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

Optimal. Leaf size=212 \[ -\frac {2 (f+g x)^{5/2} \left (e g (-a e g-2 b d g+3 b e f)-c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )}{5 g^5}+\frac {2 \sqrt {f+g x} (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}{g^5}-\frac {2 (f+g x)^{3/2} (e f-d g) (2 c f (2 e f-d g)-g (-2 a e g-b d g+3 b e f))}{3 g^5}-\frac {2 e (f+g x)^{7/2} (-b e g-2 c d g+4 c e f)}{7 g^5}+\frac {2 c e^2 (f+g x)^{9/2}}{9 g^5} \]

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Rubi [A]  time = 0.34, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {897, 1153} \begin {gather*} -\frac {2 (f+g x)^{5/2} \left (e g (-a e g-2 b d g+3 b e f)-c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )}{5 g^5}+\frac {2 \sqrt {f+g x} (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}{g^5}-\frac {2 (f+g x)^{3/2} (e f-d g) (2 c f (2 e f-d g)-g (-2 a e g-b d g+3 b e f))}{3 g^5}-\frac {2 e (f+g x)^{7/2} (-b e g-2 c d g+4 c e f)}{7 g^5}+\frac {2 c e^2 (f+g x)^{9/2}}{9 g^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*f - d*g)^2*(c*f^2 - b*f*g + a*g^2)*Sqrt[f + g*x])/g^5 - (2*(e*f - d*g)*(2*c*f*(2*e*f - d*g) - g*(3*b*e*f
- b*d*g - 2*a*e*g))*(f + g*x)^(3/2))/(3*g^5) - (2*(e*g*(3*b*e*f - 2*b*d*g - a*e*g) - c*(6*e^2*f^2 - 6*d*e*f*g
+ d^2*g^2))*(f + g*x)^(5/2))/(5*g^5) - (2*e*(4*c*e*f - 2*c*d*g - b*e*g)*(f + g*x)^(7/2))/(7*g^5) + (2*c*e^2*(
f + g*x)^(9/2))/(9*g^5)

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2 \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2 \left (\frac {c f^2-b f g+a g^2}{g^2}-\frac {(2 c f-b g) x^2}{g^2}+\frac {c x^4}{g^2}\right ) \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {(-e f+d g)^2 \left (c f^2-b f g+a g^2\right )}{g^4}+\frac {(e f-d g) (-2 c f (2 e f-d g)+g (3 b e f-b d g-2 a e g)) x^2}{g^4}+\frac {\left (-e g (3 b e f-2 b d g-a e g)+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) x^4}{g^4}+\frac {e (-4 c e f+2 c d g+b e g) x^6}{g^4}+\frac {c e^2 x^8}{g^4}\right ) \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=\frac {2 (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}{g^5}-\frac {2 (e f-d g) (2 c f (2 e f-d g)-g (3 b e f-b d g-2 a e g)) (f+g x)^{3/2}}{3 g^5}-\frac {2 \left (e g (3 b e f-2 b d g-a e g)-c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) (f+g x)^{5/2}}{5 g^5}-\frac {2 e (4 c e f-2 c d g-b e g) (f+g x)^{7/2}}{7 g^5}+\frac {2 c e^2 (f+g x)^{9/2}}{9 g^5}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 184, normalized size = 0.87 \begin {gather*} \frac {2 \sqrt {f+g x} \left (-63 (f+g x)^2 \left (-e g (a e g+2 b d g-3 b e f)-c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )+315 (e f-d g)^2 \left (g (a g-b f)+c f^2\right )-105 (f+g x) (e f-d g) (g (2 a e g+b d g-3 b e f)+2 c f (2 e f-d g))-45 e (f+g x)^3 (-b e g-2 c d g+4 c e f)+35 c e^2 (f+g x)^4\right )}{315 g^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[f + g*x]*(315*(e*f - d*g)^2*(c*f^2 + g*(-(b*f) + a*g)) - 105*(e*f - d*g)*(2*c*f*(2*e*f - d*g) + g*(-3*
b*e*f + b*d*g + 2*a*e*g))*(f + g*x) - 63*(-(e*g*(-3*b*e*f + 2*b*d*g + a*e*g)) - c*(6*e^2*f^2 - 6*d*e*f*g + d^2
*g^2))*(f + g*x)^2 - 45*e*(4*c*e*f - 2*c*d*g - b*e*g)*(f + g*x)^3 + 35*c*e^2*(f + g*x)^4))/(315*g^5)

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IntegrateAlgebraic [A]  time = 0.21, size = 368, normalized size = 1.74 \begin {gather*} \frac {2 \sqrt {f+g x} \left (315 a d^2 g^4+210 a d e g^3 (f+g x)-630 a d e f g^3+315 a e^2 f^2 g^2-210 a e^2 f g^2 (f+g x)+63 a e^2 g^2 (f+g x)^2+105 b d^2 g^3 (f+g x)-315 b d^2 f g^3+630 b d e f^2 g^2-420 b d e f g^2 (f+g x)+126 b d e g^2 (f+g x)^2-315 b e^2 f^3 g+315 b e^2 f^2 g (f+g x)-189 b e^2 f g (f+g x)^2+45 b e^2 g (f+g x)^3+315 c d^2 f^2 g^2-210 c d^2 f g^2 (f+g x)+63 c d^2 g^2 (f+g x)^2-630 c d e f^3 g+630 c d e f^2 g (f+g x)-378 c d e f g (f+g x)^2+90 c d e g (f+g x)^3+315 c e^2 f^4-420 c e^2 f^3 (f+g x)+378 c e^2 f^2 (f+g x)^2-180 c e^2 f (f+g x)^3+35 c e^2 (f+g x)^4\right )}{315 g^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[f + g*x]*(315*c*e^2*f^4 - 630*c*d*e*f^3*g - 315*b*e^2*f^3*g + 315*c*d^2*f^2*g^2 + 630*b*d*e*f^2*g^2 +
315*a*e^2*f^2*g^2 - 315*b*d^2*f*g^3 - 630*a*d*e*f*g^3 + 315*a*d^2*g^4 - 420*c*e^2*f^3*(f + g*x) + 630*c*d*e*f^
2*g*(f + g*x) + 315*b*e^2*f^2*g*(f + g*x) - 210*c*d^2*f*g^2*(f + g*x) - 420*b*d*e*f*g^2*(f + g*x) - 210*a*e^2*
f*g^2*(f + g*x) + 105*b*d^2*g^3*(f + g*x) + 210*a*d*e*g^3*(f + g*x) + 378*c*e^2*f^2*(f + g*x)^2 - 378*c*d*e*f*
g*(f + g*x)^2 - 189*b*e^2*f*g*(f + g*x)^2 + 63*c*d^2*g^2*(f + g*x)^2 + 126*b*d*e*g^2*(f + g*x)^2 + 63*a*e^2*g^
2*(f + g*x)^2 - 180*c*e^2*f*(f + g*x)^3 + 90*c*d*e*g*(f + g*x)^3 + 45*b*e^2*g*(f + g*x)^3 + 35*c*e^2*(f + g*x)
^4))/(315*g^5)

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fricas [A]  time = 0.54, size = 260, normalized size = 1.23 \begin {gather*} \frac {2 \, {\left (35 \, c e^{2} g^{4} x^{4} + 128 \, c e^{2} f^{4} + 315 \, a d^{2} g^{4} - 144 \, {\left (2 \, c d e + b e^{2}\right )} f^{3} g + 168 \, {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} f^{2} g^{2} - 210 \, {\left (b d^{2} + 2 \, a d e\right )} f g^{3} - 5 \, {\left (8 \, c e^{2} f g^{3} - 9 \, {\left (2 \, c d e + b e^{2}\right )} g^{4}\right )} x^{3} + 3 \, {\left (16 \, c e^{2} f^{2} g^{2} - 18 \, {\left (2 \, c d e + b e^{2}\right )} f g^{3} + 21 \, {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} g^{4}\right )} x^{2} - {\left (64 \, c e^{2} f^{3} g - 72 \, {\left (2 \, c d e + b e^{2}\right )} f^{2} g^{2} + 84 \, {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} f g^{3} - 105 \, {\left (b d^{2} + 2 \, a d e\right )} g^{4}\right )} x\right )} \sqrt {g x + f}}{315 \, g^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*c*e^2*g^4*x^4 + 128*c*e^2*f^4 + 315*a*d^2*g^4 - 144*(2*c*d*e + b*e^2)*f^3*g + 168*(c*d^2 + 2*b*d*e +
a*e^2)*f^2*g^2 - 210*(b*d^2 + 2*a*d*e)*f*g^3 - 5*(8*c*e^2*f*g^3 - 9*(2*c*d*e + b*e^2)*g^4)*x^3 + 3*(16*c*e^2*
f^2*g^2 - 18*(2*c*d*e + b*e^2)*f*g^3 + 21*(c*d^2 + 2*b*d*e + a*e^2)*g^4)*x^2 - (64*c*e^2*f^3*g - 72*(2*c*d*e +
b*e^2)*f^2*g^2 + 84*(c*d^2 + 2*b*d*e + a*e^2)*f*g^3 - 105*(b*d^2 + 2*a*d*e)*g^4)*x)*sqrt(g*x + f)/g^5

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giac [A]  time = 0.23, size = 363, normalized size = 1.71 \begin {gather*} \frac {2 \, {\left (315 \, \sqrt {g x + f} a d^{2} + \frac {105 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} b d^{2}}{g} + \frac {210 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} a d e}{g} + \frac {21 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} c d^{2}}{g^{2}} + \frac {42 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} b d e}{g^{2}} + \frac {21 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} a e^{2}}{g^{2}} + \frac {18 \, {\left (5 \, {\left (g x + f\right )}^{\frac {7}{2}} - 21 \, {\left (g x + f\right )}^{\frac {5}{2}} f + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} f^{2} - 35 \, \sqrt {g x + f} f^{3}\right )} c d e}{g^{3}} + \frac {9 \, {\left (5 \, {\left (g x + f\right )}^{\frac {7}{2}} - 21 \, {\left (g x + f\right )}^{\frac {5}{2}} f + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} f^{2} - 35 \, \sqrt {g x + f} f^{3}\right )} b e^{2}}{g^{3}} + \frac {{\left (35 \, {\left (g x + f\right )}^{\frac {9}{2}} - 180 \, {\left (g x + f\right )}^{\frac {7}{2}} f + 378 \, {\left (g x + f\right )}^{\frac {5}{2}} f^{2} - 420 \, {\left (g x + f\right )}^{\frac {3}{2}} f^{3} + 315 \, \sqrt {g x + f} f^{4}\right )} c e^{2}}{g^{4}}\right )}}{315 \, g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(315*sqrt(g*x + f)*a*d^2 + 105*((g*x + f)^(3/2) - 3*sqrt(g*x + f)*f)*b*d^2/g + 210*((g*x + f)^(3/2) - 3*
sqrt(g*x + f)*f)*a*d*e/g + 21*(3*(g*x + f)^(5/2) - 10*(g*x + f)^(3/2)*f + 15*sqrt(g*x + f)*f^2)*c*d^2/g^2 + 42
*(3*(g*x + f)^(5/2) - 10*(g*x + f)^(3/2)*f + 15*sqrt(g*x + f)*f^2)*b*d*e/g^2 + 21*(3*(g*x + f)^(5/2) - 10*(g*x
+ f)^(3/2)*f + 15*sqrt(g*x + f)*f^2)*a*e^2/g^2 + 18*(5*(g*x + f)^(7/2) - 21*(g*x + f)^(5/2)*f + 35*(g*x + f)^
(3/2)*f^2 - 35*sqrt(g*x + f)*f^3)*c*d*e/g^3 + 9*(5*(g*x + f)^(7/2) - 21*(g*x + f)^(5/2)*f + 35*(g*x + f)^(3/2)
*f^2 - 35*sqrt(g*x + f)*f^3)*b*e^2/g^3 + (35*(g*x + f)^(9/2) - 180*(g*x + f)^(7/2)*f + 378*(g*x + f)^(5/2)*f^2
- 420*(g*x + f)^(3/2)*f^3 + 315*sqrt(g*x + f)*f^4)*c*e^2/g^4)/g

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maple [A]  time = 0.01, size = 315, normalized size = 1.49 \begin {gather*} \frac {2 \sqrt {g x +f}\, \left (35 e^{2} c \,x^{4} g^{4}+45 b \,e^{2} g^{4} x^{3}+90 c d e \,g^{4} x^{3}-40 c \,e^{2} f \,g^{3} x^{3}+63 a \,e^{2} g^{4} x^{2}+126 b d e \,g^{4} x^{2}-54 b \,e^{2} f \,g^{3} x^{2}+63 c \,d^{2} g^{4} x^{2}-108 c d e f \,g^{3} x^{2}+48 c \,e^{2} f^{2} g^{2} x^{2}+210 a d e \,g^{4} x -84 a \,e^{2} f \,g^{3} x +105 b \,d^{2} g^{4} x -168 b d e f \,g^{3} x +72 b \,e^{2} f^{2} g^{2} x -84 c \,d^{2} f \,g^{3} x +144 c d e \,f^{2} g^{2} x -64 c \,e^{2} f^{3} g x +315 a \,d^{2} g^{4}-420 a d e f \,g^{3}+168 a \,e^{2} f^{2} g^{2}-210 b \,d^{2} f \,g^{3}+336 b d e \,f^{2} g^{2}-144 b \,e^{2} f^{3} g +168 c \,d^{2} f^{2} g^{2}-288 c d e \,f^{3} g +128 c \,e^{2} f^{4}\right )}{315 g^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(g*x+f)^(1/2)*(35*c*e^2*g^4*x^4+45*b*e^2*g^4*x^3+90*c*d*e*g^4*x^3-40*c*e^2*f*g^3*x^3+63*a*e^2*g^4*x^2+12
6*b*d*e*g^4*x^2-54*b*e^2*f*g^3*x^2+63*c*d^2*g^4*x^2-108*c*d*e*f*g^3*x^2+48*c*e^2*f^2*g^2*x^2+210*a*d*e*g^4*x-8
4*a*e^2*f*g^3*x+105*b*d^2*g^4*x-168*b*d*e*f*g^3*x+72*b*e^2*f^2*g^2*x-84*c*d^2*f*g^3*x+144*c*d*e*f^2*g^2*x-64*c
*e^2*f^3*g*x+315*a*d^2*g^4-420*a*d*e*f*g^3+168*a*e^2*f^2*g^2-210*b*d^2*f*g^3+336*b*d*e*f^2*g^2-144*b*e^2*f^3*g
+168*c*d^2*f^2*g^2-288*c*d*e*f^3*g+128*c*e^2*f^4)/g^5

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maxima [A]  time = 0.45, size = 261, normalized size = 1.23 \begin {gather*} \frac {2 \, {\left (35 \, {\left (g x + f\right )}^{\frac {9}{2}} c e^{2} - 45 \, {\left (4 \, c e^{2} f - {\left (2 \, c d e + b e^{2}\right )} g\right )} {\left (g x + f\right )}^{\frac {7}{2}} + 63 \, {\left (6 \, c e^{2} f^{2} - 3 \, {\left (2 \, c d e + b e^{2}\right )} f g + {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} g^{2}\right )} {\left (g x + f\right )}^{\frac {5}{2}} - 105 \, {\left (4 \, c e^{2} f^{3} - 3 \, {\left (2 \, c d e + b e^{2}\right )} f^{2} g + 2 \, {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} f g^{2} - {\left (b d^{2} + 2 \, a d e\right )} g^{3}\right )} {\left (g x + f\right )}^{\frac {3}{2}} + 315 \, {\left (c e^{2} f^{4} + a d^{2} g^{4} - {\left (2 \, c d e + b e^{2}\right )} f^{3} g + {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} f^{2} g^{2} - {\left (b d^{2} + 2 \, a d e\right )} f g^{3}\right )} \sqrt {g x + f}\right )}}{315 \, g^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*(g*x + f)^(9/2)*c*e^2 - 45*(4*c*e^2*f - (2*c*d*e + b*e^2)*g)*(g*x + f)^(7/2) + 63*(6*c*e^2*f^2 - 3*(
2*c*d*e + b*e^2)*f*g + (c*d^2 + 2*b*d*e + a*e^2)*g^2)*(g*x + f)^(5/2) - 105*(4*c*e^2*f^3 - 3*(2*c*d*e + b*e^2)
*f^2*g + 2*(c*d^2 + 2*b*d*e + a*e^2)*f*g^2 - (b*d^2 + 2*a*d*e)*g^3)*(g*x + f)^(3/2) + 315*(c*e^2*f^4 + a*d^2*g
^4 - (2*c*d*e + b*e^2)*f^3*g + (c*d^2 + 2*b*d*e + a*e^2)*f^2*g^2 - (b*d^2 + 2*a*d*e)*f*g^3)*sqrt(g*x + f))/g^5

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mupad [B]  time = 3.17, size = 204, normalized size = 0.96 \begin {gather*} \frac {{\left (f+g\,x\right )}^{7/2}\,\left (2\,b\,e^2\,g-8\,c\,e^2\,f+4\,c\,d\,e\,g\right )}{7\,g^5}+\frac {{\left (f+g\,x\right )}^{5/2}\,\left (2\,c\,d^2\,g^2-12\,c\,d\,e\,f\,g+4\,b\,d\,e\,g^2+12\,c\,e^2\,f^2-6\,b\,e^2\,f\,g+2\,a\,e^2\,g^2\right )}{5\,g^5}+\frac {2\,{\left (f+g\,x\right )}^{3/2}\,\left (d\,g-e\,f\right )\,\left (2\,a\,e\,g^2+b\,d\,g^2+4\,c\,e\,f^2-3\,b\,e\,f\,g-2\,c\,d\,f\,g\right )}{3\,g^5}+\frac {2\,\sqrt {f+g\,x}\,{\left (d\,g-e\,f\right )}^2\,\left (c\,f^2-b\,f\,g+a\,g^2\right )}{g^5}+\frac {2\,c\,e^2\,{\left (f+g\,x\right )}^{9/2}}{9\,g^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((f + g*x)^(7/2)*(2*b*e^2*g - 8*c*e^2*f + 4*c*d*e*g))/(7*g^5) + ((f + g*x)^(5/2)*(2*a*e^2*g^2 + 2*c*d^2*g^2 +
12*c*e^2*f^2 + 4*b*d*e*g^2 - 6*b*e^2*f*g - 12*c*d*e*f*g))/(5*g^5) + (2*(f + g*x)^(3/2)*(d*g - e*f)*(2*a*e*g^2
+ b*d*g^2 + 4*c*e*f^2 - 3*b*e*f*g - 2*c*d*f*g))/(3*g^5) + (2*(f + g*x)^(1/2)*(d*g - e*f)^2*(a*g^2 + c*f^2 - b*
f*g))/g^5 + (2*c*e^2*(f + g*x)^(9/2))/(9*g^5)

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sympy [A]  time = 105.54, size = 1001, normalized size = 4.72

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-2*a*d**2*f/sqrt(f + g*x) - 2*a*d**2*(-f/sqrt(f + g*x) - sqrt(f + g*x)) - 4*a*d*e*f*(-f/sqrt(f + g
*x) - sqrt(f + g*x))/g - 4*a*d*e*(f**2/sqrt(f + g*x) + 2*f*sqrt(f + g*x) - (f + g*x)**(3/2)/3)/g - 2*a*e**2*f*
(f**2/sqrt(f + g*x) + 2*f*sqrt(f + g*x) - (f + g*x)**(3/2)/3)/g**2 - 2*a*e**2*(-f**3/sqrt(f + g*x) - 3*f**2*sq
rt(f + g*x) + f*(f + g*x)**(3/2) - (f + g*x)**(5/2)/5)/g**2 - 2*b*d**2*f*(-f/sqrt(f + g*x) - sqrt(f + g*x))/g
- 2*b*d**2*(f**2/sqrt(f + g*x) + 2*f*sqrt(f + g*x) - (f + g*x)**(3/2)/3)/g - 4*b*d*e*f*(f**2/sqrt(f + g*x) + 2
*f*sqrt(f + g*x) - (f + g*x)**(3/2)/3)/g**2 - 4*b*d*e*(-f**3/sqrt(f + g*x) - 3*f**2*sqrt(f + g*x) + f*(f + g*x
)**(3/2) - (f + g*x)**(5/2)/5)/g**2 - 2*b*e**2*f*(-f**3/sqrt(f + g*x) - 3*f**2*sqrt(f + g*x) + f*(f + g*x)**(3
/2) - (f + g*x)**(5/2)/5)/g**3 - 2*b*e**2*(f**4/sqrt(f + g*x) + 4*f**3*sqrt(f + g*x) - 2*f**2*(f + g*x)**(3/2)
+ 4*f*(f + g*x)**(5/2)/5 - (f + g*x)**(7/2)/7)/g**3 - 2*c*d**2*f*(f**2/sqrt(f + g*x) + 2*f*sqrt(f + g*x) - (f
+ g*x)**(3/2)/3)/g**2 - 2*c*d**2*(-f**3/sqrt(f + g*x) - 3*f**2*sqrt(f + g*x) + f*(f + g*x)**(3/2) - (f + g*x)
**(5/2)/5)/g**2 - 4*c*d*e*f*(-f**3/sqrt(f + g*x) - 3*f**2*sqrt(f + g*x) + f*(f + g*x)**(3/2) - (f + g*x)**(5/2
)/5)/g**3 - 4*c*d*e*(f**4/sqrt(f + g*x) + 4*f**3*sqrt(f + g*x) - 2*f**2*(f + g*x)**(3/2) + 4*f*(f + g*x)**(5/2
)/5 - (f + g*x)**(7/2)/7)/g**3 - 2*c*e**2*f*(f**4/sqrt(f + g*x) + 4*f**3*sqrt(f + g*x) - 2*f**2*(f + g*x)**(3/
2) + 4*f*(f + g*x)**(5/2)/5 - (f + g*x)**(7/2)/7)/g**4 - 2*c*e**2*(-f**5/sqrt(f + g*x) - 5*f**4*sqrt(f + g*x)
+ 10*f**3*(f + g*x)**(3/2)/3 - 2*f**2*(f + g*x)**(5/2) + 5*f*(f + g*x)**(7/2)/7 - (f + g*x)**(9/2)/9)/g**4)/g,
Ne(g, 0)), ((a*d**2*x + c*e**2*x**5/5 + x**4*(b*e**2 + 2*c*d*e)/4 + x**3*(a*e**2 + 2*b*d*e + c*d**2)/3 + x**2
*(2*a*d*e + b*d**2)/2)/sqrt(f), True))

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