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

Optimal. Leaf size=287 \[ -\frac {3 (2 c d-b e) (-5 b e g+6 c d g+4 c e f) \tan ^{-1}\left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{8 c^{7/2} e^2}+\frac {3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-5 b e g+6 c d g+4 c e f)}{4 c^3 e^2}+\frac {(d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-5 b e g+6 c d g+4 c e f)}{2 c^2 e^2 (2 c d-b e)}+\frac {2 (d+e x)^3 (-b e g+c d g+c e f)}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}} \]

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Rubi [A]  time = 0.44, antiderivative size = 287, 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 = {788, 670, 640, 621, 204} \begin {gather*} \frac {(d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-5 b e g+6 c d g+4 c e f)}{2 c^2 e^2 (2 c d-b e)}+\frac {3 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-5 b e g+6 c d g+4 c e f)}{4 c^3 e^2}-\frac {3 (2 c d-b e) (-5 b e g+6 c d g+4 c e f) \tan ^{-1}\left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{8 c^{7/2} e^2}+\frac {2 (d+e x)^3 (-b e g+c d g+c e f)}{c e^2 (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)^3*(f + g*x))/(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2),x]

[Out]

(2*(c*e*f + c*d*g - b*e*g)*(d + e*x)^3)/(c*e^2*(2*c*d - b*e)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]) + (3*(
4*c*e*f + 6*c*d*g - 5*b*e*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(4*c^3*e^2) + ((4*c*e*f + 6*c*d*g - 5*
b*e*g)*(d + e*x)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(2*c^2*e^2*(2*c*d - b*e)) - (3*(2*c*d - b*e)*(4*c*
e*f + 6*c*d*g - 5*b*e*g)*ArcTan[(e*(b + 2*c*x))/(2*Sqrt[c]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])])/(8*c^(
7/2)*e^2)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

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

Rule 670

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[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 788

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^3 (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx &=\frac {2 (c e f+c d g-b e g) (d+e x)^3}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(4 c e f+6 c d g-5 b e g) \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{c e (2 c d-b e)}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^3}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {(4 c e f+6 c d g-5 b e g) (d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 c^2 e^2 (2 c d-b e)}-\frac {(3 (4 c e f+6 c d g-5 b e g)) \int \frac {d+e x}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{4 c^2 e}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^3}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {3 (4 c e f+6 c d g-5 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c^3 e^2}+\frac {(4 c e f+6 c d g-5 b e g) (d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 c^2 e^2 (2 c d-b e)}-\frac {(3 (2 c d-b e) (4 c e f+6 c d g-5 b e g)) \int \frac {1}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{8 c^3 e}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^3}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {3 (4 c e f+6 c d g-5 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c^3 e^2}+\frac {(4 c e f+6 c d g-5 b e g) (d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 c^2 e^2 (2 c d-b e)}-\frac {(3 (2 c d-b e) (4 c e f+6 c d g-5 b e g)) \operatorname {Subst}\left (\int \frac {1}{-4 c e^2-x^2} \, dx,x,\frac {-b e^2-2 c e^2 x}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}\right )}{4 c^3 e}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^3}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {3 (4 c e f+6 c d g-5 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c^3 e^2}+\frac {(4 c e f+6 c d g-5 b e g) (d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 c^2 e^2 (2 c d-b e)}-\frac {3 (2 c d-b e) (4 c e f+6 c d g-5 b e g) \tan ^{-1}\left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{8 c^{7/2} e^2}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 250, normalized size = 0.87 \begin {gather*} \frac {\sqrt {c} \sqrt {e} (d+e x) \sqrt {e (2 c d-b e)} \left (15 b^2 e^2 g+b c e (-43 d g-12 e f+5 e g x)+2 c^2 \left (14 d^2 g+5 d e (2 f-g x)-e^2 x (2 f+g x)\right )\right )-3 e \sqrt {d+e x} (b e-2 c d)^2 \sqrt {\frac {b e-c d+c e x}{b e-2 c d}} (-5 b e g+6 c d g+4 c e f) \sin ^{-1}\left (\frac {\sqrt {c} \sqrt {e} \sqrt {d+e x}}{\sqrt {e (2 c d-b e)}}\right )}{4 c^{7/2} e^{5/2} \sqrt {e (2 c d-b e)} \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*Sqrt[e]*Sqrt[e*(2*c*d - b*e)]*(d + e*x)*(15*b^2*e^2*g + b*c*e*(-12*e*f - 43*d*g + 5*e*g*x) + 2*c^2*(1
4*d^2*g + 5*d*e*(2*f - g*x) - e^2*x*(2*f + g*x))) - 3*e*(-2*c*d + b*e)^2*(4*c*e*f + 6*c*d*g - 5*b*e*g)*Sqrt[d
+ e*x]*Sqrt[(-(c*d) + b*e + c*e*x)/(-2*c*d + b*e)]*ArcSin[(Sqrt[c]*Sqrt[e]*Sqrt[d + e*x])/Sqrt[e*(2*c*d - b*e)
]])/(4*c^(7/2)*e^(5/2)*Sqrt[e*(2*c*d - b*e)]*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])

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IntegrateAlgebraic [F]  time = 180.19, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 1.33, size = 745, normalized size = 2.60 \begin {gather*} \left [\frac {3 \, {\left (4 \, {\left (2 \, c^{3} d^{2} e - 3 \, b c^{2} d e^{2} + b^{2} c e^{3}\right )} f + {\left (12 \, c^{3} d^{3} - 28 \, b c^{2} d^{2} e + 21 \, b^{2} c d e^{2} - 5 \, b^{3} e^{3}\right )} g - {\left (4 \, {\left (2 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} f + {\left (12 \, c^{3} d^{2} e - 16 \, b c^{2} d e^{2} + 5 \, b^{2} c e^{3}\right )} g\right )} x\right )} \sqrt {-c} \log \left (8 \, c^{2} e^{2} x^{2} + 8 \, b c e^{2} x - 4 \, c^{2} d^{2} + 4 \, b c d e + b^{2} e^{2} + 4 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {-c}\right ) + 4 \, {\left (2 \, c^{3} e^{2} g x^{2} - 4 \, {\left (5 \, c^{3} d e - 3 \, b c^{2} e^{2}\right )} f - {\left (28 \, c^{3} d^{2} - 43 \, b c^{2} d e + 15 \, b^{2} c e^{2}\right )} g + {\left (4 \, c^{3} e^{2} f + 5 \, {\left (2 \, c^{3} d e - b c^{2} e^{2}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}{16 \, {\left (c^{5} e^{3} x - c^{5} d e^{2} + b c^{4} e^{3}\right )}}, -\frac {3 \, {\left (4 \, {\left (2 \, c^{3} d^{2} e - 3 \, b c^{2} d e^{2} + b^{2} c e^{3}\right )} f + {\left (12 \, c^{3} d^{3} - 28 \, b c^{2} d^{2} e + 21 \, b^{2} c d e^{2} - 5 \, b^{3} e^{3}\right )} g - {\left (4 \, {\left (2 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} f + {\left (12 \, c^{3} d^{2} e - 16 \, b c^{2} d e^{2} + 5 \, b^{2} c e^{3}\right )} g\right )} x\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {c}}{2 \, {\left (c^{2} e^{2} x^{2} + b c e^{2} x - c^{2} d^{2} + b c d e\right )}}\right ) - 2 \, {\left (2 \, c^{3} e^{2} g x^{2} - 4 \, {\left (5 \, c^{3} d e - 3 \, b c^{2} e^{2}\right )} f - {\left (28 \, c^{3} d^{2} - 43 \, b c^{2} d e + 15 \, b^{2} c e^{2}\right )} g + {\left (4 \, c^{3} e^{2} f + 5 \, {\left (2 \, c^{3} d e - b c^{2} e^{2}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}{8 \, {\left (c^{5} e^{3} x - c^{5} d e^{2} + b c^{4} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(4*(2*c^3*d^2*e - 3*b*c^2*d*e^2 + b^2*c*e^3)*f + (12*c^3*d^3 - 28*b*c^2*d^2*e + 21*b^2*c*d*e^2 - 5*b^
3*e^3)*g - (4*(2*c^3*d*e^2 - b*c^2*e^3)*f + (12*c^3*d^2*e - 16*b*c^2*d*e^2 + 5*b^2*c*e^3)*g)*x)*sqrt(-c)*log(8
*c^2*e^2*x^2 + 8*b*c*e^2*x - 4*c^2*d^2 + 4*b*c*d*e + b^2*e^2 + 4*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2
*c*e*x + b*e)*sqrt(-c)) + 4*(2*c^3*e^2*g*x^2 - 4*(5*c^3*d*e - 3*b*c^2*e^2)*f - (28*c^3*d^2 - 43*b*c^2*d*e + 15
*b^2*c*e^2)*g + (4*c^3*e^2*f + 5*(2*c^3*d*e - b*c^2*e^2)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e))/(c^
5*e^3*x - c^5*d*e^2 + b*c^4*e^3), -1/8*(3*(4*(2*c^3*d^2*e - 3*b*c^2*d*e^2 + b^2*c*e^3)*f + (12*c^3*d^3 - 28*b*
c^2*d^2*e + 21*b^2*c*d*e^2 - 5*b^3*e^3)*g - (4*(2*c^3*d*e^2 - b*c^2*e^3)*f + (12*c^3*d^2*e - 16*b*c^2*d*e^2 +
5*b^2*c*e^3)*g)*x)*sqrt(c)*arctan(1/2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(c)/(c^2*
e^2*x^2 + b*c*e^2*x - c^2*d^2 + b*c*d*e)) - 2*(2*c^3*e^2*g*x^2 - 4*(5*c^3*d*e - 3*b*c^2*e^2)*f - (28*c^3*d^2 -
43*b*c^2*d*e + 15*b^2*c*e^2)*g + (4*c^3*e^2*f + 5*(2*c^3*d*e - b*c^2*e^2)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c
*d^2 - b*d*e))/(c^5*e^3*x - c^5*d*e^2 + b*c^4*e^3)]

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giac [B]  time = 0.60, size = 623, normalized size = 2.17 \begin {gather*} \frac {\sqrt {-c x^{2} e^{2} + c d^{2} - b x e^{2} - b d e} {\left ({\left ({\left (\frac {2 \, {\left (4 \, c^{4} d^{2} g e^{5} - 4 \, b c^{3} d g e^{6} + b^{2} c^{2} g e^{7}\right )} x}{4 \, c^{5} d^{2} e^{4} - 4 \, b c^{4} d e^{5} + b^{2} c^{3} e^{6}} + \frac {48 \, c^{4} d^{3} g e^{4} + 16 \, c^{4} d^{2} f e^{5} - 68 \, b c^{3} d^{2} g e^{5} - 16 \, b c^{3} d f e^{6} + 32 \, b^{2} c^{2} d g e^{6} + 4 \, b^{2} c^{2} f e^{7} - 5 \, b^{3} c g e^{7}}{4 \, c^{5} d^{2} e^{4} - 4 \, b c^{4} d e^{5} + b^{2} c^{3} e^{6}}\right )} x - \frac {72 \, c^{4} d^{4} g e^{3} + 64 \, c^{4} d^{3} f e^{4} - 224 \, b c^{3} d^{3} g e^{4} - 112 \, b c^{3} d^{2} f e^{5} + 230 \, b^{2} c^{2} d^{2} g e^{5} + 64 \, b^{2} c^{2} d f e^{6} - 98 \, b^{3} c d g e^{6} - 12 \, b^{3} c f e^{7} + 15 \, b^{4} g e^{7}}{4 \, c^{5} d^{2} e^{4} - 4 \, b c^{4} d e^{5} + b^{2} c^{3} e^{6}}\right )} x - \frac {112 \, c^{4} d^{5} g e^{2} + 80 \, c^{4} d^{4} f e^{3} - 284 \, b c^{3} d^{4} g e^{3} - 128 \, b c^{3} d^{3} f e^{4} + 260 \, b^{2} c^{2} d^{3} g e^{4} + 68 \, b^{2} c^{2} d^{2} f e^{5} - 103 \, b^{3} c d^{2} g e^{5} - 12 \, b^{3} c d f e^{6} + 15 \, b^{4} d g e^{6}}{4 \, c^{5} d^{2} e^{4} - 4 \, b c^{4} d e^{5} + b^{2} c^{3} e^{6}}\right )}}{4 \, {\left (c x^{2} e^{2} - c d^{2} + b x e^{2} + b d e\right )}} - \frac {3 \, {\left (12 \, c^{2} d^{2} g + 8 \, c^{2} d f e - 16 \, b c d g e - 4 \, b c f e^{2} + 5 \, b^{2} g e^{2}\right )} \sqrt {-c e^{2}} e^{\left (-3\right )} \log \left ({\left | -2 \, {\left (\sqrt {-c e^{2}} x - \sqrt {-c x^{2} e^{2} + c d^{2} - b x e^{2} - b d e}\right )} c - \sqrt {-c e^{2}} b \right |}\right )}{8 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(-c*x^2*e^2 + c*d^2 - b*x*e^2 - b*d*e)*(((2*(4*c^4*d^2*g*e^5 - 4*b*c^3*d*g*e^6 + b^2*c^2*g*e^7)*x/(4*c
^5*d^2*e^4 - 4*b*c^4*d*e^5 + b^2*c^3*e^6) + (48*c^4*d^3*g*e^4 + 16*c^4*d^2*f*e^5 - 68*b*c^3*d^2*g*e^5 - 16*b*c
^3*d*f*e^6 + 32*b^2*c^2*d*g*e^6 + 4*b^2*c^2*f*e^7 - 5*b^3*c*g*e^7)/(4*c^5*d^2*e^4 - 4*b*c^4*d*e^5 + b^2*c^3*e^
6))*x - (72*c^4*d^4*g*e^3 + 64*c^4*d^3*f*e^4 - 224*b*c^3*d^3*g*e^4 - 112*b*c^3*d^2*f*e^5 + 230*b^2*c^2*d^2*g*e
^5 + 64*b^2*c^2*d*f*e^6 - 98*b^3*c*d*g*e^6 - 12*b^3*c*f*e^7 + 15*b^4*g*e^7)/(4*c^5*d^2*e^4 - 4*b*c^4*d*e^5 + b
^2*c^3*e^6))*x - (112*c^4*d^5*g*e^2 + 80*c^4*d^4*f*e^3 - 284*b*c^3*d^4*g*e^3 - 128*b*c^3*d^3*f*e^4 + 260*b^2*c
^2*d^3*g*e^4 + 68*b^2*c^2*d^2*f*e^5 - 103*b^3*c*d^2*g*e^5 - 12*b^3*c*d*f*e^6 + 15*b^4*d*g*e^6)/(4*c^5*d^2*e^4
- 4*b*c^4*d*e^5 + b^2*c^3*e^6))/(c*x^2*e^2 - c*d^2 + b*x*e^2 + b*d*e) - 3/8*(12*c^2*d^2*g + 8*c^2*d*f*e - 16*b
*c*d*g*e - 4*b*c*f*e^2 + 5*b^2*g*e^2)*sqrt(-c*e^2)*e^(-3)*log(abs(-2*(sqrt(-c*e^2)*x - sqrt(-c*x^2*e^2 + c*d^2
- b*x*e^2 - b*d*e))*c - sqrt(-c*e^2)*b))/c^4

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maple [B]  time = 0.06, size = 2032, normalized size = 7.08

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/8*e*g*b^2/c^3*x/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)-15/16*e^5*g*b^5/c^4/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*
e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)-15/8*e*g*b^2/c^3/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+1/2*b/c)/(-
c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2))-43/4/e*g*b/c^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d^2-6*g/c^2*x/(-c*
e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*b*d+9/2/e*g/c*x/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d^2+6*g/c^2/(c*e^2)^
(1/2)*arctan((c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2))*b*d-9/2/e*g/c/(c*e^2)^(1/2)*arc
tan((c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2))*d^2+5/4*e*g*b/c^2*x^2/(-c*e^2*x^2-b*e^2*
x-b*d*e+c*d^2)^(1/2)+3/2*b/c^2*e/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^
2)^(1/2))*f+3/4*b^4/c^3/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*e^5*f-3/2*
b/c^2*x*e/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*f+11*e^4*g*b^3/c^2/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e
^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*x*d-7*e^4/c*b^2/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d
*e+c*d^2)^(1/2)*x*d*f-43/2*e^3*g*b^2/c/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(
1/2)*x*d^2+7/e^2/c/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d^3*g+5/e/c/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d
^2*f+3*x/c/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d*f-3/c/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e
^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2))*d*f-43/4*e^3*g*b^3/c^2/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e
^2*x-b*d*e+c*d^2)^(1/2)*d^2+3/2*b^3/c^2/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^
(1/2)*e^5*x*f+2*d^3*f*(-2*c*e^2*x-b*e^2)/(-4*c*e^2*(-b*d*e+c*d^2)-b^2*e^4)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1
/2)-1/2*e*g*x^3/c/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)-15/16*e*g*b^3/c^4/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1
/2)+11/2*g*b^2/c^3/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d-3*x^2/c/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d*g
-x^2/c*e/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*f+3/4*b^2/c^3*e/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*f-7/2/c
^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*b*d*f+5/c*b^2/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2
*x-b*d*e+c*d^2)^(1/2)*d^2*e^3*f-7/2*e^4/c^2*b^3/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e
+c*d^2)^(1/2)*d*f+7/c*b^2/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d^3*e^2*
g-15/8*e^5*g*b^4/c^3/(-b^2*e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*x+14*b/(-b^2*
e^4+4*b*c*d*e^3-4*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*x*d^3*e^2*g+10*b/(-b^2*e^4+4*b*c*d*e^3-4
*c^2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*x*d^2*e^3*f+11/2*e^4*g*b^4/c^3/(-b^2*e^4+4*b*c*d*e^3-4*c^
2*d^2*e^2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*d

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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((e*x+d)^3*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(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(b*e-2*c*d>0)', see `assume?` f
or more details)Is b*e-2*c*d zero or nonzero?

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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 )}^3}{{\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)^3)/(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)^3)/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2), 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 )^{3} \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)**3*(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)**3*(f + g*x)/(-(d + e*x)*(b*e - c*d + c*e*x))**(3/2), x)

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