2} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2}+\frac{(2 (2 c d-b e) (5 c e f+3 c d g-4 b e g)) \int \frac{\sqrt{d+e x}}{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{15 c^2 e}\\ &=-\frac{4 (2 c d-b e) (5 c e f+3 c d g-4 b e g) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^3 e^2 \sqrt{d+e x}}-\frac{2 (5 c e f+3 c d g-4 b e g) \sqrt{d+e x} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^2 e^2}-\frac{2 g (d+e x)^{3/2} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2}\\ \end{align*}

Mathematica [A]  time = 0.101212, size = 118, normalized size = 0.61 \[ \frac{2 \sqrt{d+e x} (b e-c d+c e x) \left (8 b^2 e^2 g-2 b c e (13 d g+5 e f+2 e g x)+c^2 \left (18 d^2 g+d e (25 f+9 g x)+e^2 x (5 f+3 g x)\right )\right )}{15 c^3 e^2 \sqrt{(d+e x) (c (d-e x)-b e)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(-(c*d) + b*e + c*e*x)*(8*b^2*e^2*g - 2*b*c*e*(5*e*f + 13*d*g + 2*e*g*x) + c^2*(18*d^2*g + e^
2*x*(5*f + 3*g*x) + d*e*(25*f + 9*g*x))))/(15*c^3*e^2*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])

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Maple [A]  time = 0.003, size = 139, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2\,cex+2\,be-2\,cd \right ) \left ( 3\,g{x}^{2}{c}^{2}{e}^{2}-4\,bc{e}^{2}gx+9\,{c}^{2}degx+5\,{c}^{2}{e}^{2}fx+8\,{b}^{2}{e}^{2}g-26\,bcdeg-10\,bc{e}^{2}f+18\,{c}^{2}{d}^{2}g+25\,{c}^{2}def \right ) }{15\,{c}^{3}{e}^{2}}\sqrt{ex+d}{\frac{1}{\sqrt{-c{e}^{2}{x}^{2}-b{e}^{2}x-bde+c{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x)

[Out]

2/15*(c*e*x+b*e-c*d)*(3*c^2*e^2*g*x^2-4*b*c*e^2*g*x+9*c^2*d*e*g*x+5*c^2*e^2*f*x+8*b^2*e^2*g-26*b*c*d*e*g-10*b*
c*e^2*f+18*c^2*d^2*g+25*c^2*d*e*f)*(e*x+d)^(1/2)/c^3/e^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)

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Maxima [A]  time = 1.24237, size = 271, normalized size = 1.4 \begin{align*} \frac{2 \,{\left (c^{2} e^{2} x^{2} - 5 \, c^{2} d^{2} + 7 \, b c d e - 2 \, b^{2} e^{2} +{\left (4 \, c^{2} d e - b c e^{2}\right )} x\right )} f}{3 \, \sqrt{-c e x + c d - b e} c^{2} e} + \frac{2 \,{\left (3 \, c^{3} e^{3} x^{3} - 18 \, c^{3} d^{3} + 44 \, b c^{2} d^{2} e - 34 \, b^{2} c d e^{2} + 8 \, b^{3} e^{3} +{\left (6 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} +{\left (9 \, c^{3} d^{2} e - 13 \, b c^{2} d e^{2} + 4 \, b^{2} c e^{3}\right )} x\right )} g}{15 \, \sqrt{-c e x + c d - b e} c^{3} e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

2/3*(c^2*e^2*x^2 - 5*c^2*d^2 + 7*b*c*d*e - 2*b^2*e^2 + (4*c^2*d*e - b*c*e^2)*x)*f/(sqrt(-c*e*x + c*d - b*e)*c^
2*e) + 2/15*(3*c^3*e^3*x^3 - 18*c^3*d^3 + 44*b*c^2*d^2*e - 34*b^2*c*d*e^2 + 8*b^3*e^3 + (6*c^3*d*e^2 - b*c^2*e
^3)*x^2 + (9*c^3*d^2*e - 13*b*c^2*d*e^2 + 4*b^2*c*e^3)*x)*g/(sqrt(-c*e*x + c*d - b*e)*c^3*e^2)

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Fricas [A]  time = 1.48913, size = 304, normalized size = 1.58 \begin{align*} -\frac{2 \,{\left (3 \, c^{2} e^{2} g x^{2} + 5 \,{\left (5 \, c^{2} d e - 2 \, b c e^{2}\right )} f + 2 \,{\left (9 \, c^{2} d^{2} - 13 \, b c d e + 4 \, b^{2} e^{2}\right )} g +{\left (5 \, c^{2} e^{2} f +{\left (9 \, c^{2} d e - 4 \, b c e^{2}\right )} g\right )} x\right )} \sqrt{-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt{e x + d}}{15 \,{\left (c^{3} e^{3} x + c^{3} d e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(3*c^2*e^2*g*x^2 + 5*(5*c^2*d*e - 2*b*c*e^2)*f + 2*(9*c^2*d^2 - 13*b*c*d*e + 4*b^2*e^2)*g + (5*c^2*e^2*f
+ (9*c^2*d*e - 4*b*c*e^2)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d)/(c^3*e^3*x + c^3*d*e
^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{\frac{3}{2}} \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)**(3/2)*(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)**(3/2)*(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*} \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)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x