3.21.15 \(\int \frac {(f+g x) (c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=250 \[ \frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (d+e x)^{3/2}}+\frac {2 (2 c d-b e) (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 \sqrt {d+e x}}-\frac {2 (2 c d-b e)^{3/2} (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 c e^2 (d+e x)^{5/2}} \]

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Rubi [A]  time = 0.45, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {794, 664, 660, 208} \begin {gather*} \frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (d+e x)^{3/2}}+\frac {2 (2 c d-b e) (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 \sqrt {d+e x}}-\frac {2 (2 c d-b e)^{3/2} (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 c e^2 (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(2*(2*c*d - b*e)*(e*f - d*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(e^2*Sqrt[d + e*x]) + (2*(e*f - d*g)*(
d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(3*e^2*(d + e*x)^(3/2)) - (2*g*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^
2)^(5/2))/(5*c*e^2*(d + e*x)^(5/2)) - (2*(2*c*d - b*e)^(3/2)*(e*f - d*g)*ArcTanh[Sqrt[d*(c*d - b*e) - b*e^2*x
- c*e^2*x^2]/(Sqrt[2*c*d - b*e]*Sqrt[d + e*x])])/e^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 660

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

Rule 664

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 199, normalized size = 0.80 \begin {gather*} -\frac {2 \sqrt {d+e x} \sqrt {c (d-e x)-b e} \left (\sqrt {c (d-e x)-b e} \left (3 b^2 e^2 g+2 b c e (-13 d g+10 e f+3 e g x)+c^2 \left (38 d^2 g-d e (35 f+11 g x)+e^2 x (5 f+3 g x)\right )\right )-15 c (2 c d-b e)^{3/2} (d g-e f) \tanh ^{-1}\left (\frac {\sqrt {-b e+c d-c e x}}{\sqrt {2 c d-b e}}\right )\right )}{15 c e^2 \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(-2*Sqrt[d + e*x]*Sqrt[-(b*e) + c*(d - e*x)]*(Sqrt[-(b*e) + c*(d - e*x)]*(3*b^2*e^2*g + 2*b*c*e*(10*e*f - 13*d
*g + 3*e*g*x) + c^2*(38*d^2*g + e^2*x*(5*f + 3*g*x) - d*e*(35*f + 11*g*x))) - 15*c*(2*c*d - b*e)^(3/2)*(-(e*f)
+ d*g)*ArcTanh[Sqrt[c*d - b*e - c*e*x]/Sqrt[2*c*d - b*e]]))/(15*c*e^2*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])

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IntegrateAlgebraic [A]  time = 4.13, size = 238, normalized size = 0.95 \begin {gather*} -\frac {2 \sqrt {-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)} \left (3 b^2 e^2 g+6 b c e g (d+e x)-32 b c d e g+20 b c e^2 f+52 c^2 d^2 g+5 c^2 e f (d+e x)-40 c^2 d e f+3 c^2 g (d+e x)^2-17 c^2 d g (d+e x)\right )}{15 c e^2 \sqrt {d+e x}}-\frac {2 (b e-2 c d)^{3/2} (d g-e f) \tan ^{-1}\left (\frac {\sqrt {b e-2 c d} \sqrt {(d+e x) (2 c d-b e)-c (d+e x)^2}}{\sqrt {d+e x} (b e+c (d+e x)-2 c d)}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(-2*Sqrt[2*c*d*(d + e*x) - b*e*(d + e*x) - c*(d + e*x)^2]*(-40*c^2*d*e*f + 20*b*c*e^2*f + 52*c^2*d^2*g - 32*b*
c*d*e*g + 3*b^2*e^2*g + 5*c^2*e*f*(d + e*x) - 17*c^2*d*g*(d + e*x) + 6*b*c*e*g*(d + e*x) + 3*c^2*g*(d + e*x)^2
))/(15*c*e^2*Sqrt[d + e*x]) - (2*(-2*c*d + b*e)^(3/2)*(-(e*f) + d*g)*ArcTan[(Sqrt[-2*c*d + b*e]*Sqrt[(2*c*d -
b*e)*(d + e*x) - c*(d + e*x)^2])/(Sqrt[d + e*x]*(-2*c*d + b*e + c*(d + e*x)))])/e^2

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fricas [A]  time = 0.44, size = 656, normalized size = 2.62 \begin {gather*} \left [\frac {15 \, \sqrt {2 \, c d - b e} {\left ({\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} f - {\left (2 \, c^{2} d^{3} - b c d^{2} e\right )} g + {\left ({\left (2 \, c^{2} d e^{2} - b c e^{3}\right )} f - {\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} g\right )} x\right )} \log \left (-\frac {c e^{2} x^{2} - 3 \, c d^{2} + 2 \, b d e - 2 \, {\left (c d e - b e^{2}\right )} x + 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {2 \, c d - b e} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, {\left (3 \, c^{2} e^{2} g x^{2} - 5 \, {\left (7 \, c^{2} d e - 4 \, b c e^{2}\right )} f + {\left (38 \, c^{2} d^{2} - 26 \, b c d e + 3 \, b^{2} e^{2}\right )} g + {\left (5 \, c^{2} e^{2} f - {\left (11 \, c^{2} d e - 6 \, 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 e^{3} x + c d e^{2}\right )}}, -\frac {2 \, {\left (15 \, \sqrt {-2 \, c d + b e} {\left ({\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} f - {\left (2 \, c^{2} d^{3} - b c d^{2} e\right )} g + {\left ({\left (2 \, c^{2} d e^{2} - b c e^{3}\right )} f - {\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} g\right )} x\right )} \arctan \left (\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {-2 \, c d + b e} \sqrt {e x + d}}{c e^{2} x^{2} + b e^{2} x - c d^{2} + b d e}\right ) + {\left (3 \, c^{2} e^{2} g x^{2} - 5 \, {\left (7 \, c^{2} d e - 4 \, b c e^{2}\right )} f + {\left (38 \, c^{2} d^{2} - 26 \, b c d e + 3 \, b^{2} e^{2}\right )} g + {\left (5 \, c^{2} e^{2} f - {\left (11 \, c^{2} d e - 6 \, 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}\right )}}{15 \, {\left (c e^{3} x + c d e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

[1/15*(15*sqrt(2*c*d - b*e)*((2*c^2*d^2*e - b*c*d*e^2)*f - (2*c^2*d^3 - b*c*d^2*e)*g + ((2*c^2*d*e^2 - b*c*e^3
)*f - (2*c^2*d^2*e - b*c*d*e^2)*g)*x)*log(-(c*e^2*x^2 - 3*c*d^2 + 2*b*d*e - 2*(c*d*e - b*e^2)*x + 2*sqrt(-c*e^
2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(2*c*d - b*e)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - 2*(3*c^2*e^2*g*
x^2 - 5*(7*c^2*d*e - 4*b*c*e^2)*f + (38*c^2*d^2 - 26*b*c*d*e + 3*b^2*e^2)*g + (5*c^2*e^2*f - (11*c^2*d*e - 6*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*e^3*x + c*d*e^2), -2/15*(15*sqrt(-2
*c*d + b*e)*((2*c^2*d^2*e - b*c*d*e^2)*f - (2*c^2*d^3 - b*c*d^2*e)*g + ((2*c^2*d*e^2 - b*c*e^3)*f - (2*c^2*d^2
*e - b*c*d*e^2)*g)*x)*arctan(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(-2*c*d + b*e)*sqrt(e*x + d)/(c*e^
2*x^2 + b*e^2*x - c*d^2 + b*d*e)) + (3*c^2*e^2*g*x^2 - 5*(7*c^2*d*e - 4*b*c*e^2)*f + (38*c^2*d^2 - 26*b*c*d*e
+ 3*b^2*e^2)*g + (5*c^2*e^2*f - (11*c^2*d*e - 6*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*e^3*x + c*d*e^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:inde
x.cc index_m operator + Error: Bad Argument Valueindex.cc index_m operator + Error: Bad Argument Valueindex.cc
index_m operator + Error: Bad Argument ValueEvaluation time: 2.99Done

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maple [B]  time = 0.08, size = 601, normalized size = 2.40 \begin {gather*} -\frac {2 \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (15 b^{2} c d \,e^{2} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-15 b^{2} c \,e^{3} f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-60 b \,c^{2} d^{2} e g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+60 b \,c^{2} d \,e^{2} f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+60 c^{3} d^{3} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-60 c^{3} d^{2} e f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+3 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c^{2} e^{2} g \,x^{2}+6 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, b c \,e^{2} g x -11 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c^{2} d e g x +5 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c^{2} e^{2} f x +3 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, b^{2} e^{2} g -26 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, b c d e g +20 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, b c \,e^{2} f +38 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c^{2} d^{2} g -35 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c^{2} d e f \right )}{15 \sqrt {e x +d}\, \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x)

[Out]

-2/15*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*(3*x^2*c^2*e^2*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+15*arct
an((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b^2*c*d*e^2*g-15*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))
*b^2*c*e^3*f-60*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c^2*d^2*e*g+60*arctan((-c*e*x-b*e+c*d)^(1/2
)/(b*e-2*c*d)^(1/2))*b*c^2*d*e^2*f+60*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*d^3*g-60*arctan((-c
*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*d^2*e*f+6*x*b*c*e^2*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-11*x
*c^2*d*e*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+5*x*c^2*e^2*f*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+3*b
^2*e^2*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-26*b*c*d*e*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+20*b*c
*e^2*f*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+38*c^2*d^2*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-35*c^2*d
*e*f*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2))/(e*x+d)^(1/2)/(-c*e*x-b*e+c*d)^(1/2)/c/e^2/(b*e-2*c*d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}} {\left (g x + f\right )}}{{\left (e x + d\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)^(3/2)*(g*x + f)/(e*x + d)^(5/2), x)

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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 (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{3/2}}{{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2))/(d + e*x)^(5/2),x)

[Out]

int(((f + g*x)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2))/(d + e*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}} \left (f + g x\right )}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(3/2)/(e*x+d)**(5/2),x)

[Out]

Integral((-(d + e*x)*(b*e - c*d + c*e*x))**(3/2)*(f + g*x)/(d + e*x)**(5/2), x)

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