3.23.45 \(\int \frac {(f+g x) (c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\) [2245]

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

[Out]

2/3*(-d*g+e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(3/2)/e^2/(e*x+d)^(3/2)-2/5*g*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)
^(5/2)/c/e^2/(e*x+d)^(5/2)-2*(-b*e+2*c*d)^(3/2)*(-d*g+e*f)*arctanh((d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/(-b*
e+2*c*d)^(1/2)/(e*x+d)^(1/2))/e^2+2*(-b*e+2*c*d)*(-d*g+e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/e^2/(e*x+d)
^(1/2)

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Rubi [A]
time = 0.27, 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 = {808, 678, 674, 214} \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 214

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

Rule 674

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 678

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 808

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 ) \text {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 = 0.37, size = 194, normalized size = 0.78 \begin {gather*} \frac {2 ((d+e x) (-b e+c (d-e x)))^{3/2} \left (\frac {-3 b^2 e^2 g-2 b c e (10 e f-13 d g+3 e g x)+c^2 \left (-38 d^2 g-e^2 x (5 f+3 g x)+d e (35 f+11 g x)\right )}{c (-b e+c (d-e x))}-\frac {15 (-2 c d+b e)^{3/2} (-e f+d g) \tan ^{-1}\left (\frac {\sqrt {-b e+c (d-e x)}}{\sqrt {-2 c d+b e}}\right )}{(-b e+c (d-e x))^{3/2}}\right )}{15 e^2 (d+e x)^{3/2}} \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*((d + e*x)*(-(b*e) + c*(d - e*x)))^(3/2)*((-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)))/(c*(-(b*e) + c*(d - e*x))) - (15*(-2*c*d + b*e)^(3/2)*(-(e*f
) + d*g)*ArcTan[Sqrt[-(b*e) + c*(d - e*x)]/Sqrt[-2*c*d + b*e]])/(-(b*e) + c*(d - e*x))^(3/2)))/(15*e^2*(d + e*
x)^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(592\) vs. \(2(226)=452\).
time = 0.04, size = 593, normalized size = 2.37

method	result	size
default	\(-\frac {2 \sqrt {-\left (e x +d \right ) \left (c e x +b e -c d \right )}\, \left (3 c^{2} e^{2} g \,x^{2} \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}+15 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b^{2} c d \,e^{2} g -15 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b^{2} c \,e^{3} f -60 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,c^{2} d^{2} e g +60 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,c^{2} d \,e^{2} f +60 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{3} g -60 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{2} e f +6 b c \,e^{2} g x \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}-11 c^{2} d e g x \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}+5 c^{2} e^{2} f x \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}+3 b^{2} e^{2} g \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}-26 b c d e g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+20 b c \,e^{2} f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+38 c^{2} d^{2} g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-35 c^{2} d e f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\right )}{15 \sqrt {e x +d}\, \sqrt {-c e x -b e +c d}\, c \,e^{2} \sqrt {b e -2 c d}}\)	\(593\)

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,method=_RETURNVERBOSE)

[Out]

-2/15*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)*(3*c^2*e^2*g*x^2*(b*e-2*c*d)^(1/2)*(-c*e*x-b*e+c*d)^(1/2)+15*arctan((-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*b*c*e^2*g*x*(b*e-2*c*d)^(1/2)*(-c*e*x-b*e+c*d)^(1/2)-11*c^2*d*e
*g*x*(b*e-2*c*d)^(1/2)*(-c*e*x-b*e+c*d)^(1/2)+5*c^2*e^2*f*x*(b*e-2*c*d)^(1/2)*(-c*e*x-b*e+c*d)^(1/2)+3*b^2*e^2
*g*(b*e-2*c*d)^(1/2)*(-c*e*x-b*e+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*} \text {Failed to integrate} \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*x^2*e^2 + c*d^2 - b*x*e^2 - b*d*e)^(3/2)*(g*x + f)/(x*e + d)^(5/2), x)

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Fricas [A]
time = 3.69, size = 580, normalized size = 2.32 \begin {gather*} \left [-\frac {15 \, {\left (2 \, c^{2} d^{3} g + b c f x e^{3} + {\left (b c d f - {\left (2 \, c^{2} d f + b c d g\right )} x\right )} e^{2} + {\left (2 \, c^{2} d^{2} g x - 2 \, c^{2} d^{2} f - b c d^{2} g\right )} e\right )} \sqrt {2 \, c d - b e} \log \left (\frac {3 \, c d^{2} - {\left (c x^{2} + 2 \, b x\right )} e^{2} + 2 \, {\left (c d x - b d\right )} e - 2 \, \sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}} \sqrt {2 \, c d - b e} \sqrt {x e + d}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, {\left (38 \, c^{2} d^{2} g + {\left (3 \, c^{2} g x^{2} + 20 \, b c f + 3 \, b^{2} g + {\left (5 \, c^{2} f + 6 \, b c g\right )} x\right )} e^{2} - {\left (11 \, c^{2} d g x + 35 \, c^{2} d f + 26 \, b c d g\right )} e\right )} \sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}} \sqrt {x e + d}}{15 \, {\left (c x e^{3} + c d e^{2}\right )}}, \frac {2 \, {\left (15 \, {\left (2 \, c^{2} d^{3} g + b c f x e^{3} + {\left (b c d f - {\left (2 \, c^{2} d f + b c d g\right )} x\right )} e^{2} + {\left (2 \, c^{2} d^{2} g x - 2 \, c^{2} d^{2} f - b c d^{2} g\right )} e\right )} \sqrt {-2 \, c d + b e} \arctan \left (-\frac {\sqrt {-2 \, c d + b e} \sqrt {x e + d}}{\sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}}}\right ) - {\left (38 \, c^{2} d^{2} g + {\left (3 \, c^{2} g x^{2} + 20 \, b c f + 3 \, b^{2} g + {\left (5 \, c^{2} f + 6 \, b c g\right )} x\right )} e^{2} - {\left (11 \, c^{2} d g x + 35 \, c^{2} d f + 26 \, b c d g\right )} e\right )} \sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}} \sqrt {x e + d}\right )}}{15 \, {\left (c x e^{3} + 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*(2*c^2*d^3*g + b*c*f*x*e^3 + (b*c*d*f - (2*c^2*d*f + b*c*d*g)*x)*e^2 + (2*c^2*d^2*g*x - 2*c^2*d^2*f
- b*c*d^2*g)*e)*sqrt(2*c*d - b*e)*log((3*c*d^2 - (c*x^2 + 2*b*x)*e^2 + 2*(c*d*x - b*d)*e - 2*sqrt(c*d^2 - b*d
*e - (c*x^2 + b*x)*e^2)*sqrt(2*c*d - b*e)*sqrt(x*e + d))/(x^2*e^2 + 2*d*x*e + d^2)) + 2*(38*c^2*d^2*g + (3*c^2
*g*x^2 + 20*b*c*f + 3*b^2*g + (5*c^2*f + 6*b*c*g)*x)*e^2 - (11*c^2*d*g*x + 35*c^2*d*f + 26*b*c*d*g)*e)*sqrt(c*
d^2 - b*d*e - (c*x^2 + b*x)*e^2)*sqrt(x*e + d))/(c*x*e^3 + c*d*e^2), 2/15*(15*(2*c^2*d^3*g + b*c*f*x*e^3 + (b*
c*d*f - (2*c^2*d*f + b*c*d*g)*x)*e^2 + (2*c^2*d^2*g*x - 2*c^2*d^2*f - b*c*d^2*g)*e)*sqrt(-2*c*d + b*e)*arctan(
-sqrt(-2*c*d + b*e)*sqrt(x*e + d)/sqrt(c*d^2 - b*d*e - (c*x^2 + b*x)*e^2)) - (38*c^2*d^2*g + (3*c^2*g*x^2 + 20
*b*c*f + 3*b^2*g + (5*c^2*f + 6*b*c*g)*x)*e^2 - (11*c^2*d*g*x + 35*c^2*d*f + 26*b*c*d*g)*e)*sqrt(c*d^2 - b*d*e
- (c*x^2 + b*x)*e^2)*sqrt(x*e + d))/(c*x*e^3 + c*d*e^2)]

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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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 723 vs. \(2 (228) = 456\).
time = 0.89, size = 723, normalized size = 2.89 \begin {gather*} -\frac {2}{15} \, {\left (\frac {15 \, {\left (4 \, c^{2} d^{3} g - 4 \, c^{2} d^{2} f e - 4 \, b c d^{2} g e + 4 \, b c d f e^{2} + b^{2} d g e^{2} - b^{2} f e^{3}\right )} \arctan \left (\frac {\sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{\sqrt {-2 \, c d + b e}} + \frac {30 \, \sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e} c^{6} d^{2} g - 30 \, \sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e} c^{6} d f e - 15 \, \sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e} b c^{5} d g e + 5 \, {\left (-{\left (x e + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{5} d g + 15 \, \sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e} b c^{5} f e^{2} - 5 \, {\left (-{\left (x e + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{5} f e + 3 \, {\left ({\left (x e + d\right )} c - 2 \, c d + b e\right )}^{2} \sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e} c^{4} g}{c^{5}}\right )} e^{\left (-2\right )} + \frac {2 \, {\left (60 \, c^{3} d^{3} g \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) - 60 \, c^{3} d^{2} f \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) e - 60 \, b c^{2} d^{2} g \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) e + 60 \, b c^{2} d f \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) e^{2} + 15 \, b^{2} c d g \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) e^{2} + 52 \, \sqrt {2 \, c d - b e} \sqrt {-2 \, c d + b e} c^{2} d^{2} g - 15 \, b^{2} c f \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) e^{3} - 40 \, \sqrt {2 \, c d - b e} \sqrt {-2 \, c d + b e} c^{2} d f e - 32 \, \sqrt {2 \, c d - b e} \sqrt {-2 \, c d + b e} b c d g e + 20 \, \sqrt {2 \, c d - b e} \sqrt {-2 \, c d + b e} b c f e^{2} + 3 \, \sqrt {2 \, c d - b e} \sqrt {-2 \, c d + b e} b^{2} g e^{2}\right )} e^{\left (-2\right )}}{15 \, \sqrt {-2 \, c d + b e} c} \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]

-2/15*(15*(4*c^2*d^3*g - 4*c^2*d^2*f*e - 4*b*c*d^2*g*e + 4*b*c*d*f*e^2 + b^2*d*g*e^2 - b^2*f*e^3)*arctan(sqrt(
-(x*e + d)*c + 2*c*d - b*e)/sqrt(-2*c*d + b*e))/sqrt(-2*c*d + b*e) + (30*sqrt(-(x*e + d)*c + 2*c*d - b*e)*c^6*
d^2*g - 30*sqrt(-(x*e + d)*c + 2*c*d - b*e)*c^6*d*f*e - 15*sqrt(-(x*e + d)*c + 2*c*d - b*e)*b*c^5*d*g*e + 5*(-
(x*e + d)*c + 2*c*d - b*e)^(3/2)*c^5*d*g + 15*sqrt(-(x*e + d)*c + 2*c*d - b*e)*b*c^5*f*e^2 - 5*(-(x*e + d)*c +
2*c*d - b*e)^(3/2)*c^5*f*e + 3*((x*e + d)*c - 2*c*d + b*e)^2*sqrt(-(x*e + d)*c + 2*c*d - b*e)*c^4*g)/c^5)*e^(
-2) + 2/15*(60*c^3*d^3*g*arctan(sqrt(2*c*d - b*e)/sqrt(-2*c*d + b*e)) - 60*c^3*d^2*f*arctan(sqrt(2*c*d - b*e)/
sqrt(-2*c*d + b*e))*e - 60*b*c^2*d^2*g*arctan(sqrt(2*c*d - b*e)/sqrt(-2*c*d + b*e))*e + 60*b*c^2*d*f*arctan(sq
rt(2*c*d - b*e)/sqrt(-2*c*d + b*e))*e^2 + 15*b^2*c*d*g*arctan(sqrt(2*c*d - b*e)/sqrt(-2*c*d + b*e))*e^2 + 52*s
qrt(2*c*d - b*e)*sqrt(-2*c*d + b*e)*c^2*d^2*g - 15*b^2*c*f*arctan(sqrt(2*c*d - b*e)/sqrt(-2*c*d + b*e))*e^3 -
40*sqrt(2*c*d - b*e)*sqrt(-2*c*d + b*e)*c^2*d*f*e - 32*sqrt(2*c*d - b*e)*sqrt(-2*c*d + b*e)*b*c*d*g*e + 20*sqr
t(2*c*d - b*e)*sqrt(-2*c*d + b*e)*b*c*f*e^2 + 3*sqrt(2*c*d - b*e)*sqrt(-2*c*d + b*e)*b^2*g*e^2)*e^(-2)/(sqrt(-
2*c*d + b*e)*c)

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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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