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

Optimal. Leaf size=206 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (e g (-3 a e g-b d g+4 b e f)-c \left (3 d^2 g^2-8 d e f g+8 e^2 f^2\right )\right )}{4 e^{5/2} (e f-d g)^{5/2}}-\frac {\sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{2 (d+e x)^2 (e f-d g)}+\frac {\sqrt {f+g x} (c d (8 e f-5 d g)-e (-3 a e g-b d g+4 b e f))}{4 e^2 (d+e x) (e f-d g)^2} \]

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Rubi [A]  time = 0.39, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {897, 1157, 385, 208} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (e g (-3 a e g-b d g+4 b e f)-c \left (3 d^2 g^2-8 d e f g+8 e^2 f^2\right )\right )}{4 e^{5/2} (e f-d g)^{5/2}}-\frac {\sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{2 (d+e x)^2 (e f-d g)}+\frac {\sqrt {f+g x} (c d (8 e f-5 d g)-e (-3 a e g-b d g+4 b e f))}{4 e^2 (d+e x) (e f-d g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + (d*(c*d - b*e))/e^2)*Sqrt[f + g*x])/(2*(e*f - d*g)*(d + e*x)^2) + ((c*d*(8*e*f - 5*d*g) - e*(4*b*e*f -
b*d*g - 3*a*e*g))*Sqrt[f + g*x])/(4*e^2*(e*f - d*g)^2*(d + e*x)) + ((e*g*(4*b*e*f - b*d*g - 3*a*e*g) - c*(8*e^
2*f^2 - 8*d*e*f*g + 3*d^2*g^2))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(4*e^(5/2)*(e*f - d*g)^(5/2)
)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{(d+e x)^3 \sqrt {f+g x}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\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}}{\left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^3} \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e^2 (e f-d g) (d+e x)^2}+\frac {\operatorname {Subst}\left (\int \frac {-3 a+\frac {c d^2}{e^2}-\frac {b d}{e}-\frac {4 c f^2}{g^2}+\frac {4 b f}{g}+\frac {4 c (e f-d g) x^2}{e g^2}}{\left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2} \, dx,x,\sqrt {f+g x}\right )}{2 (e f-d g)}\\ &=-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e^2 (e f-d g) (d+e x)^2}+\frac {(c d (8 e f-5 d g)-e (4 b e f-b d g-3 a e g)) \sqrt {f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}-\frac {\left (e g (4 b e f-b d g-3 a e g)-c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {-e f+d g}{g}+\frac {e x^2}{g}} \, dx,x,\sqrt {f+g x}\right )}{4 e^2 g (e f-d g)^2}\\ &=-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e^2 (e f-d g) (d+e x)^2}+\frac {(c d (8 e f-5 d g)-e (4 b e f-b d g-3 a e g)) \sqrt {f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}+\frac {\left (e g (4 b e f-b d g-3 a e g)-c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.65, size = 297, normalized size = 1.44 \begin {gather*} \frac {-\frac {2 \sqrt {e} \sqrt {f+g x} \left (e (a e-b d)+c d^2\right )}{(d+e x)^2 (e f-d g)}-\frac {3 g \left (e (a e-b d)+c d^2\right ) \left (g (d+e x) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )-\sqrt {e} \sqrt {f+g x} \sqrt {e f-d g}\right )}{(d+e x) (e f-d g)^{5/2}}-\frac {4 \sqrt {e} \sqrt {f+g x} (b e-2 c d)}{(d+e x) (e f-d g)}-\frac {4 g (2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{(e f-d g)^{3/2}}-\frac {8 c \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e f-d g}}}{4 e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[e]*(c*d^2 + e*(-(b*d) + a*e))*Sqrt[f + g*x])/((e*f - d*g)*(d + e*x)^2) - (4*Sqrt[e]*(-2*c*d + b*e)*S
qrt[f + g*x])/((e*f - d*g)*(d + e*x)) - (4*(2*c*d - b*e)*g*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(
e*f - d*g)^(3/2) - (8*c*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/Sqrt[e*f - d*g] - (3*(c*d^2 + e*(-(b
*d) + a*e))*g*(-(Sqrt[e]*Sqrt[e*f - d*g]*Sqrt[f + g*x]) + g*(d + e*x)*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f
- d*g]]))/((e*f - d*g)^(5/2)*(d + e*x)))/(4*e^(5/2))

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IntegrateAlgebraic [A]  time = 0.90, size = 293, normalized size = 1.42 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x} \sqrt {d g-e f}}{e f-d g}\right ) \left (-3 a e^2 g^2-b d e g^2+4 b e^2 f g-3 c d^2 g^2+8 c d e f g-8 c e^2 f^2\right )}{4 e^{5/2} (d g-e f)^{5/2}}-\frac {g \sqrt {f+g x} \left (-5 a d e^2 g^2-3 a e^3 g (f+g x)+5 a e^3 f g+b d^2 e g^2-b d e^2 g (f+g x)+3 b d e^2 f g-4 b e^3 f^2+4 b e^3 f (f+g x)+3 c d^3 g^2+5 c d^2 e g (f+g x)-11 c d^2 e f g+8 c d e^2 f^2-8 c d e^2 f (f+g x)\right )}{4 e^2 (e f-d g)^2 (-d g-e (f+g x)+e f)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(g*Sqrt[f + g*x]*(8*c*d*e^2*f^2 - 4*b*e^3*f^2 - 11*c*d^2*e*f*g + 3*b*d*e^2*f*g + 5*a*e^3*f*g + 3*c*d^3*g^
2 + b*d^2*e*g^2 - 5*a*d*e^2*g^2 - 8*c*d*e^2*f*(f + g*x) + 4*b*e^3*f*(f + g*x) + 5*c*d^2*e*g*(f + g*x) - b*d*e^
2*g*(f + g*x) - 3*a*e^3*g*(f + g*x)))/(e^2*(e*f - d*g)^2*(e*f - d*g - e*(f + g*x))^2) + ((-8*c*e^2*f^2 + 8*c*d
*e*f*g + 4*b*e^2*f*g - 3*c*d^2*g^2 - b*d*e*g^2 - 3*a*e^2*g^2)*ArcTan[(Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[f + g*x]
)/(e*f - d*g)])/(4*e^(5/2)*(-(e*f) + d*g)^(5/2))

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fricas [B]  time = 0.66, size = 1096, normalized size = 5.32 \begin {gather*} \left [\frac {{\left (8 \, c d^{2} e^{2} f^{2} - 4 \, {\left (2 \, c d^{3} e + b d^{2} e^{2}\right )} f g + {\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2}\right )} g^{2} + {\left (8 \, c e^{4} f^{2} - 4 \, {\left (2 \, c d e^{3} + b e^{4}\right )} f g + {\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} g^{2}\right )} x^{2} + 2 \, {\left (8 \, c d e^{3} f^{2} - 4 \, {\left (2 \, c d^{2} e^{2} + b d e^{3}\right )} f g + {\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} g^{2}\right )} x\right )} \sqrt {e^{2} f - d e g} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {e^{2} f - d e g} \sqrt {g x + f}}{e x + d}\right ) + 2 \, {\left (2 \, {\left (3 \, c d^{2} e^{3} - b d e^{4} - a e^{5}\right )} f^{2} - {\left (9 \, c d^{3} e^{2} - b d^{2} e^{3} - 7 \, a d e^{4}\right )} f g + {\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} g^{2} + {\left (4 \, {\left (2 \, c d e^{4} - b e^{5}\right )} f^{2} - {\left (13 \, c d^{2} e^{3} - 5 \, b d e^{4} - 3 \, a e^{5}\right )} f g + {\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} g^{2}\right )} x\right )} \sqrt {g x + f}}{8 \, {\left (d^{2} e^{6} f^{3} - 3 \, d^{3} e^{5} f^{2} g + 3 \, d^{4} e^{4} f g^{2} - d^{5} e^{3} g^{3} + {\left (e^{8} f^{3} - 3 \, d e^{7} f^{2} g + 3 \, d^{2} e^{6} f g^{2} - d^{3} e^{5} g^{3}\right )} x^{2} + 2 \, {\left (d e^{7} f^{3} - 3 \, d^{2} e^{6} f^{2} g + 3 \, d^{3} e^{5} f g^{2} - d^{4} e^{4} g^{3}\right )} x\right )}}, \frac {{\left (8 \, c d^{2} e^{2} f^{2} - 4 \, {\left (2 \, c d^{3} e + b d^{2} e^{2}\right )} f g + {\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2}\right )} g^{2} + {\left (8 \, c e^{4} f^{2} - 4 \, {\left (2 \, c d e^{3} + b e^{4}\right )} f g + {\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} g^{2}\right )} x^{2} + 2 \, {\left (8 \, c d e^{3} f^{2} - 4 \, {\left (2 \, c d^{2} e^{2} + b d e^{3}\right )} f g + {\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} g^{2}\right )} x\right )} \sqrt {-e^{2} f + d e g} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) + {\left (2 \, {\left (3 \, c d^{2} e^{3} - b d e^{4} - a e^{5}\right )} f^{2} - {\left (9 \, c d^{3} e^{2} - b d^{2} e^{3} - 7 \, a d e^{4}\right )} f g + {\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} g^{2} + {\left (4 \, {\left (2 \, c d e^{4} - b e^{5}\right )} f^{2} - {\left (13 \, c d^{2} e^{3} - 5 \, b d e^{4} - 3 \, a e^{5}\right )} f g + {\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} g^{2}\right )} x\right )} \sqrt {g x + f}}{4 \, {\left (d^{2} e^{6} f^{3} - 3 \, d^{3} e^{5} f^{2} g + 3 \, d^{4} e^{4} f g^{2} - d^{5} e^{3} g^{3} + {\left (e^{8} f^{3} - 3 \, d e^{7} f^{2} g + 3 \, d^{2} e^{6} f g^{2} - d^{3} e^{5} g^{3}\right )} x^{2} + 2 \, {\left (d e^{7} f^{3} - 3 \, d^{2} e^{6} f^{2} g + 3 \, d^{3} e^{5} f g^{2} - d^{4} e^{4} g^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((8*c*d^2*e^2*f^2 - 4*(2*c*d^3*e + b*d^2*e^2)*f*g + (3*c*d^4 + b*d^3*e + 3*a*d^2*e^2)*g^2 + (8*c*e^4*f^2
- 4*(2*c*d*e^3 + b*e^4)*f*g + (3*c*d^2*e^2 + b*d*e^3 + 3*a*e^4)*g^2)*x^2 + 2*(8*c*d*e^3*f^2 - 4*(2*c*d^2*e^2 +
b*d*e^3)*f*g + (3*c*d^3*e + b*d^2*e^2 + 3*a*d*e^3)*g^2)*x)*sqrt(e^2*f - d*e*g)*log((e*g*x + 2*e*f - d*g - 2*s
qrt(e^2*f - d*e*g)*sqrt(g*x + f))/(e*x + d)) + 2*(2*(3*c*d^2*e^3 - b*d*e^4 - a*e^5)*f^2 - (9*c*d^3*e^2 - b*d^2
*e^3 - 7*a*d*e^4)*f*g + (3*c*d^4*e + b*d^3*e^2 - 5*a*d^2*e^3)*g^2 + (4*(2*c*d*e^4 - b*e^5)*f^2 - (13*c*d^2*e^3
- 5*b*d*e^4 - 3*a*e^5)*f*g + (5*c*d^3*e^2 - b*d^2*e^3 - 3*a*d*e^4)*g^2)*x)*sqrt(g*x + f))/(d^2*e^6*f^3 - 3*d^
3*e^5*f^2*g + 3*d^4*e^4*f*g^2 - d^5*e^3*g^3 + (e^8*f^3 - 3*d*e^7*f^2*g + 3*d^2*e^6*f*g^2 - d^3*e^5*g^3)*x^2 +
2*(d*e^7*f^3 - 3*d^2*e^6*f^2*g + 3*d^3*e^5*f*g^2 - d^4*e^4*g^3)*x), 1/4*((8*c*d^2*e^2*f^2 - 4*(2*c*d^3*e + b*d
^2*e^2)*f*g + (3*c*d^4 + b*d^3*e + 3*a*d^2*e^2)*g^2 + (8*c*e^4*f^2 - 4*(2*c*d*e^3 + b*e^4)*f*g + (3*c*d^2*e^2
+ b*d*e^3 + 3*a*e^4)*g^2)*x^2 + 2*(8*c*d*e^3*f^2 - 4*(2*c*d^2*e^2 + b*d*e^3)*f*g + (3*c*d^3*e + b*d^2*e^2 + 3*
a*d*e^3)*g^2)*x)*sqrt(-e^2*f + d*e*g)*arctan(sqrt(-e^2*f + d*e*g)*sqrt(g*x + f)/(e*g*x + e*f)) + (2*(3*c*d^2*e
^3 - b*d*e^4 - a*e^5)*f^2 - (9*c*d^3*e^2 - b*d^2*e^3 - 7*a*d*e^4)*f*g + (3*c*d^4*e + b*d^3*e^2 - 5*a*d^2*e^3)*
g^2 + (4*(2*c*d*e^4 - b*e^5)*f^2 - (13*c*d^2*e^3 - 5*b*d*e^4 - 3*a*e^5)*f*g + (5*c*d^3*e^2 - b*d^2*e^3 - 3*a*d
*e^4)*g^2)*x)*sqrt(g*x + f))/(d^2*e^6*f^3 - 3*d^3*e^5*f^2*g + 3*d^4*e^4*f*g^2 - d^5*e^3*g^3 + (e^8*f^3 - 3*d*e
^7*f^2*g + 3*d^2*e^6*f*g^2 - d^3*e^5*g^3)*x^2 + 2*(d*e^7*f^3 - 3*d^2*e^6*f^2*g + 3*d^3*e^5*f*g^2 - d^4*e^4*g^3
)*x)]

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giac [B]  time = 0.20, size = 373, normalized size = 1.81 \begin {gather*} \frac {{\left (3 \, c d^{2} g^{2} - 8 \, c d f g e + b d g^{2} e + 8 \, c f^{2} e^{2} - 4 \, b f g e^{2} + 3 \, a g^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{4 \, {\left (d^{2} g^{2} e^{2} - 2 \, d f g e^{3} + f^{2} e^{4}\right )} \sqrt {d g e - f e^{2}}} - \frac {3 \, \sqrt {g x + f} c d^{3} g^{3} + 5 \, {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} g^{2} e - 11 \, \sqrt {g x + f} c d^{2} f g^{2} e + \sqrt {g x + f} b d^{2} g^{3} e - 8 \, {\left (g x + f\right )}^{\frac {3}{2}} c d f g e^{2} + 8 \, \sqrt {g x + f} c d f^{2} g e^{2} - {\left (g x + f\right )}^{\frac {3}{2}} b d g^{2} e^{2} + 3 \, \sqrt {g x + f} b d f g^{2} e^{2} - 5 \, \sqrt {g x + f} a d g^{3} e^{2} + 4 \, {\left (g x + f\right )}^{\frac {3}{2}} b f g e^{3} - 4 \, \sqrt {g x + f} b f^{2} g e^{3} - 3 \, {\left (g x + f\right )}^{\frac {3}{2}} a g^{2} e^{3} + 5 \, \sqrt {g x + f} a f g^{2} e^{3}}{4 \, {\left (d^{2} g^{2} e^{2} - 2 \, d f g e^{3} + f^{2} e^{4}\right )} {\left (d g + {\left (g x + f\right )} e - f e\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*c*d^2*g^2 - 8*c*d*f*g*e + b*d*g^2*e + 8*c*f^2*e^2 - 4*b*f*g*e^2 + 3*a*g^2*e^2)*arctan(sqrt(g*x + f)*e/s
qrt(d*g*e - f*e^2))/((d^2*g^2*e^2 - 2*d*f*g*e^3 + f^2*e^4)*sqrt(d*g*e - f*e^2)) - 1/4*(3*sqrt(g*x + f)*c*d^3*g
^3 + 5*(g*x + f)^(3/2)*c*d^2*g^2*e - 11*sqrt(g*x + f)*c*d^2*f*g^2*e + sqrt(g*x + f)*b*d^2*g^3*e - 8*(g*x + f)^
(3/2)*c*d*f*g*e^2 + 8*sqrt(g*x + f)*c*d*f^2*g*e^2 - (g*x + f)^(3/2)*b*d*g^2*e^2 + 3*sqrt(g*x + f)*b*d*f*g^2*e^
2 - 5*sqrt(g*x + f)*a*d*g^3*e^2 + 4*(g*x + f)^(3/2)*b*f*g*e^3 - 4*sqrt(g*x + f)*b*f^2*g*e^3 - 3*(g*x + f)^(3/2
)*a*g^2*e^3 + 5*sqrt(g*x + f)*a*f*g^2*e^3)/((d^2*g^2*e^2 - 2*d*f*g*e^3 + f^2*e^4)*(d*g + (g*x + f)*e - f*e)^2)

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maple [B]  time = 0.02, size = 538, normalized size = 2.61 \begin {gather*} \frac {3 a \,g^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}}+\frac {b d \,g^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}\, e}-\frac {b f g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}}+\frac {3 c \,d^{2} g^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}\, e^{2}}-\frac {2 c d f g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}\, e}+\frac {2 c \,f^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \sqrt {\left (d g -e f \right ) e}}+\frac {\frac {\left (3 a \,e^{2} g +b d e g -4 b \,e^{2} f -5 c \,d^{2} g +8 d e c f \right ) \left (g x +f \right )^{\frac {3}{2}} g}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) e}+\frac {\left (5 a \,e^{2} g -b d e g -4 b \,e^{2} f -3 c \,d^{2} g +8 d e c f \right ) \sqrt {g x +f}\, g}{4 \left (d g -e f \right ) e^{2}}}{\left (d g -e f +\left (g x +f \right ) e \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f positive or negative?

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mupad [B]  time = 0.28, size = 270, normalized size = 1.31 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (3\,c\,d^2\,g^2-8\,c\,d\,e\,f\,g+b\,d\,e\,g^2+8\,c\,e^2\,f^2-4\,b\,e^2\,f\,g+3\,a\,e^2\,g^2\right )}{4\,e^{5/2}\,{\left (d\,g-e\,f\right )}^{5/2}}-\frac {\frac {\sqrt {f+g\,x}\,\left (3\,c\,d^2\,g^2+b\,d\,e\,g^2-8\,c\,f\,d\,e\,g-5\,a\,e^2\,g^2+4\,b\,f\,e^2\,g\right )}{4\,e^2\,\left (d\,g-e\,f\right )}-\frac {{\left (f+g\,x\right )}^{3/2}\,\left (-5\,c\,d^2\,g^2+b\,d\,e\,g^2+8\,c\,f\,d\,e\,g+3\,a\,e^2\,g^2-4\,b\,f\,e^2\,g\right )}{4\,e\,{\left (d\,g-e\,f\right )}^2}}{e^2\,{\left (f+g\,x\right )}^2-\left (f+g\,x\right )\,\left (2\,e^2\,f-2\,d\,e\,g\right )+d^2\,g^2+e^2\,f^2-2\,d\,e\,f\,g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)/(e*x+d)**3/(g*x+f)**(1/2),x)

[Out]

Timed out

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