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

Optimal. Leaf size=248 \[ -\frac {\sqrt {f+g x} \left (a e^2-b d e+c d^2\right )}{2 e (d+e x)^2 (e f-d g)^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (3 e g (5 a e g-b (d g+4 e f))+c \left (-d^2 g^2+8 d e f g+8 e^2 f^2\right )\right )}{4 e^{3/2} (e f-d g)^{7/2}}+\frac {2 \left (a g^2-b f g+c f^2\right )}{\sqrt {f+g x} (e f-d g)^3}+\frac {\sqrt {f+g x} (c d (8 e f-d g)-e (-7 a e g+3 b d g+4 b e f))}{4 e (d+e x) (e f-d g)^3} \]

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Rubi [A]  time = 0.63, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {897, 1259, 456, 453, 208} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (3 e g (5 a e g-b (d g+4 e f))+c \left (-d^2 g^2+8 d e f g+8 e^2 f^2\right )\right )}{4 e^{3/2} (e f-d g)^{7/2}}-\frac {\sqrt {f+g x} \left (a e^2-b d e+c d^2\right )}{2 e (d+e x)^2 (e f-d g)^2}+\frac {2 \left (a g^2-b f g+c f^2\right )}{\sqrt {f+g x} (e f-d g)^3}+\frac {\sqrt {f+g x} (c d (8 e f-d g)-e (-7 a e g+3 b d g+4 b e f))}{4 e (d+e x) (e f-d g)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*f^2 - b*f*g + a*g^2))/((e*f - d*g)^3*Sqrt[f + g*x]) - ((c*d^2 - b*d*e + a*e^2)*Sqrt[f + g*x])/(2*e*(e*f
- d*g)^2*(d + e*x)^2) + ((c*d*(8*e*f - d*g) - e*(4*b*e*f + 3*b*d*g - 7*a*e*g))*Sqrt[f + g*x])/(4*e*(e*f - d*g)
^3*(d + e*x)) - ((c*(8*e^2*f^2 + 8*d*e*f*g - d^2*g^2) + 3*e*g*(5*a*e*g - b*(4*e*f + d*g)))*ArcTanh[(Sqrt[e]*Sq
rt[f + g*x])/Sqrt[e*f - d*g]])/(4*e^(3/2)*(e*f - d*g)^(7/2))

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 453

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

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 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 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, 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}}{x^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 (e f-d g)^2 (d+e x)^2}-\frac {g^3 \operatorname {Subst}\left (\int \frac {\frac {4 e^2 (e f-d g) \left (c f^2-b f g+a g^2\right )}{g^5}-\frac {e \left (3 e (b d-a e) g^2+c \left (4 e^2 f^2-8 d e f g+d^2 g^2\right )\right ) x^2}{g^5}}{x^2 \left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2} \, dx,x,\sqrt {f+g x}\right )}{2 e^2 (e f-d g)^2}\\ &=-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}+\frac {(c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) \sqrt {f+g x}}{4 e (e f-d g)^3 (d+e x)}+\frac {g^3 \operatorname {Subst}\left (\int \frac {\frac {8 e^2 \left (c f^2-b f g+a g^2\right )}{g^4}+\frac {e (c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) x^2}{g^3 (e f-d g)}}{x^2 \left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )} \, dx,x,\sqrt {f+g x}\right )}{4 e^2 (e f-d g)^2}\\ &=\frac {2 \left (c f^2-b f g+a g^2\right )}{(e f-d g)^3 \sqrt {f+g x}}-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}+\frac {(c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) \sqrt {f+g x}}{4 e (e f-d g)^3 (d+e x)}+\frac {\left (c \left (8 e^2 f^2+8 d e f g-d^2 g^2\right )+3 e g (5 a e g-b (4 e f+d g))\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 g (e f-d g)^3}\\ &=\frac {2 \left (c f^2-b f g+a g^2\right )}{(e f-d g)^3 \sqrt {f+g x}}-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}+\frac {(c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) \sqrt {f+g x}}{4 e (e f-d g)^3 (d+e x)}-\frac {\left (c \left (8 e^2 f^2+8 d e f g-d^2 g^2\right )+3 e g (5 a e g-b (4 e f+d g))\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{3/2} (e f-d g)^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.10, size = 290, normalized size = 1.17 \begin {gather*} \frac {1}{4} \left (-\frac {2 \sqrt {f+g x} \left (e (a e-b d)+c d^2\right )}{e (d+e x)^2 (e f-d g)^2}+\frac {g \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) (e (-7 a e g+3 b d g+4 b e f)+c d (d g-8 e f))}{e^{3/2} (e f-d g)^{7/2}}+\frac {8 \left (g (a g-b f)+c f^2\right )}{\sqrt {f+g x} (e f-d g)^3}-\frac {8 \sqrt {e} \left (g (a g-b f)+c f^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{(e f-d g)^{7/2}}-\frac {\sqrt {f+g x} (e (-7 a e g+3 b d g+4 b e f)+c d (d g-8 e f))}{e (d+e x) (e f-d g)^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*(c*f^2 + g*(-(b*f) + a*g)))/((e*f - d*g)^3*Sqrt[f + g*x]) - (2*(c*d^2 + e*(-(b*d) + a*e))*Sqrt[f + g*x])/(
e*(e*f - d*g)^2*(d + e*x)^2) - ((c*d*(-8*e*f + d*g) + e*(4*b*e*f + 3*b*d*g - 7*a*e*g))*Sqrt[f + g*x])/(e*(e*f
- d*g)^3*(d + e*x)) - (8*Sqrt[e]*(c*f^2 + g*(-(b*f) + a*g))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/
(e*f - d*g)^(7/2) + (g*(c*d*(-8*e*f + d*g) + e*(4*b*e*f + 3*b*d*g - 7*a*e*g))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/
Sqrt[e*f - d*g]])/(e^(3/2)*(e*f - d*g)^(7/2)))/4

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IntegrateAlgebraic [B]  time = 1.20, size = 497, normalized size = 2.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x} \sqrt {d g-e f}}{e f-d g}\right ) \left (-15 a e^2 g^2+3 b d e g^2+12 b e^2 f g+c d^2 g^2-8 c d e f g-8 c e^2 f^2\right )}{4 e^{3/2} (e f-d g)^3 \sqrt {d g-e f}}+\frac {8 a d^2 e g^4+25 a d e^2 g^3 (f+g x)-16 a d e^2 f g^3+8 a e^3 f^2 g^2-25 a e^3 f g^2 (f+g x)+15 a e^3 g^2 (f+g x)^2-5 b d^2 e g^3 (f+g x)-8 b d^2 e f g^3+16 b d e^2 f^2 g^2-15 b d e^2 f g^2 (f+g x)-3 b d e^2 g^2 (f+g x)^2-8 b e^3 f^3 g+20 b e^3 f^2 g (f+g x)-12 b e^3 f g (f+g x)^2+c d^3 g^3 (f+g x)+8 c d^2 e f^2 g^2+7 c d^2 e f g^2 (f+g x)-c d^2 e g^2 (f+g x)^2-16 c d e^2 f^3 g+8 c d e^2 f^2 g (f+g x)+8 c d e^2 f g (f+g x)^2+8 c e^3 f^4-16 c e^3 f^3 (f+g x)+8 c e^3 f^2 (f+g x)^2}{4 e \sqrt {f+g x} (e f-d g)^3 (-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*(f + g*x)^(3/2)),x]

[Out]

(8*c*e^3*f^4 - 16*c*d*e^2*f^3*g - 8*b*e^3*f^3*g + 8*c*d^2*e*f^2*g^2 + 16*b*d*e^2*f^2*g^2 + 8*a*e^3*f^2*g^2 - 8
*b*d^2*e*f*g^3 - 16*a*d*e^2*f*g^3 + 8*a*d^2*e*g^4 - 16*c*e^3*f^3*(f + g*x) + 8*c*d*e^2*f^2*g*(f + g*x) + 20*b*
e^3*f^2*g*(f + g*x) + 7*c*d^2*e*f*g^2*(f + g*x) - 15*b*d*e^2*f*g^2*(f + g*x) - 25*a*e^3*f*g^2*(f + g*x) + c*d^
3*g^3*(f + g*x) - 5*b*d^2*e*g^3*(f + g*x) + 25*a*d*e^2*g^3*(f + g*x) + 8*c*e^3*f^2*(f + g*x)^2 + 8*c*d*e^2*f*g
*(f + g*x)^2 - 12*b*e^3*f*g*(f + g*x)^2 - c*d^2*e*g^2*(f + g*x)^2 - 3*b*d*e^2*g^2*(f + g*x)^2 + 15*a*e^3*g^2*(
f + g*x)^2)/(4*e*(e*f - d*g)^3*Sqrt[f + g*x]*(e*f - d*g - e*(f + g*x))^2) + ((-8*c*e^2*f^2 - 8*c*d*e*f*g + 12*
b*e^2*f*g + c*d^2*g^2 + 3*b*d*e*g^2 - 15*a*e^2*g^2)*ArcTan[(Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[f + g*x])/(e*f - d
*g)])/(4*e^(3/2)*(e*f - d*g)^3*Sqrt[-(e*f) + d*g])

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fricas [B]  time = 0.50, size = 1883, normalized size = 7.59

result too large to display

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)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*((8*c*d^2*e^2*f^3 + 4*(2*c*d^3*e - 3*b*d^2*e^2)*f^2*g - (c*d^4 + 3*b*d^3*e - 15*a*d^2*e^2)*f*g^2 + (8*c*
e^4*f^2*g + 4*(2*c*d*e^3 - 3*b*e^4)*f*g^2 - (c*d^2*e^2 + 3*b*d*e^3 - 15*a*e^4)*g^3)*x^3 + (8*c*e^4*f^3 + 12*(2
*c*d*e^3 - b*e^4)*f^2*g + 3*(5*c*d^2*e^2 - 9*b*d*e^3 + 5*a*e^4)*f*g^2 - 2*(c*d^3*e + 3*b*d^2*e^2 - 15*a*d*e^3)
*g^3)*x^2 + (16*c*d*e^3*f^3 + 24*(c*d^2*e^2 - b*d*e^3)*f^2*g + 6*(c*d^3*e - 3*b*d^2*e^2 + 5*a*d*e^3)*f*g^2 - (
c*d^4 + 3*b*d^3*e - 15*a*d^2*e^2)*g^3)*x)*sqrt(e^2*f - d*e*g)*log((e*g*x + 2*e*f - d*g + 2*sqrt(e^2*f - d*e*g)
*sqrt(g*x + f))/(e*x + d)) + 2*(8*a*d^3*e^2*g^3 - 2*(7*c*d^2*e^3 - b*d*e^4 - a*e^5)*f^3 + (13*c*d^3*e^2 + 11*b
*d^2*e^3 - 11*a*d*e^4)*f^2*g + (c*d^4*e - 13*b*d^3*e^2 + a*d^2*e^3)*f*g^2 - (8*c*e^5*f^3 - 12*b*e^5*f^2*g - 3*
(3*c*d^2*e^3 - 3*b*d*e^4 - 5*a*e^5)*f*g^2 + (c*d^3*e^2 + 3*b*d^2*e^3 - 15*a*d*e^4)*g^3)*x^2 - (4*(6*c*d*e^4 -
b*e^5)*f^3 - (19*c*d^2*e^3 + 17*b*d*e^4 - 5*a*e^5)*f^2*g - 4*(c*d^3*e^2 - 4*b*d^2*e^3 - 5*a*d*e^4)*f*g^2 - (c*
d^4*e - 5*b*d^3*e^2 + 25*a*d^2*e^3)*g^3)*x)*sqrt(g*x + f))/(d^2*e^6*f^5 - 4*d^3*e^5*f^4*g + 6*d^4*e^4*f^3*g^2
- 4*d^5*e^3*f^2*g^3 + d^6*e^2*f*g^4 + (e^8*f^4*g - 4*d*e^7*f^3*g^2 + 6*d^2*e^6*f^2*g^3 - 4*d^3*e^5*f*g^4 + d^4
*e^4*g^5)*x^3 + (e^8*f^5 - 2*d*e^7*f^4*g - 2*d^2*e^6*f^3*g^2 + 8*d^3*e^5*f^2*g^3 - 7*d^4*e^4*f*g^4 + 2*d^5*e^3
*g^5)*x^2 + (2*d*e^7*f^5 - 7*d^2*e^6*f^4*g + 8*d^3*e^5*f^3*g^2 - 2*d^4*e^4*f^2*g^3 - 2*d^5*e^3*f*g^4 + d^6*e^2
*g^5)*x), 1/4*((8*c*d^2*e^2*f^3 + 4*(2*c*d^3*e - 3*b*d^2*e^2)*f^2*g - (c*d^4 + 3*b*d^3*e - 15*a*d^2*e^2)*f*g^2
+ (8*c*e^4*f^2*g + 4*(2*c*d*e^3 - 3*b*e^4)*f*g^2 - (c*d^2*e^2 + 3*b*d*e^3 - 15*a*e^4)*g^3)*x^3 + (8*c*e^4*f^3
+ 12*(2*c*d*e^3 - b*e^4)*f^2*g + 3*(5*c*d^2*e^2 - 9*b*d*e^3 + 5*a*e^4)*f*g^2 - 2*(c*d^3*e + 3*b*d^2*e^2 - 15*
a*d*e^3)*g^3)*x^2 + (16*c*d*e^3*f^3 + 24*(c*d^2*e^2 - b*d*e^3)*f^2*g + 6*(c*d^3*e - 3*b*d^2*e^2 + 5*a*d*e^3)*f
*g^2 - (c*d^4 + 3*b*d^3*e - 15*a*d^2*e^2)*g^3)*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)) - (8*a*d^3*e^2*g^3 - 2*(7*c*d^2*e^3 - b*d*e^4 - a*e^5)*f^3 + (13*c*d^3*e^2 + 11*b*d^2*e^3 -
11*a*d*e^4)*f^2*g + (c*d^4*e - 13*b*d^3*e^2 + a*d^2*e^3)*f*g^2 - (8*c*e^5*f^3 - 12*b*e^5*f^2*g - 3*(3*c*d^2*e^
3 - 3*b*d*e^4 - 5*a*e^5)*f*g^2 + (c*d^3*e^2 + 3*b*d^2*e^3 - 15*a*d*e^4)*g^3)*x^2 - (4*(6*c*d*e^4 - b*e^5)*f^3
- (19*c*d^2*e^3 + 17*b*d*e^4 - 5*a*e^5)*f^2*g - 4*(c*d^3*e^2 - 4*b*d^2*e^3 - 5*a*d*e^4)*f*g^2 - (c*d^4*e - 5*b
*d^3*e^2 + 25*a*d^2*e^3)*g^3)*x)*sqrt(g*x + f))/(d^2*e^6*f^5 - 4*d^3*e^5*f^4*g + 6*d^4*e^4*f^3*g^2 - 4*d^5*e^3
*f^2*g^3 + d^6*e^2*f*g^4 + (e^8*f^4*g - 4*d*e^7*f^3*g^2 + 6*d^2*e^6*f^2*g^3 - 4*d^3*e^5*f*g^4 + d^4*e^4*g^5)*x
^3 + (e^8*f^5 - 2*d*e^7*f^4*g - 2*d^2*e^6*f^3*g^2 + 8*d^3*e^5*f^2*g^3 - 7*d^4*e^4*f*g^4 + 2*d^5*e^3*g^5)*x^2 +
(2*d*e^7*f^5 - 7*d^2*e^6*f^4*g + 8*d^3*e^5*f^3*g^2 - 2*d^4*e^4*f^2*g^3 - 2*d^5*e^3*f*g^4 + d^6*e^2*g^5)*x)]

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giac [B]  time = 0.26, size = 462, normalized size = 1.86 \begin {gather*} \frac {{\left (c d^{2} g^{2} - 8 \, c d f g e + 3 \, b d g^{2} e - 8 \, c f^{2} e^{2} + 12 \, b f g e^{2} - 15 \, 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^{3} g^{3} e - 3 \, d^{2} f g^{2} e^{2} + 3 \, d f^{2} g e^{3} - f^{3} e^{4}\right )} \sqrt {d g e - f e^{2}}} - \frac {2 \, {\left (c f^{2} - b f g + a g^{2}\right )}}{{\left (d^{3} g^{3} - 3 \, d^{2} f g^{2} e + 3 \, d f^{2} g e^{2} - f^{3} e^{3}\right )} \sqrt {g x + f}} - \frac {\sqrt {g x + f} c d^{3} g^{3} - {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} g^{2} e + 7 \, \sqrt {g x + f} c d^{2} f g^{2} e - 5 \, \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} - 3 \, {\left (g x + f\right )}^{\frac {3}{2}} b d g^{2} e^{2} + \sqrt {g x + f} b d f g^{2} e^{2} + 9 \, \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} + 7 \, {\left (g x + f\right )}^{\frac {3}{2}} a g^{2} e^{3} - 9 \, \sqrt {g x + f} a f g^{2} e^{3}}{4 \, {\left (d^{3} g^{3} e - 3 \, d^{2} f g^{2} e^{2} + 3 \, d f^{2} g e^{3} - f^{3} 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)^(3/2),x, algorithm="giac")

[Out]

1/4*(c*d^2*g^2 - 8*c*d*f*g*e + 3*b*d*g^2*e - 8*c*f^2*e^2 + 12*b*f*g*e^2 - 15*a*g^2*e^2)*arctan(sqrt(g*x + f)*e
/sqrt(d*g*e - f*e^2))/((d^3*g^3*e - 3*d^2*f*g^2*e^2 + 3*d*f^2*g*e^3 - f^3*e^4)*sqrt(d*g*e - f*e^2)) - 2*(c*f^2
- b*f*g + a*g^2)/((d^3*g^3 - 3*d^2*f*g^2*e + 3*d*f^2*g*e^2 - f^3*e^3)*sqrt(g*x + f)) - 1/4*(sqrt(g*x + f)*c*d
^3*g^3 - (g*x + f)^(3/2)*c*d^2*g^2*e + 7*sqrt(g*x + f)*c*d^2*f*g^2*e - 5*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 - 3*(g*x + f)^(3/2)*b*d*g^2*e^2 + sqrt(g*x + f)*b*d*f*g^2
*e^2 + 9*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 + 7*(g*x + f)^(
3/2)*a*g^2*e^3 - 9*sqrt(g*x + f)*a*f*g^2*e^3)/((d^3*g^3*e - 3*d^2*f*g^2*e^2 + 3*d*f^2*g*e^3 - f^3*e^4)*(d*g +
(g*x + f)*e - f*e)^2)

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maple [B]  time = 0.03, size = 847, normalized size = 3.42 \begin {gather*} -\frac {9 \sqrt {g x +f}\, a d e \,g^{3}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}+\frac {9 \sqrt {g x +f}\, a \,e^{2} f \,g^{2}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}+\frac {5 \sqrt {g x +f}\, b \,d^{2} g^{3}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}-\frac {\sqrt {g x +f}\, b d e f \,g^{2}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}-\frac {\sqrt {g x +f}\, b \,e^{2} f^{2} g}{\left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}-\frac {\sqrt {g x +f}\, c \,d^{3} g^{3}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2} e}-\frac {7 \sqrt {g x +f}\, c \,d^{2} f \,g^{2}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}+\frac {2 \sqrt {g x +f}\, c d e \,f^{2} g}{\left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}-\frac {7 \left (g x +f \right )^{\frac {3}{2}} a \,e^{2} g^{2}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}-\frac {15 a e \,g^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d g -e f \right )^{3} \sqrt {\left (d g -e f \right ) e}}+\frac {3 \left (g x +f \right )^{\frac {3}{2}} b d e \,g^{2}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}+\frac {3 b d \,g^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d g -e f \right )^{3} \sqrt {\left (d g -e f \right ) e}}+\frac {\left (g x +f \right )^{\frac {3}{2}} b \,e^{2} f g}{\left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}+\frac {3 b e f g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{3} \sqrt {\left (d g -e f \right ) e}}+\frac {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 g -e f \right )^{3} \sqrt {\left (d g -e f \right ) e}\, e}+\frac {\left (g x +f \right )^{\frac {3}{2}} c \,d^{2} g^{2}}{4 \left (d g -e f \right )^{3} \left (e g x +d g \right )^{2}}-\frac {2 \left (g x +f \right )^{\frac {3}{2}} c d e f g}{\left (d g -e f \right )^{3} \left (e g x +d g \right )^{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 g -e f \right )^{3} \sqrt {\left (d g -e f \right ) e}}-\frac {2 c e \,f^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{3} \sqrt {\left (d g -e f \right ) e}}-\frac {2 a \,g^{2}}{\left (d g -e f \right )^{3} \sqrt {g x +f}}+\frac {2 b f g}{\left (d g -e f \right )^{3} \sqrt {g x +f}}-\frac {2 c \,f^{2}}{\left (d g -e f \right )^{3} \sqrt {g x +f}} \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)^(3/2),x)

[Out]

-7/4/(d*g-e*f)^3/(e*g*x+d*g)^2*(g*x+f)^(3/2)*a*e^2*g^2+3/4/(d*g-e*f)^3/(e*g*x+d*g)^2*(g*x+f)^(3/2)*b*d*e*g^2+1
/(d*g-e*f)^3/(e*g*x+d*g)^2*(g*x+f)^(3/2)*b*e^2*f*g+1/4/(d*g-e*f)^3/(e*g*x+d*g)^2*(g*x+f)^(3/2)*c*d^2*g^2-2/(d*
g-e*f)^3/(e*g*x+d*g)^2*(g*x+f)^(3/2)*c*d*e*f*g-9/4/(d*g-e*f)^3/(e*g*x+d*g)^2*g^3*e*(g*x+f)^(1/2)*a*d+9/4/(d*g-
e*f)^3/(e*g*x+d*g)^2*g^2*e^2*(g*x+f)^(1/2)*a*f+5/4/(d*g-e*f)^3/(e*g*x+d*g)^2*g^3*(g*x+f)^(1/2)*b*d^2-1/4/(d*g-
e*f)^3/(e*g*x+d*g)^2*g^2*e*(g*x+f)^(1/2)*f*d*b-1/(d*g-e*f)^3/(e*g*x+d*g)^2*g*e^2*(g*x+f)^(1/2)*b*f^2-1/4/(d*g-
e*f)^3/(e*g*x+d*g)^2*g^3/e*(g*x+f)^(1/2)*c*d^3-7/4/(d*g-e*f)^3/(e*g*x+d*g)^2*g^2*(g*x+f)^(1/2)*c*d^2*f+2/(d*g-
e*f)^3/(e*g*x+d*g)^2*g*e*(g*x+f)^(1/2)*c*d*f^2-15/4/(d*g-e*f)^3*e/((d*g-e*f)*e)^(1/2)*arctan((g*x+f)^(1/2)/((d
*g-e*f)*e)^(1/2)*e)*a*g^2+3/4/(d*g-e*f)^3/((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+3/(d*g-e*f)^3*e/((d*g-e*f)*e)^(1/2)*arctan((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*b*f*g+1/4/(d*g-e*f)^3/e/((
d*g-e*f)*e)^(1/2)*arctan((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*c*d^2*g^2-2/(d*g-e*f)^3/((d*g-e*f)*e)^(1/2)*arct
an((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*c*d*f*g-2/(d*g-e*f)^3*e/((d*g-e*f)*e)^(1/2)*arctan((g*x+f)^(1/2)/((d*g
-e*f)*e)^(1/2)*e)*c*f^2-2/(d*g-e*f)^3/(g*x+f)^(1/2)*a*g^2+2/(d*g-e*f)^3/(g*x+f)^(1/2)*b*f*g-2/(d*g-e*f)^3/(g*x
+f)^(1/2)*c*f^2

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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)^(3/2),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 = 3.41, size = 363, normalized size = 1.46 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {f+g\,x}\,\left (-d^3\,e\,g^3+3\,d^2\,e^2\,f\,g^2-3\,d\,e^3\,f^2\,g+e^4\,f^3\right )}{\sqrt {e}\,{\left (d\,g-e\,f\right )}^{7/2}}\right )\,\left (-c\,d^2\,g^2+8\,c\,d\,e\,f\,g-3\,b\,d\,e\,g^2+8\,c\,e^2\,f^2-12\,b\,e^2\,f\,g+15\,a\,e^2\,g^2\right )}{4\,e^{3/2}\,{\left (d\,g-e\,f\right )}^{7/2}}-\frac {\frac {2\,\left (c\,f^2-b\,f\,g+a\,g^2\right )}{d\,g-e\,f}+\frac {{\left (f+g\,x\right )}^2\,\left (-c\,d^2\,g^2+8\,c\,d\,e\,f\,g-3\,b\,d\,e\,g^2+8\,c\,e^2\,f^2-12\,b\,e^2\,f\,g+15\,a\,e^2\,g^2\right )}{4\,{\left (d\,g-e\,f\right )}^3}+\frac {\left (f+g\,x\right )\,\left (c\,d^2\,g^2+8\,c\,d\,e\,f\,g-5\,b\,d\,e\,g^2+16\,c\,e^2\,f^2-20\,b\,e^2\,f\,g+25\,a\,e^2\,g^2\right )}{4\,e\,{\left (d\,g-e\,f\right )}^2}}{e^2\,{\left (f+g\,x\right )}^{5/2}-{\left (f+g\,x\right )}^{3/2}\,\left (2\,e^2\,f-2\,d\,e\,g\right )+\sqrt {f+g\,x}\,\left (d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan(((f + g*x)^(1/2)*(e^4*f^3 - d^3*e*g^3 + 3*d^2*e^2*f*g^2 - 3*d*e^3*f^2*g))/(e^(1/2)*(d*g - e*f)^(7/2)))*(
15*a*e^2*g^2 - c*d^2*g^2 + 8*c*e^2*f^2 - 3*b*d*e*g^2 - 12*b*e^2*f*g + 8*c*d*e*f*g))/(4*e^(3/2)*(d*g - e*f)^(7/
2)) - ((2*(a*g^2 + c*f^2 - b*f*g))/(d*g - e*f) + ((f + g*x)^2*(15*a*e^2*g^2 - c*d^2*g^2 + 8*c*e^2*f^2 - 3*b*d*
e*g^2 - 12*b*e^2*f*g + 8*c*d*e*f*g))/(4*(d*g - e*f)^3) + ((f + g*x)*(25*a*e^2*g^2 + c*d^2*g^2 + 16*c*e^2*f^2 -
5*b*d*e*g^2 - 20*b*e^2*f*g + 8*c*d*e*f*g))/(4*e*(d*g - e*f)^2))/(e^2*(f + g*x)^(5/2) - (f + g*x)^(3/2)*(2*e^2
*f - 2*d*e*g) + (f + g*x)^(1/2)*(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)**(3/2),x)

[Out]

Timed out

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