3.9.70 \(\int \frac {(f+g x)^4}{(d+e x) \sqrt {a+b x+c x^2}} \, dx\) [870]

Optimal. Leaf size=431 \[ \frac {g^2 \left (15 b^2 e^2 g^2-4 c e g (18 b e f-7 b d g+4 a e g)+4 c^2 \left (36 e^2 f^2-36 d e f g+11 d^2 g^2\right )\right ) \sqrt {a+b x+c x^2}}{24 c^3 e^3}+\frac {g^3 (24 c e f-14 c d g-5 b e g) (d+e x) \sqrt {a+b x+c x^2}}{12 c^2 e^3}+\frac {g^4 (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c e^3}-\frac {g \left (5 b^3 e^3 g^3-6 b c e^2 g^2 (4 b e f-b d g+2 a e g)-16 c^3 \left (4 e^3 f^3-6 d e^2 f^2 g+4 d^2 e f g^2-d^3 g^3\right )+8 c^2 e g \left (a e g (4 e f-d g)+b \left (6 e^2 f^2-4 d e f g+d^2 g^2\right )\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{7/2} e^4}+\frac {(e f-d g)^4 \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^4 \sqrt {c d^2-b d e+a e^2}} \]

[Out]

-1/16*g*(5*b^3*e^3*g^3-6*b*c*e^2*g^2*(2*a*e*g-b*d*g+4*b*e*f)-16*c^3*(-d^3*g^3+4*d^2*e*f*g^2-6*d*e^2*f^2*g+4*e^
3*f^3)+8*c^2*e*g*(a*e*g*(-d*g+4*e*f)+b*(d^2*g^2-4*d*e*f*g+6*e^2*f^2)))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*
x+a)^(1/2))/c^(7/2)/e^4+(-d*g+e*f)^4*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^4/(a*e^2-b*d*e+c*d^2)^(1/2)+1/24*g^2*(15*b^2*e^2*g^2-4*c*e*g*(4*a*e*g-7*b*d*g+18*b*e*f)+4*c^2*(
11*d^2*g^2-36*d*e*f*g+36*e^2*f^2))*(c*x^2+b*x+a)^(1/2)/c^3/e^3+1/12*g^3*(-5*b*e*g-14*c*d*g+24*c*e*f)*(e*x+d)*(
c*x^2+b*x+a)^(1/2)/c^2/e^3+1/3*g^4*(e*x+d)^2*(c*x^2+b*x+a)^(1/2)/c/e^3

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Rubi [A]
time = 0.86, antiderivative size = 431, normalized size of antiderivative = 1.00, number of steps used = 8, 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 (8 c^2 e g \left (a e g (4 e f-d g)+b \left (d^2 g^2-4 d e f g+6 e^2 f^2\right )\right )-6 b c e^2 g^2 (2 a e g-b d g+4 b e f)+5 b^3 e^3 g^3-16 c^3 \left (-d^3 g^3+4 d^2 e f g^2-6 d e^2 f^2 g+4 e^3 f^3\right )\right )}{16 c^{7/2} e^4}+\frac {g^2 \sqrt {a+b x+c x^2} \left (-4 c e g (4 a e g-7 b d g+18 b e f)+15 b^2 e^2 g^2+4 c^2 \left (11 d^2 g^2-36 d e f g+36 e^2 f^2\right )\right )}{24 c^3 e^3}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2} (-5 b e g-14 c d g+24 c e f)}{12 c^2 e^3}+\frac {(e f-d g)^4 \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^4 \sqrt {a e^2-b d e+c d^2}}+\frac {g^4 (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^4/((d + e*x)*Sqrt[a + b*x + c*x^2]),x]

[Out]

(g^2*(15*b^2*e^2*g^2 - 4*c*e*g*(18*b*e*f - 7*b*d*g + 4*a*e*g) + 4*c^2*(36*e^2*f^2 - 36*d*e*f*g + 11*d^2*g^2))*
Sqrt[a + b*x + c*x^2])/(24*c^3*e^3) + (g^3*(24*c*e*f - 14*c*d*g - 5*b*e*g)*(d + e*x)*Sqrt[a + b*x + c*x^2])/(1
2*c^2*e^3) + (g^4*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(3*c*e^3) - (g*(5*b^3*e^3*g^3 - 6*b*c*e^2*g^2*(4*b*e*f -
b*d*g + 2*a*e*g) - 16*c^3*(4*e^3*f^3 - 6*d*e^2*f^2*g + 4*d^2*e*f*g^2 - d^3*g^3) + 8*c^2*e*g*(a*e*g*(4*e*f - d*
g) + b*(6*e^2*f^2 - 4*d*e*f*g + d^2*g^2)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(7/2)
*e^4) + ((e*f - d*g)^4*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^4*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)^4}{(d+e x) \sqrt {a+b x+c x^2}} \, dx &=\frac {g^4 (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c e^3}+\frac {\int \frac {\frac {1}{2} e \left (6 c e^3 f^4-d^2 (b d+4 a e) g^4\right )-\frac {1}{2} e g \left (d e (7 b d+8 a e) g^3-c \left (24 e^3 f^3-2 d^3 g^3\right )\right ) x-\frac {1}{2} e^2 g^2 \left (e (11 b d+4 a e) g^2-c \left (36 e^2 f^2-10 d^2 g^2\right )\right ) x^2+\frac {1}{2} e^3 g^3 (24 c e f-14 c d g-5 b e g) x^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{3 c e^4}\\ &=\frac {g^3 (24 c e f-14 c d g-5 b e g) (d+e x) \sqrt {a+b x+c x^2}}{12 c^2 e^3}+\frac {g^4 (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c e^3}+\frac {\int \frac {\frac {1}{4} e^4 \left (24 c^2 e^3 f^4+5 b d e (b d+2 a e) g^4-2 c d g^3 (b d (12 e f-5 d g)+6 a e (4 e f-d g))\right )+\frac {1}{2} e^4 g \left (5 b e^2 (2 b d+a e) g^3+2 c^2 \left (24 e^3 f^3-12 d^2 e f g^2+5 d^3 g^3\right )-c e g^2 (b d (48 e f-19 d g)+2 a e (12 e f+d g))\right ) x+\frac {1}{4} e^5 g^2 \left (15 b^2 e^2 g^2-4 c e g (18 b e f-7 b d g+4 a e g)+4 c^2 \left (36 e^2 f^2-36 d e f g+11 d^2 g^2\right )\right ) x^2}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{6 c^2 e^7}\\ &=\frac {g^2 \left (15 b^2 e^2 g^2-4 c e g (18 b e f-7 b d g+4 a e g)+4 c^2 \left (36 e^2 f^2-36 d e f g+11 d^2 g^2\right )\right ) \sqrt {a+b x+c x^2}}{24 c^3 e^3}+\frac {g^3 (24 c e f-14 c d g-5 b e g) (d+e x) \sqrt {a+b x+c x^2}}{12 c^2 e^3}+\frac {g^4 (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c e^3}+\frac {\int \frac {\frac {3}{8} e^6 \left (16 c^3 e^3 f^4-5 b^3 d e^2 g^4+6 b c d e g^3 (4 b e f-b d g+2 a e g)-8 c^2 d g^2 \left (a e g (4 e f-d g)+b \left (6 e^2 f^2-4 d e f g+d^2 g^2\right )\right )\right )-\frac {3}{8} e^6 g \left (5 b^3 e^3 g^3-6 b c e^2 g^2 (4 b e f-b d g+2 a e g)-16 c^3 \left (4 e^3 f^3-6 d e^2 f^2 g+4 d^2 e f g^2-d^3 g^3\right )+8 c^2 e g \left (a e g (4 e f-d g)+b \left (6 e^2 f^2-4 d e f g+d^2 g^2\right )\right )\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{6 c^3 e^9}\\ &=\frac {g^2 \left (15 b^2 e^2 g^2-4 c e g (18 b e f-7 b d g+4 a e g)+4 c^2 \left (36 e^2 f^2-36 d e f g+11 d^2 g^2\right )\right ) \sqrt {a+b x+c x^2}}{24 c^3 e^3}+\frac {g^3 (24 c e f-14 c d g-5 b e g) (d+e x) \sqrt {a+b x+c x^2}}{12 c^2 e^3}+\frac {g^4 (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c e^3}+\frac {(e f-d g)^4 \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e^4}-\frac {\left (g \left (5 b^3 e^3 g^3-6 b c e^2 g^2 (4 b e f-b d g+2 a e g)-16 c^3 \left (4 e^3 f^3-6 d e^2 f^2 g+4 d^2 e f g^2-d^3 g^3\right )+8 c^2 e g \left (a e g (4 e f-d g)+b \left (6 e^2 f^2-4 d e f g+d^2 g^2\right )\right )\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^3 e^4}\\ &=\frac {g^2 \left (15 b^2 e^2 g^2-4 c e g (18 b e f-7 b d g+4 a e g)+4 c^2 \left (36 e^2 f^2-36 d e f g+11 d^2 g^2\right )\right ) \sqrt {a+b x+c x^2}}{24 c^3 e^3}+\frac {g^3 (24 c e f-14 c d g-5 b e g) (d+e x) \sqrt {a+b x+c x^2}}{12 c^2 e^3}+\frac {g^4 (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c e^3}-\frac {\left (2 (e f-d g)^4\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^4}-\frac {\left (g \left (5 b^3 e^3 g^3-6 b c e^2 g^2 (4 b e f-b d g+2 a e g)-16 c^3 \left (4 e^3 f^3-6 d e^2 f^2 g+4 d^2 e f g^2-d^3 g^3\right )+8 c^2 e g \left (a e g (4 e f-d g)+b \left (6 e^2 f^2-4 d e f g+d^2 g^2\right )\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 )}{8 c^3 e^4}\\ &=\frac {g^2 \left (15 b^2 e^2 g^2-4 c e g (18 b e f-7 b d g+4 a e g)+4 c^2 \left (36 e^2 f^2-36 d e f g+11 d^2 g^2\right )\right ) \sqrt {a+b x+c x^2}}{24 c^3 e^3}+\frac {g^3 (24 c e f-14 c d g-5 b e g) (d+e x) \sqrt {a+b x+c x^2}}{12 c^2 e^3}+\frac {g^4 (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c e^3}-\frac {g \left (5 b^3 e^3 g^3-6 b c e^2 g^2 (4 b e f-b d g+2 a e g)-16 c^3 \left (4 e^3 f^3-6 d e^2 f^2 g+4 d^2 e f g^2-d^3 g^3\right )+8 c^2 e g \left (a e g (4 e f-d g)+b \left (6 e^2 f^2-4 d e f g+d^2 g^2\right )\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{7/2} e^4}+\frac {(e f-d g)^4 \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^4 \sqrt {c d^2-b d e+a e^2}}\\ \end {align*}

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Mathematica [A]
time = 2.17, size = 380, normalized size = 0.88 \begin {gather*} \frac {\frac {2 e g^2 \sqrt {a+x (b+c x)} \left (15 b^2 e^2 g^2-2 c e g (8 a e g+b (36 e f-9 d g+5 e g x))+4 c^2 \left (6 d^2 g^2-3 d e g (8 f+g x)+2 e^2 \left (18 f^2+6 f g x+g^2 x^2\right )\right )\right )}{c^3}+\frac {96 \sqrt {-c d^2+b d e-a e^2} (e f-d g)^4 \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 {3 g \left (5 b^3 e^3 g^3-6 b c e^2 g^2 (4 b e f-b d g+2 a e g)-16 c^3 \left (4 e^3 f^3-6 d e^2 f^2 g+4 d^2 e f g^2-d^3 g^3\right )+8 c^2 e g \left (a e g (4 e f-d g)+b \left (6 e^2 f^2-4 d e f g+d^2 g^2\right )\right )\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{c^{7/2}}}{48 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^4/((d + e*x)*Sqrt[a + b*x + c*x^2]),x]

[Out]

((2*e*g^2*Sqrt[a + x*(b + c*x)]*(15*b^2*e^2*g^2 - 2*c*e*g*(8*a*e*g + b*(36*e*f - 9*d*g + 5*e*g*x)) + 4*c^2*(6*
d^2*g^2 - 3*d*e*g*(8*f + g*x) + 2*e^2*(18*f^2 + 6*f*g*x + g^2*x^2))))/c^3 + (96*Sqrt[-(c*d^2) + b*d*e - a*e^2]
*(e*f - d*g)^4*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)) + (3*g*(5*b^3*e^3*g^3 - 6*b*c*e^2*g^2*(4*b*e*f - b*d*g + 2*a*e*g) - 16*c^3*(4*e^3*f^3 - 6*d*
e^2*f^2*g + 4*d^2*e*f*g^2 - d^3*g^3) + 8*c^2*e*g*(a*e*g*(4*e*f - d*g) + b*(6*e^2*f^2 - 4*d*e*f*g + d^2*g^2)))*
Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/c^(7/2))/(48*e^4)

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Maple [A]
time = 0.17, size = 755, normalized size = 1.75

method	result	size
default	\(-\frac {g \left (-e^{3} g^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \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 )}{6 c}-\frac {2 a \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 )}{3 c}\right )+\left (d \,e^{2} g^{3}-4 e^{3} f \,g^{2}\right ) \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^{2} e \,g^{3}+4 d \,e^{2} f \,g^{2}-6 e^{3} f^{2} 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^{3} g^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {4 d^{2} e f \,g^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+\frac {6 d \,e^{2} f^{2} g \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {4 e^{3} f^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}\right )}{e^{4}}-\frac {\left (d^{4} g^{4}-4 d^{3} e f \,g^{3}+6 d^{2} e^{2} f^{2} g^{2}-4 d \,e^{3} f^{3} g +e^{4} f^{4}\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^{5} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\)	\(755\)
risch	\(\text {Expression too large to display}\)	\(1461\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^4/(e*x+d)/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-g/e^4*(-e^3*g^3*(1/3*x^2/c*(c*x^2+b*x+a)^(1/2)-5/6*b/c*(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)))-2/3*a/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))))+(
d*e^2*g^3-4*e^3*f*g^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^2*e*g^3+4*d*e
^2*f*g^2-6*e^3*f^2*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^3*
g^3*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-4*d^2*e*f*g^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/
2))/c^(1/2)+6*d*e^2*f^2*g*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-4*e^3*f^3*ln((1/2*b+c*x)/c^(1/2)
+(c*x^2+b*x+a)^(1/2))/c^(1/2))-(d^4*g^4-4*d^3*e*f*g^3+6*d^2*e^2*f^2*g^2-4*d*e^3*f^3*g+e^4*f^4)/e^5/((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)^4/(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)^4/(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 )^{4}}{\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)**4/(e*x+d)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((f + g*x)**4/((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)^4/(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 )}^4}{\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)^4/((d + e*x)*(a + b*x + c*x^2)^(1/2)),x)

[Out]

int((f + g*x)^4/((d + e*x)*(a + b*x + c*x^2)^(1/2)), x)

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