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

Optimal. Leaf size=296 \[ \frac {\left (48 c^2 e^2 \left (a^2 e^2+8 a b d e+6 b^2 d^2\right )-40 b^2 c e^3 (3 a e+4 b d)-128 c^3 d^2 e (3 a e+2 b d)+35 b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{9/2}}+\frac {e \sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (9 a e+26 b d)+35 b^2 e^2+104 c^2 d^2\right )-8 c^2 d e (64 a e+101 b d)+20 b c e^2 (11 a e+24 b d)-105 b^3 e^3+608 c^3 d^3\right )}{192 c^4}+\frac {7 e (d+e x)^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{24 c^2}+\frac {e (d+e x)^3 \sqrt {a+b x+c x^2}}{4 c} \]

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Rubi [A]  time = 0.38, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {742, 832, 779, 621, 206} \begin {gather*} \frac {\left (48 c^2 e^2 \left (a^2 e^2+8 a b d e+6 b^2 d^2\right )-40 b^2 c e^3 (3 a e+4 b d)-128 c^3 d^2 e (3 a e+2 b d)+35 b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{9/2}}+\frac {e \sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (9 a e+26 b d)+35 b^2 e^2+104 c^2 d^2\right )-8 c^2 d e (64 a e+101 b d)+20 b c e^2 (11 a e+24 b d)-105 b^3 e^3+608 c^3 d^3\right )}{192 c^4}+\frac {7 e (d+e x)^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{24 c^2}+\frac {e (d+e x)^3 \sqrt {a+b x+c x^2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*e*(2*c*d - b*e)*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(24*c^2) + (e*(d + e*x)^3*Sqrt[a + b*x + c*x^2])/(4*c) +
(e*(608*c^3*d^3 - 105*b^3*e^3 + 20*b*c*e^2*(24*b*d + 11*a*e) - 8*c^2*d*e*(101*b*d + 64*a*e) + 2*c*e*(104*c^2*
d^2 + 35*b^2*e^2 - 4*c*e*(26*b*d + 9*a*e))*x)*Sqrt[a + b*x + c*x^2])/(192*c^4) + ((128*c^4*d^4 + 35*b^4*e^4 -
128*c^3*d^2*e*(2*b*d + 3*a*e) - 40*b^2*c*e^3*(4*b*d + 3*a*e) + 48*c^2*e^2*(6*b^2*d^2 + 8*a*b*d*e + a^2*e^2))*A
rcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(9/2))

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 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, 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]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 832

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^4}{\sqrt {a+b x+c x^2}} \, dx &=\frac {e (d+e x)^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\int \frac {(d+e x)^2 \left (\frac {1}{2} \left (8 c d^2-e (b d+6 a e)\right )+\frac {7}{2} e (2 c d-b e) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{4 c}\\ &=\frac {7 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {e (d+e x)^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\int \frac {(d+e x) \left (\frac {1}{4} \left (48 c^2 d^3+7 b e^2 (b d+4 a e)-4 c d e (5 b d+23 a e)\right )+\frac {1}{4} e \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{12 c^2}\\ &=\frac {7 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {e (d+e x)^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {e \left (608 c^3 d^3-105 b^3 e^3+20 b c e^2 (24 b d+11 a e)-8 c^2 d e (101 b d+64 a e)+2 c e \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (\frac {3}{8} b^2 e^2 \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right )-\frac {1}{2} a c e^2 \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right )+2 c \left (\frac {1}{2} c d \left (48 c^2 d^3+7 b e^2 (b d+4 a e)-4 c d e (5 b d+23 a e)\right )-b \left (\frac {1}{4} d e \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right )+\frac {1}{4} e \left (48 c^2 d^3+7 b e^2 (b d+4 a e)-4 c d e (5 b d+23 a e)\right )\right )\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{48 c^4}\\ &=\frac {7 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {e (d+e x)^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {e \left (608 c^3 d^3-105 b^3 e^3+20 b c e^2 (24 b d+11 a e)-8 c^2 d e (101 b d+64 a e)+2 c e \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (\frac {3}{8} b^2 e^2 \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right )-\frac {1}{2} a c e^2 \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right )+2 c \left (\frac {1}{2} c d \left (48 c^2 d^3+7 b e^2 (b d+4 a e)-4 c d e (5 b d+23 a e)\right )-b \left (\frac {1}{4} d e \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right )+\frac {1}{4} e \left (48 c^2 d^3+7 b e^2 (b d+4 a e)-4 c d e (5 b d+23 a e)\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 )}{24 c^4}\\ &=\frac {7 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {e (d+e x)^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {e \left (608 c^3 d^3-105 b^3 e^3+20 b c e^2 (24 b d+11 a e)-8 c^2 d e (101 b d+64 a e)+2 c e \left (104 c^2 d^2+35 b^2 e^2-4 c e (26 b d+9 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (128 c^4 d^4+35 b^4 e^4-128 c^3 d^2 e (2 b d+3 a e)-40 b^2 c e^3 (4 b d+3 a e)+48 c^2 e^2 \left (6 b^2 d^2+8 a b d e+a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 352, normalized size = 1.19 \begin {gather*} \frac {\left (48 c^2 e^2 \left (a^2 e^2+8 a b d e+6 b^2 d^2\right )-40 b^2 c e^3 (3 a e+4 b d)-128 c^3 d^2 e (3 a e+2 b d)+35 b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{128 c^{9/2}}+\frac {e \left (-4 a^2 c e^2 (2 c (64 d+9 e x)-55 b e)+a \left (-105 b^3 e^3+10 b^2 c e^2 (48 d+29 e x)+4 b c^2 e \left (-216 d^2-208 d e x+23 e^2 x^2\right )+8 c^3 \left (96 d^3+72 d^2 e x-32 d e^2 x^2-3 e^3 x^3\right )\right )+x (b+c x) \left (-105 b^3 e^3+10 b^2 c e^2 (48 d+7 e x)-8 b c^2 e \left (108 d^2+40 d e x+7 e^2 x^2\right )+16 c^3 \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )\right )}{192 c^4 \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(-4*a^2*c*e^2*(-55*b*e + 2*c*(64*d + 9*e*x)) + a*(-105*b^3*e^3 + 10*b^2*c*e^2*(48*d + 29*e*x) + 4*b*c^2*e*(
-216*d^2 - 208*d*e*x + 23*e^2*x^2) + 8*c^3*(96*d^3 + 72*d^2*e*x - 32*d*e^2*x^2 - 3*e^3*x^3)) + x*(b + c*x)*(-1
05*b^3*e^3 + 10*b^2*c*e^2*(48*d + 7*e*x) - 8*b*c^2*e*(108*d^2 + 40*d*e*x + 7*e^2*x^2) + 16*c^3*(48*d^3 + 36*d^
2*e*x + 16*d*e^2*x^2 + 3*e^3*x^3))))/(192*c^4*Sqrt[a + x*(b + c*x)]) + ((128*c^4*d^4 + 35*b^4*e^4 - 128*c^3*d^
2*e*(2*b*d + 3*a*e) - 40*b^2*c*e^3*(4*b*d + 3*a*e) + 48*c^2*e^2*(6*b^2*d^2 + 8*a*b*d*e + a^2*e^2))*ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(128*c^(9/2))

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IntegrateAlgebraic [A]  time = 0.93, size = 290, normalized size = 0.98 \begin {gather*} \frac {\left (-48 a^2 c^2 e^4+120 a b^2 c e^4-384 a b c^2 d e^3+384 a c^3 d^2 e^2-35 b^4 e^4+160 b^3 c d e^3-288 b^2 c^2 d^2 e^2+256 b c^3 d^3 e-128 c^4 d^4\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{128 c^{9/2}}+\frac {\sqrt {a+b x+c x^2} \left (220 a b c e^4-512 a c^2 d e^3-72 a c^2 e^4 x-105 b^3 e^4+480 b^2 c d e^3+70 b^2 c e^4 x-864 b c^2 d^2 e^2-320 b c^2 d e^3 x-56 b c^2 e^4 x^2+768 c^3 d^3 e+576 c^3 d^2 e^2 x+256 c^3 d e^3 x^2+48 c^3 e^4 x^3\right )}{192 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(768*c^3*d^3*e - 864*b*c^2*d^2*e^2 + 480*b^2*c*d*e^3 - 512*a*c^2*d*e^3 - 105*b^3*e^4 +
220*a*b*c*e^4 + 576*c^3*d^2*e^2*x - 320*b*c^2*d*e^3*x + 70*b^2*c*e^4*x - 72*a*c^2*e^4*x + 256*c^3*d*e^3*x^2 -
56*b*c^2*e^4*x^2 + 48*c^3*e^4*x^3))/(192*c^4) + ((-128*c^4*d^4 + 256*b*c^3*d^3*e - 288*b^2*c^2*d^2*e^2 + 384*a
*c^3*d^2*e^2 + 160*b^3*c*d*e^3 - 384*a*b*c^2*d*e^3 - 35*b^4*e^4 + 120*a*b^2*c*e^4 - 48*a^2*c^2*e^4)*Log[b + 2*
c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(128*c^(9/2))

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fricas [A]  time = 0.53, size = 599, normalized size = 2.02 \begin {gather*} \left [\frac {3 \, {\left (128 \, c^{4} d^{4} - 256 \, b c^{3} d^{3} e + 96 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e^{2} - 32 \, {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d e^{3} + {\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} e^{4}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} e^{4} x^{3} + 768 \, c^{4} d^{3} e - 864 \, b c^{3} d^{2} e^{2} + 32 \, {\left (15 \, b^{2} c^{2} - 16 \, a c^{3}\right )} d e^{3} - 5 \, {\left (21 \, b^{3} c - 44 \, a b c^{2}\right )} e^{4} + 8 \, {\left (32 \, c^{4} d e^{3} - 7 \, b c^{3} e^{4}\right )} x^{2} + 2 \, {\left (288 \, c^{4} d^{2} e^{2} - 160 \, b c^{3} d e^{3} + {\left (35 \, b^{2} c^{2} - 36 \, a c^{3}\right )} e^{4}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{5}}, -\frac {3 \, {\left (128 \, c^{4} d^{4} - 256 \, b c^{3} d^{3} e + 96 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e^{2} - 32 \, {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d e^{3} + {\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} e^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (48 \, c^{4} e^{4} x^{3} + 768 \, c^{4} d^{3} e - 864 \, b c^{3} d^{2} e^{2} + 32 \, {\left (15 \, b^{2} c^{2} - 16 \, a c^{3}\right )} d e^{3} - 5 \, {\left (21 \, b^{3} c - 44 \, a b c^{2}\right )} e^{4} + 8 \, {\left (32 \, c^{4} d e^{3} - 7 \, b c^{3} e^{4}\right )} x^{2} + 2 \, {\left (288 \, c^{4} d^{2} e^{2} - 160 \, b c^{3} d e^{3} + {\left (35 \, b^{2} c^{2} - 36 \, a c^{3}\right )} e^{4}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(128*c^4*d^4 - 256*b*c^3*d^3*e + 96*(3*b^2*c^2 - 4*a*c^3)*d^2*e^2 - 32*(5*b^3*c - 12*a*b*c^2)*d*e^3
+ (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*e^4)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2
*c*x + b)*sqrt(c) - 4*a*c) + 4*(48*c^4*e^4*x^3 + 768*c^4*d^3*e - 864*b*c^3*d^2*e^2 + 32*(15*b^2*c^2 - 16*a*c^3
)*d*e^3 - 5*(21*b^3*c - 44*a*b*c^2)*e^4 + 8*(32*c^4*d*e^3 - 7*b*c^3*e^4)*x^2 + 2*(288*c^4*d^2*e^2 - 160*b*c^3*
d*e^3 + (35*b^2*c^2 - 36*a*c^3)*e^4)*x)*sqrt(c*x^2 + b*x + a))/c^5, -1/384*(3*(128*c^4*d^4 - 256*b*c^3*d^3*e +
96*(3*b^2*c^2 - 4*a*c^3)*d^2*e^2 - 32*(5*b^3*c - 12*a*b*c^2)*d*e^3 + (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*e^4)
*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(48*c^4*e^4*x^3 +
768*c^4*d^3*e - 864*b*c^3*d^2*e^2 + 32*(15*b^2*c^2 - 16*a*c^3)*d*e^3 - 5*(21*b^3*c - 44*a*b*c^2)*e^4 + 8*(32*
c^4*d*e^3 - 7*b*c^3*e^4)*x^2 + 2*(288*c^4*d^2*e^2 - 160*b*c^3*d*e^3 + (35*b^2*c^2 - 36*a*c^3)*e^4)*x)*sqrt(c*x
^2 + b*x + a))/c^5]

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giac [A]  time = 0.66, size = 276, normalized size = 0.93 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, x {\left (\frac {6 \, x e^{4}}{c} + \frac {32 \, c^{3} d e^{3} - 7 \, b c^{2} e^{4}}{c^{4}}\right )} + \frac {288 \, c^{3} d^{2} e^{2} - 160 \, b c^{2} d e^{3} + 35 \, b^{2} c e^{4} - 36 \, a c^{2} e^{4}}{c^{4}}\right )} x + \frac {768 \, c^{3} d^{3} e - 864 \, b c^{2} d^{2} e^{2} + 480 \, b^{2} c d e^{3} - 512 \, a c^{2} d e^{3} - 105 \, b^{3} e^{4} + 220 \, a b c e^{4}}{c^{4}}\right )} - \frac {{\left (128 \, c^{4} d^{4} - 256 \, b c^{3} d^{3} e + 288 \, b^{2} c^{2} d^{2} e^{2} - 384 \, a c^{3} d^{2} e^{2} - 160 \, b^{3} c d e^{3} + 384 \, a b c^{2} d e^{3} + 35 \, b^{4} e^{4} - 120 \, a b^{2} c e^{4} + 48 \, a^{2} c^{2} e^{4}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*x*(6*x*e^4/c + (32*c^3*d*e^3 - 7*b*c^2*e^4)/c^4) + (288*c^3*d^2*e^2 - 160*b*
c^2*d*e^3 + 35*b^2*c*e^4 - 36*a*c^2*e^4)/c^4)*x + (768*c^3*d^3*e - 864*b*c^2*d^2*e^2 + 480*b^2*c*d*e^3 - 512*a
*c^2*d*e^3 - 105*b^3*e^4 + 220*a*b*c*e^4)/c^4) - 1/128*(128*c^4*d^4 - 256*b*c^3*d^3*e + 288*b^2*c^2*d^2*e^2 -
384*a*c^3*d^2*e^2 - 160*b^3*c*d*e^3 + 384*a*b*c^2*d*e^3 + 35*b^4*e^4 - 120*a*b^2*c*e^4 + 48*a^2*c^2*e^4)*log(a
bs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)

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maple [B]  time = 0.06, size = 627, normalized size = 2.12 \begin {gather*} \frac {\sqrt {c \,x^{2}+b x +a}\, e^{4} x^{3}}{4 c}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, b \,e^{4} x^{2}}{24 c^{2}}+\frac {4 \sqrt {c \,x^{2}+b x +a}\, d \,e^{3} x^{2}}{3 c}+\frac {3 a^{2} e^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}-\frac {15 a \,b^{2} e^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}+\frac {3 a b d \,e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {5}{2}}}-\frac {3 a \,d^{2} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {35 b^{4} e^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {9}{2}}}-\frac {5 b^{3} d \,e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {7}{2}}}+\frac {9 b^{2} d^{2} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {5}{2}}}-\frac {2 b \,d^{3} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {d^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, a \,e^{4} x}{8 c^{2}}+\frac {35 \sqrt {c \,x^{2}+b x +a}\, b^{2} e^{4} x}{96 c^{3}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, b d \,e^{3} x}{3 c^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, d^{2} e^{2} x}{c}+\frac {55 \sqrt {c \,x^{2}+b x +a}\, a b \,e^{4}}{48 c^{3}}-\frac {8 \sqrt {c \,x^{2}+b x +a}\, a d \,e^{3}}{3 c^{2}}-\frac {35 \sqrt {c \,x^{2}+b x +a}\, b^{3} e^{4}}{64 c^{4}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{2} d \,e^{3}}{2 c^{3}}-\frac {9 \sqrt {c \,x^{2}+b x +a}\, b \,d^{2} e^{2}}{2 c^{2}}+\frac {4 \sqrt {c \,x^{2}+b x +a}\, d^{3} e}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^4*x^3/c*(c*x^2+b*x+a)^(1/2)-7/24*e^4*b/c^2*x^2*(c*x^2+b*x+a)^(1/2)+35/96*e^4*b^2/c^3*x*(c*x^2+b*x+a)^(1/
2)-35/64*e^4*b^3/c^4*(c*x^2+b*x+a)^(1/2)+35/128*e^4*b^4/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-15
/16*e^4*b^2/c^(7/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+55/48*e^4*b/c^3*a*(c*x^2+b*x+a)^(1/2)-3/8*e^
4*a/c^2*x*(c*x^2+b*x+a)^(1/2)+3/8*e^4*a^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+4/3*d*e^3*x^2/c*
(c*x^2+b*x+a)^(1/2)-5/3*d*e^3*b/c^2*x*(c*x^2+b*x+a)^(1/2)+5/2*d*e^3*b^2/c^3*(c*x^2+b*x+a)^(1/2)-5/4*d*e^3*b^3/
c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3*d*e^3*b/c^(5/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(
1/2))-8/3*d*e^3*a/c^2*(c*x^2+b*x+a)^(1/2)+3*d^2*e^2*x/c*(c*x^2+b*x+a)^(1/2)-9/2*d^2*e^2*b/c^2*(c*x^2+b*x+a)^(1
/2)+9/4*d^2*e^2*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3*d^2*e^2*a/c^(3/2)*ln((c*x+1/2*b)/c^(
1/2)+(c*x^2+b*x+a)^(1/2))+4*d^3*e/c*(c*x^2+b*x+a)^(1/2)-2*d^3*e*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)
^(1/2))+d^4*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((e*x+d)^4/(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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^4}{\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d + e*x)^4/(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 (d + e x\right )^{4}}{\sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((d + e*x)**4/sqrt(a + b*x + c*x**2), x)

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