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

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

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Rubi [A]  time = 1.37, 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 = {1653, 843, 621, 206, 724} \begin {gather*} \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 \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 (4 d^2 e f g^2-d^3 g^3-6 d e^2 f^2 g+4 e^3 f^3\right )\right )}{16 c^{7/2} e^4}+\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 206

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

Rule 621

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 724

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 843

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 1653

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 ) \operatorname {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 ) \operatorname {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 = 0.89, size = 553, normalized size = 1.28 \begin {gather*} \frac {\frac {6 e^2 g (e f-d g) \left (\left (-4 c g (a g+2 b f)+3 b^2 g^2+8 c^2 f^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+6 \sqrt {c} g \sqrt {a+x (b+c x)} (2 c f-b g)\right )}{c^{5/2}}+\frac {e^3 g \left (\frac {2 g \sqrt {a+x (b+c x)} \left (-2 c g (8 a g+27 b f+5 b g x)+15 b^2 g^2+4 c^2 f (16 f+5 g x)\right )}{c^2}+\frac {3 (2 c f-b g) \left (-4 c g (3 a g+2 b f)+5 b^2 g^2+8 c^2 f^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{c^{5/2}}\right )}{c}+\frac {24 e g (2 c f-b g) (e f-d g)^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{c^{3/2}}+\frac {48 (e f-d g)^4 \tanh ^{-1}\left (\frac {-2 a e+b (d-e x)+2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{\sqrt {e (a e-b d)+c d^2}}+\frac {24 e^2 g^2 (f+g x) \sqrt {a+x (b+c x)} (e f-d g)}{c}+\frac {48 e g^2 \sqrt {a+x (b+c x)} (e f-d g)^2}{c}+\frac {48 g (e f-d g)^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}+\frac {16 e^3 g^2 (f+g x)^2 \sqrt {a+x (b+c x)}}{c}}{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]

((48*e*g^2*(e*f - d*g)^2*Sqrt[a + x*(b + c*x)])/c + (24*e^2*g^2*(e*f - d*g)*(f + g*x)*Sqrt[a + x*(b + c*x)])/c
+ (16*e^3*g^2*(f + g*x)^2*Sqrt[a + x*(b + c*x)])/c + (24*e*g*(2*c*f - b*g)*(e*f - d*g)^2*ArcTanh[(b + 2*c*x)/
(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/c^(3/2) + (48*g*(e*f - d*g)^3*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b
+ c*x)])])/Sqrt[c] + (6*e^2*g*(e*f - d*g)*(6*Sqrt[c]*g*(2*c*f - b*g)*Sqrt[a + x*(b + c*x)] + (8*c^2*f^2 + 3*b
^2*g^2 - 4*c*g*(2*b*f + a*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/c^(5/2) + (e^3*g*((2*g*
Sqrt[a + x*(b + c*x)]*(15*b^2*g^2 + 4*c^2*f*(16*f + 5*g*x) - 2*c*g*(27*b*f + 8*a*g + 5*b*g*x)))/c^2 + (3*(2*c*
f - b*g)*(8*c^2*f^2 + 5*b^2*g^2 - 4*c*g*(2*b*f + 3*a*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])
])/c^(5/2)))/c + (48*(e*f - d*g)^4*ArcTanh[(-2*a*e + 2*c*d*x + b*(d - e*x))/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*
Sqrt[a + x*(b + c*x)])])/Sqrt[c*d^2 + e*(-(b*d) + a*e)])/(48*e^4)

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IntegrateAlgebraic [A]  time = 2.71, size = 509, normalized size = 1.18 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-16 a c e^2 g^4+15 b^2 e^2 g^4+18 b c d e g^4-72 b c e^2 f g^3-10 b c e^2 g^4 x+24 c^2 d^2 g^4-96 c^2 d e f g^3-12 c^2 d e g^4 x+144 c^2 e^2 f^2 g^2+48 c^2 e^2 f g^3 x+8 c^2 e^2 g^4 x^2\right )}{24 c^3 e^3}+\frac {\log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right ) \left (-12 a b c e^3 g^4-8 a c^2 d e^2 g^4+32 a c^2 e^3 f g^3+5 b^3 e^3 g^4+6 b^2 c d e^2 g^4-24 b^2 c e^3 f g^3+8 b c^2 d^2 e g^4-32 b c^2 d e^2 f g^3+48 b c^2 e^3 f^2 g^2+16 c^3 d^3 g^4-64 c^3 d^2 e f g^3+96 c^3 d e^2 f^2 g^2-64 c^3 e^3 f^3 g\right )}{16 c^{7/2} e^4}+\frac {2 \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 ) \sqrt {-a e^2+b d e-c d^2} \tan ^{-1}\left (\frac {-e \sqrt {a+b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2+b d e-c d^2}}\right )}{e^4 \left (a e^2-b d e+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(144*c^2*e^2*f^2*g^2 - 96*c^2*d*e*f*g^3 - 72*b*c*e^2*f*g^3 + 24*c^2*d^2*g^4 + 18*b*c*d*
e*g^4 + 15*b^2*e^2*g^4 - 16*a*c*e^2*g^4 + 48*c^2*e^2*f*g^3*x - 12*c^2*d*e*g^4*x - 10*b*c*e^2*g^4*x + 8*c^2*e^2
*g^4*x^2))/(24*c^3*e^3) + (2*Sqrt[-(c*d^2) + b*d*e - a*e^2]*(e^4*f^4 - 4*d*e^3*f^3*g + 6*d^2*e^2*f^2*g^2 - 4*d
^3*e*f*g^3 + d^4*g^4)*ArcTan[(Sqrt[c]*d + Sqrt[c]*e*x - e*Sqrt[a + b*x + c*x^2])/Sqrt[-(c*d^2) + b*d*e - a*e^2
]])/(e^4*(c*d^2 - b*d*e + a*e^2)) + ((-64*c^3*e^3*f^3*g + 96*c^3*d*e^2*f^2*g^2 + 48*b*c^2*e^3*f^2*g^2 - 64*c^3
*d^2*e*f*g^3 - 32*b*c^2*d*e^2*f*g^3 - 24*b^2*c*e^3*f*g^3 + 32*a*c^2*e^3*f*g^3 + 16*c^3*d^3*g^4 + 8*b*c^2*d^2*e
*g^4 + 6*b^2*c*d*e^2*g^4 - 8*a*c^2*d*e^2*g^4 + 5*b^3*e^3*g^4 - 12*a*b*c*e^3*g^4)*Log[b + 2*c*x - 2*Sqrt[c]*Sqr
t[a + b*x + c*x^2]])/(16*c^(7/2)*e^4)

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fricas [F(-1)]  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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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,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.03, size = 1597, normalized size = 3.71

result too large to display

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)

[Out]

-1/e/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2
)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*f^4+2*g^3/e^2*b/c^(3/
2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*f+2*g^3/e*x/c*(c*x^2+b*x+a)^(1/2)*f+3/4*g^4/e^2*b/c^2*(c*x^2+
b*x+a)^(1/2)*d-3*g^3/e*b/c^2*(c*x^2+b*x+a)^(1/2)*f-3/8*g^4/e^2*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a
)^(1/2))*d+3/2*g^3/e*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*f+1/2*g^4/e^2*a/c^(3/2)*ln((c*x+1
/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d-2*g^3/e*a/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*f-4*g^3/e^2
/c*(c*x^2+b*x+a)^(1/2)*d*f-1/2*g^4/e^3*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d^2-3*g^2/e*b/c^(
3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*f^2+4*g^3/e^3*d^2*f*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2
))/c^(1/2)-6*g^2/e^2*d*f^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+4/e^4/((a*e^2-b*d*e+c*d^2)/e^2)
^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e
-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*g^3*f*d^3-6/e^3/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(
((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x
+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*d^2*f^2*g^2+4/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*
c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+
(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*d*f^3*g-1/e^5/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)
/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e
+c*d^2)/e^2)^(1/2))/(x+d/e))*g^4*d^4+1/3*g^4/e*x^2/c*(c*x^2+b*x+a)^(1/2)-5/12*g^4/e*b/c^2*x*(c*x^2+b*x+a)^(1/2
)+3/4*g^4/e*b/c^(5/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/2*g^4/e^2*x/c*(c*x^2+b*x+a)^(1/2)*d+5/8*
g^4/e*b^2/c^3*(c*x^2+b*x+a)^(1/2)-5/16*g^4/e*b^3/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2/3*g^4/e
/c^2*a*(c*x^2+b*x+a)^(1/2)+g^4/e^3/c*(c*x^2+b*x+a)^(1/2)*d^2+6*g^2/e/c*(c*x^2+b*x+a)^(1/2)*f^2-g^4/e^4*d^3*ln(
(c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+4*g/e*f^3*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)

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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((b/e-(2*c*d)/e^2)^2>0)', see `
assume?` for more details)Is (b/e-(2*c*d)/e^2)^2    -(4*c       *((-(b*d)/e)        +(c*d^2)/e^2+a))     /e^2
zero or nonzero?

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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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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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