3.2274 \(\int \frac{f+g x}{(d+e x)^{3/2} (c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=308 \[ -\frac{e f-d g}{2 e^2 (d+e x)^{3/2} (2 c d-b e) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac{3 c \sqrt{d+e x} (-4 b e g+3 c d g+5 c e f)}{4 e^2 (2 c d-b e)^3 \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac{-4 b e g+3 c d g+5 c e f}{4 e^2 \sqrt{d+e x} (2 c d-b e)^2 \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac{3 c (-4 b e g+3 c d g+5 c e f) \tanh ^{-1}\left (\frac{\sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt{d+e x} \sqrt{2 c d-b e}}\right )}{4 e^2 (2 c d-b e)^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.466718, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.109, Rules used = {792, 672, 666, 660, 208} \[ -\frac{e f-d g}{2 e^2 (d+e x)^{3/2} (2 c d-b e) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac{3 c \sqrt{d+e x} (-4 b e g+3 c d g+5 c e f)}{4 e^2 (2 c d-b e)^3 \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac{-4 b e g+3 c d g+5 c e f}{4 e^2 \sqrt{d+e x} (2 c d-b e)^2 \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac{3 c (-4 b e g+3 c d g+5 c e f) \tanh ^{-1}\left (\frac{\sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt{d+e x} \sqrt{2 c d-b e}}\right )}{4 e^2 (2 c d-b e)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 792

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

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(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] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 666

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((2*c*d - b*e)*(d +
e*x)^m*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*c*d - b*e)*(m + 2*p + 2))/((p + 1)*
(b^2 - 4*a*c)), 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] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

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 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]

Rubi steps

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

Mathematica [C]  time = 0.120985, size = 129, normalized size = 0.42 \[ \frac{\frac{c (d+e x)^2 (-4 b e g+3 c d g+5 c e f) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{-c d+b e+c e x}{b e-2 c d}\right )}{e (b e-2 c d)^2}+\frac{d g}{e}-f}{2 e (d+e x)^{3/2} (2 c d-b e) \sqrt{(d+e x) (c (d-e x)-b e)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-f + (d*g)/e + (c*(5*c*e*f + 3*c*d*g - 4*b*e*g)*(d + e*x)^2*Hypergeometric2F1[-1/2, 2, 1/2, (-(c*d) + b*e + c
*e*x)/(-2*c*d + b*e)])/(e*(-2*c*d + b*e)^2))/(2*e*(2*c*d - b*e)*(d + e*x)^(3/2)*Sqrt[(d + e*x)*(-(b*e) + c*(d
- e*x))])

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Maple [B]  time = 0.03, size = 824, normalized size = 2.7 \begin{align*} -{\frac{1}{ \left ( 4\,cex+4\,be-4\,cd \right ){e}^{2}}\sqrt{-c{e}^{2}{x}^{2}-b{e}^{2}x-bde+c{d}^{2}} \left ( 12\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}{x}^{2}bc{e}^{3}g-9\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}{x}^{2}{c}^{2}d{e}^{2}g-15\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}{x}^{2}{c}^{2}{e}^{3}f+12\,\sqrt{be-2\,cd}{x}^{2}bc{e}^{3}g-9\,\sqrt{be-2\,cd}{x}^{2}{c}^{2}d{e}^{2}g-15\,\sqrt{be-2\,cd}{x}^{2}{c}^{2}{e}^{3}f+24\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}xbcd{e}^{2}g-18\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}x{c}^{2}{d}^{2}eg-30\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}x{c}^{2}d{e}^{2}f+4\,\sqrt{be-2\,cd}x{b}^{2}{e}^{3}g+13\,\sqrt{be-2\,cd}xbcd{e}^{2}g-5\,\sqrt{be-2\,cd}xbc{e}^{3}f-12\,\sqrt{be-2\,cd}x{c}^{2}{d}^{2}eg-20\,\sqrt{be-2\,cd}x{c}^{2}d{e}^{2}f+12\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}bc{d}^{2}eg-9\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}{c}^{2}{d}^{3}g-15\,\arctan \left ({\frac{\sqrt{-cex-be+cd}}{\sqrt{be-2\,cd}}} \right ) \sqrt{-cex-be+cd}{c}^{2}{d}^{2}ef+2\,\sqrt{be-2\,cd}{b}^{2}d{e}^{2}g+2\,\sqrt{be-2\,cd}{b}^{2}{e}^{3}f+9\,\sqrt{be-2\,cd}bc{d}^{2}eg-13\,\sqrt{be-2\,cd}bcd{e}^{2}f-11\,\sqrt{be-2\,cd}{c}^{2}{d}^{3}g+3\,\sqrt{be-2\,cd}{c}^{2}{d}^{2}ef \right ) \left ( ex+d \right ) ^{-{\frac{5}{2}}} \left ( be-2\,cd \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(e*x+d)^(5/2)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*(12*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))
*(-c*e*x-b*e+c*d)^(1/2)*x^2*b*c*e^3*g-9*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*(-c*e*x-b*e+c*d)^(1/2
)*x^2*c^2*d*e^2*g-15*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*(-c*e*x-b*e+c*d)^(1/2)*x^2*c^2*e^3*f+12*
(b*e-2*c*d)^(1/2)*x^2*b*c*e^3*g-9*(b*e-2*c*d)^(1/2)*x^2*c^2*d*e^2*g-15*(b*e-2*c*d)^(1/2)*x^2*c^2*e^3*f+24*arct
an((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*(-c*e*x-b*e+c*d)^(1/2)*x*b*c*d*e^2*g-18*arctan((-c*e*x-b*e+c*d)^(
1/2)/(b*e-2*c*d)^(1/2))*(-c*e*x-b*e+c*d)^(1/2)*x*c^2*d^2*e*g-30*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2
))*(-c*e*x-b*e+c*d)^(1/2)*x*c^2*d*e^2*f+4*(b*e-2*c*d)^(1/2)*x*b^2*e^3*g+13*(b*e-2*c*d)^(1/2)*x*b*c*d*e^2*g-5*(
b*e-2*c*d)^(1/2)*x*b*c*e^3*f-12*(b*e-2*c*d)^(1/2)*x*c^2*d^2*e*g-20*(b*e-2*c*d)^(1/2)*x*c^2*d*e^2*f+12*arctan((
-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*(-c*e*x-b*e+c*d)^(1/2)*b*c*d^2*e*g-9*arctan((-c*e*x-b*e+c*d)^(1/2)/(b
*e-2*c*d)^(1/2))*(-c*e*x-b*e+c*d)^(1/2)*c^2*d^3*g-15*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*(-c*e*x-
b*e+c*d)^(1/2)*c^2*d^2*e*f+2*(b*e-2*c*d)^(1/2)*b^2*d*e^2*g+2*(b*e-2*c*d)^(1/2)*b^2*e^3*f+9*(b*e-2*c*d)^(1/2)*b
*c*d^2*e*g-13*(b*e-2*c*d)^(1/2)*b*c*d*e^2*f-11*(b*e-2*c*d)^(1/2)*c^2*d^3*g+3*(b*e-2*c*d)^(1/2)*c^2*d^2*e*f)/(c
*e*x+b*e-c*d)/e^2/(b*e-2*c*d)^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.75919, size = 4041, normalized size = 13.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*((5*c^3*e^5*f + (3*c^3*d*e^4 - 4*b*c^2*e^5)*g)*x^4 + (5*(2*c^3*d*e^4 + b*c^2*e^5)*f + (6*c^3*d^2*e^3
- 5*b*c^2*d*e^4 - 4*b^2*c*e^5)*g)*x^3 + 3*(5*b*c^2*d*e^4*f + (3*b*c^2*d^2*e^3 - 4*b^2*c*d*e^4)*g)*x^2 - 5*(c^3
*d^4*e - b*c^2*d^3*e^2)*f - (3*c^3*d^5 - 7*b*c^2*d^4*e + 4*b^2*c*d^3*e^2)*g - (5*(2*c^3*d^3*e^2 - 3*b*c^2*d^2*
e^3)*f + (6*c^3*d^4*e - 17*b*c^2*d^3*e^2 + 12*b^2*c*d^2*e^3)*g)*x)*sqrt(2*c*d - b*e)*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*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(3*(5*(2*c^3*d*e^3 - b*c^2*e^4)*f
+ (6*c^3*d^2*e^2 - 11*b*c^2*d*e^3 + 4*b^2*c*e^4)*g)*x^2 - (6*c^3*d^3*e - 29*b*c^2*d^2*e^2 + 17*b^2*c*d*e^3 -
2*b^3*e^4)*f + (22*c^3*d^4 - 29*b*c^2*d^3*e + 5*b^2*c*d^2*e^2 + 2*b^3*d*e^3)*g + (5*(8*c^3*d^2*e^2 - 2*b*c^2*d
*e^3 - b^2*c*e^4)*f + (24*c^3*d^3*e - 38*b*c^2*d^2*e^2 + 5*b^2*c*d*e^3 + 4*b^3*e^4)*g)*x)*sqrt(e*x + d))/(16*c
^5*d^8*e^2 - 48*b*c^4*d^7*e^3 + 56*b^2*c^3*d^6*e^4 - 32*b^3*c^2*d^5*e^5 + 9*b^4*c*d^4*e^6 - b^5*d^3*e^7 - (16*
c^5*d^4*e^6 - 32*b*c^4*d^3*e^7 + 24*b^2*c^3*d^2*e^8 - 8*b^3*c^2*d*e^9 + b^4*c*e^10)*x^4 - (32*c^5*d^5*e^5 - 48
*b*c^4*d^4*e^6 + 16*b^2*c^3*d^3*e^7 + 8*b^3*c^2*d^2*e^8 - 6*b^4*c*d*e^9 + b^5*e^10)*x^3 - 3*(16*b*c^4*d^5*e^5
- 32*b^2*c^3*d^4*e^6 + 24*b^3*c^2*d^3*e^7 - 8*b^4*c*d^2*e^8 + b^5*d*e^9)*x^2 + (32*c^5*d^7*e^3 - 112*b*c^4*d^6
*e^4 + 144*b^2*c^3*d^5*e^5 - 88*b^3*c^2*d^4*e^6 + 26*b^4*c*d^3*e^7 - 3*b^5*d^2*e^8)*x), 1/4*(3*((5*c^3*e^5*f +
(3*c^3*d*e^4 - 4*b*c^2*e^5)*g)*x^4 + (5*(2*c^3*d*e^4 + b*c^2*e^5)*f + (6*c^3*d^2*e^3 - 5*b*c^2*d*e^4 - 4*b^2*
c*e^5)*g)*x^3 + 3*(5*b*c^2*d*e^4*f + (3*b*c^2*d^2*e^3 - 4*b^2*c*d*e^4)*g)*x^2 - 5*(c^3*d^4*e - b*c^2*d^3*e^2)*
f - (3*c^3*d^5 - 7*b*c^2*d^4*e + 4*b^2*c*d^3*e^2)*g - (5*(2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3)*f + (6*c^3*d^4*e -
17*b*c^2*d^3*e^2 + 12*b^2*c*d^2*e^3)*g)*x)*sqrt(-2*c*d + b*e)*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)) + sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2
- b*d*e)*(3*(5*(2*c^3*d*e^3 - b*c^2*e^4)*f + (6*c^3*d^2*e^2 - 11*b*c^2*d*e^3 + 4*b^2*c*e^4)*g)*x^2 - (6*c^3*d^
3*e - 29*b*c^2*d^2*e^2 + 17*b^2*c*d*e^3 - 2*b^3*e^4)*f + (22*c^3*d^4 - 29*b*c^2*d^3*e + 5*b^2*c*d^2*e^2 + 2*b^
3*d*e^3)*g + (5*(8*c^3*d^2*e^2 - 2*b*c^2*d*e^3 - b^2*c*e^4)*f + (24*c^3*d^3*e - 38*b*c^2*d^2*e^2 + 5*b^2*c*d*e
^3 + 4*b^3*e^4)*g)*x)*sqrt(e*x + d))/(16*c^5*d^8*e^2 - 48*b*c^4*d^7*e^3 + 56*b^2*c^3*d^6*e^4 - 32*b^3*c^2*d^5*
e^5 + 9*b^4*c*d^4*e^6 - b^5*d^3*e^7 - (16*c^5*d^4*e^6 - 32*b*c^4*d^3*e^7 + 24*b^2*c^3*d^2*e^8 - 8*b^3*c^2*d*e^
9 + b^4*c*e^10)*x^4 - (32*c^5*d^5*e^5 - 48*b*c^4*d^4*e^6 + 16*b^2*c^3*d^3*e^7 + 8*b^3*c^2*d^2*e^8 - 6*b^4*c*d*
e^9 + b^5*e^10)*x^3 - 3*(16*b*c^4*d^5*e^5 - 32*b^2*c^3*d^4*e^6 + 24*b^3*c^2*d^3*e^7 - 8*b^4*c*d^2*e^8 + b^5*d*
e^9)*x^2 + (32*c^5*d^7*e^3 - 112*b*c^4*d^6*e^4 + 144*b^2*c^3*d^5*e^5 - 88*b^3*c^2*d^4*e^6 + 26*b^4*c*d^3*e^7 -
3*b^5*d^2*e^8)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[undef, undef, undef, undef, undef, undef, undef, 1]