3.9.66 \(\int \frac {(a+b x+c x^2)^{3/2}}{(d+e x) (f+g x)} \, dx\) [866]

Optimal. Leaf size=491 \[ \frac {\left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}{e^2 (e f-d g)}-\frac {\left (4 c e f^2-g (5 b e f-b d g-4 a e g)-2 c g (e f-d g) x\right ) \sqrt {a+b x+c x^2}}{4 e g^2 (e f-d g)}-\frac {(2 c d-b e) \left (c d^2-b d e+a e^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} e^3 (e f-d g)}+\frac {\left (8 c^2 e f^3+b g^2 (3 b e f+b d g-4 a e g)-4 c g \left (3 b e f^2-a g (3 e f-d g)\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} e g^3 (e f-d g)}+\frac {\left (c d^2-b d e+a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^3 (e f-d g)}-\frac {\left (c f^2-b f g+a g^2\right )^{3/2} \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{g^3 (e f-d g)} \]

[Out]

(a*e^2-b*d*e+c*d^2)^(3/2)*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2)
)/e^3/(-d*g+e*f)-(a*g^2-b*f*g+c*f^2)^(3/2)*arctanh(1/2*(b*f-2*a*g+(-b*g+2*c*f)*x)/(a*g^2-b*f*g+c*f^2)^(1/2)/(c
*x^2+b*x+a)^(1/2))/g^3/(-d*g+e*f)-1/2*(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*
x+a)^(1/2))/e^3/(-d*g+e*f)/c^(1/2)+1/8*(8*c^2*e*f^3+b*g^2*(-4*a*e*g+b*d*g+3*b*e*f)-4*c*g*(3*b*e*f^2-a*g*(-d*g+
3*e*f)))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/e/g^3/(-d*g+e*f)/c^(1/2)+(a*e^2-b*d*e+c*d^2)*(c*x^
2+b*x+a)^(1/2)/e^2/(-d*g+e*f)-1/4*(4*c*e*f^2-g*(-4*a*e*g-b*d*g+5*b*e*f)-2*c*g*(-d*g+e*f)*x)*(c*x^2+b*x+a)^(1/2
)/e/g^2/(-d*g+e*f)

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Rubi [A]
time = 0.52, antiderivative size = 491, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {909, 748, 857, 635, 212, 738, 828} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c g \left (3 b e f^2-a g (3 e f-d g)\right )+b g^2 (-4 a e g+b d g+3 b e f)+8 c^2 e f^3\right )}{8 \sqrt {c} e g^3 (e f-d g)}+\frac {\sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}{e^2 (e f-d g)}+\frac {\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^3 (e f-d g)}-\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} e^3 (e f-d g)}-\frac {\sqrt {a+b x+c x^2} \left (-g (-4 a e g-b d g+5 b e f)-2 c g x (e f-d g)+4 c e f^2\right )}{4 e g^2 (e f-d g)}-\frac {\left (a g^2-b f g+c f^2\right )^{3/2} \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{g^3 (e f-d g)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

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

Rule 635

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 738

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

Rule 748

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/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 828

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

Rule 857

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

Rule 909

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)/(((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))), x_Symbol] :> Dist[(c
*d^2 - b*d*e + a*e^2)/(e*(e*f - d*g)), Int[(a + b*x + c*x^2)^(p - 1)/(d + e*x), x], x] - Dist[1/(e*(e*f - d*g)
), Int[Simp[c*d*f - b*e*f + a*e*g - c*(e*f - d*g)*x, x]*((a + b*x + c*x^2)^(p - 1)/(f + g*x)), x], 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] && Fra
ctionQ[p] && GtQ[p, 0]

Rubi steps

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

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Mathematica [A]
time = 10.75, size = 323, normalized size = 0.66 \begin {gather*} \frac {\frac {\left (3 b^2 e^2 g^2-12 c e g (b e f+b d g-a e g)+8 c^2 \left (e^2 f^2+d e f g+d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}+\frac {2 \left (e g (e f-d g) \sqrt {a+x (b+c x)} (5 b e g+c (-4 e f-4 d g+2 e g x))-4 \left (c d^2+e (-b d+a e)\right )^{3/2} g^3 \tanh ^{-1}\left (\frac {-b d+2 a e-2 c d x+b e x}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )+4 e^3 \left (c f^2+g (-b f+a g)\right )^{3/2} \tanh ^{-1}\left (\frac {-b f+2 a g-2 c f x+b g x}{2 \sqrt {c f^2+g (-b f+a g)} \sqrt {a+x (b+c x)}}\right )\right )}{e f-d g}}{8 e^3 g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*b^2*e^2*g^2 - 12*c*e*g*(b*e*f + b*d*g - a*e*g) + 8*c^2*(e^2*f^2 + d*e*f*g + d^2*g^2))*ArcTanh[(b + 2*c*x)
/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/Sqrt[c] + (2*(e*g*(e*f - d*g)*Sqrt[a + x*(b + c*x)]*(5*b*e*g + c*(-4*e*f
- 4*d*g + 2*e*g*x)) - 4*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*g^3*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqr
t[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])] + 4*e^3*(c*f^2 + g*(-(b*f) + a*g))^(3/2)*ArcTanh[(-(b*f) +
2*a*g - 2*c*f*x + b*g*x)/(2*Sqrt[c*f^2 + g*(-(b*f) + a*g)]*Sqrt[a + x*(b + c*x)])]))/(e*f - d*g))/(8*e^3*g^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1264\) vs. \(2(449)=898\).
time = 0.19, size = 1265, normalized size = 2.58

method	result	size
default	\(\text {Expression too large to display}\)	\(1265\)
risch	\(\text {Expression too large to display}\)	\(2348\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(d*g-e*f)*(1/3*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/2*(b*e-2*c*d)/e*(1/4*(2*
c*(x+d/e)+(b*e-2*c*d)/e)/c*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b
*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2)/c^(3/2)*ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+(b*e-2*c*d)/e
*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)))+(a*e^2-b*d*e+c*d^2)/e^2*((c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b
*d*e+c*d^2)/e^2)^(1/2)+1/2*(b*e-2*c*d)/e*ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+(b*e-2*c*d)/e*(
x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e^2-b*d*e+c*d^2)/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(
a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d
/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))))+1/(d*g-e*f)*(1/3*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*
g+c*f^2)/g^2)^(3/2)+1/2*(b*g-2*c*f)/g*(1/4*(2*c*(x+f/g)+(b*g-2*c*f)/g)/c*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a
*g^2-b*f*g+c*f^2)/g^2)^(1/2)+1/8*(4*c*(a*g^2-b*f*g+c*f^2)/g^2-(b*g-2*c*f)^2/g^2)/c^(3/2)*ln((1/2*(b*g-2*c*f)/g
+c*(x+f/g))/c^(1/2)+((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)))+(a*g^2-b*f*g+c*f^2)/g^
2*(((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)+1/2*(b*g-2*c*f)/g*ln((1/2*(b*g-2*c*f)/g+c
*(x+f/g))/c^(1/2)+((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/c^(1/2)-(a*g^2-b*f*g+c*f^
2)/g^2/((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*ln((2*(a*g^2-b*f*g+c*f^2)/g^2+(b*g-2*c*f)/g*(x+f/g)+2*((a*g^2-b*f*g+c*f
^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/(x+f/g))))

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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)^(3/2)/(e*x+d)/(g*x+f),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?` fo
r more detai

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Fricas [F(-1)] Timed out
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)^(3/2)/(e*x+d)/(g*x+f),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{\left (f+g\,x\right )\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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