3.9.71 \(\int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx\) [871]

Optimal. Leaf size=270 \[ \frac {3 g^2 (4 c e f-2 c d g-b e g) \sqrt {a+b x+c x^2}}{4 c^2 e^2}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2}+\frac {g \left (3 b^2 e^2 g^2-4 c e g (3 b e f-b d g+a e g)+8 c^2 \left (3 e^2 f^2-3 d e f g+d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} e^3}+\frac {(e f-d g)^3 \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^3 \sqrt {c d^2-b d e+a e^2}} \]

[Out]

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

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Rubi [A]
time = 0.45, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1667, 857, 635, 212, 738} \begin {gather*} \frac {g \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c e g (a e g-b d g+3 b e f)+3 b^2 e^2 g^2+8 c^2 \left (d^2 g^2-3 d e f g+3 e^2 f^2\right )\right )}{8 c^{5/2} e^3}+\frac {3 g^2 \sqrt {a+b x+c x^2} (-b e g-2 c d g+4 c e f)}{4 c^2 e^2}+\frac {(e f-d g)^3 \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 \sqrt {a e^2-b d e+c d^2}}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 1667

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m
+ q + 2*p + 1))), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

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

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Mathematica [A]
time = 1.25, size = 247, normalized size = 0.91 \begin {gather*} -\frac {-\frac {2 e g^2 \sqrt {a+x (b+c x)} (-3 b e g+2 c (6 e f-2 d g+e g x))}{c^2}+\frac {16 \sqrt {-c d^2+b d e-a e^2} (-e f+d g)^3 \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{c d^2+e (-b d+a e)}+\frac {g \left (3 b^2 e^2 g^2-4 c e g (3 b e f-b d g+a e g)+8 c^2 \left (3 e^2 f^2-3 d e f g+d^2 g^2\right )\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{c^{5/2}}}{8 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((-2*e*g^2*Sqrt[a + x*(b + c*x)]*(-3*b*e*g + 2*c*(6*e*f - 2*d*g + e*g*x)))/c^2 + (16*Sqrt[-(c*d^2) + b*d*
e - a*e^2]*(-(e*f) + d*g)^3*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + x*(b + c*x)])/Sqrt[-(c*d^2) + e*(b*d - a*e)
]])/(c*d^2 + e*(-(b*d) + a*e)) + (g*(3*b^2*e^2*g^2 - 4*c*e*g*(3*b*e*f - b*d*g + a*e*g) + 8*c^2*(3*e^2*f^2 - 3*
d*e*f*g + d^2*g^2))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/c^(5/2))/e^3

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Maple [A]
time = 0.13, size = 480, normalized size = 1.78

method	result	size
default	\(\frac {g \left (g^{2} e^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+\left (-d e \,g^{2}+3 e^{2} f g \right ) \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+\frac {d^{2} g^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {3 d e f g \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+\frac {3 e^{2} f^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}\right )}{e^{3}}-\frac {\left (-d^{3} g^{3}+3 d^{2} e f \,g^{2}-3 d \,e^{2} f^{2} g +e^{3} f^{3}\right ) \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{4} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\)	\(480\)
risch	\(-\frac {g^{2} \left (-2 c e g x +3 b e g +4 d g c -12 c e f \right ) \sqrt {c \,x^{2}+b x +a}}{4 c^{2} e^{2}}-\frac {g^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a}{2 c^{\frac {3}{2}} e}+\frac {3 g^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{2}}{8 c^{\frac {5}{2}} e}+\frac {g^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b d}{2 c^{\frac {3}{2}} e^{2}}-\frac {3 g^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b f}{2 c^{\frac {3}{2}} e}+\frac {g^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d^{2}}{\sqrt {c}\, e^{3}}-\frac {3 g^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d f}{\sqrt {c}\, e^{2}}+\frac {3 g \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) f^{2}}{\sqrt {c}\, e}+\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) d^{3} g^{3}}{e^{4} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {3 \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) d^{2} f \,g^{2}}{e^{3} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}+\frac {3 \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) d \,f^{2} g}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) f^{3}}{e \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\)	\(956\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^3/(e*x+d)/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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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((g*x+f)^3/(e*x+d)/(c*x^2+b*x+a)^(1/2),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((%e^-1*b-2*%e^-2*c*d)^2>0)', s
ee `assume?`

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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((g*x+f)^3/(e*x+d)/(c*x^2+b*x+a)^(1/2),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 (f + g x\right )^{3}}{\left (d + e x\right ) \sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**3/(e*x+d)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((f + g*x)**3/((d + e*x)*sqrt(a + b*x + c*x**2)), 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((g*x+f)^3/(e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)^3/((d + e*x)*(a + b*x + c*x^2)^(1/2)),x)

[Out]

int((f + g*x)^3/((d + e*x)*(a + b*x + c*x^2)^(1/2)), x)

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