3.14.81 \(\int \frac {(b+2 c x) (d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=245 \[ \frac {3 e \left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right )-8 c^2 d e (16 a e+5 b d)+4 b c e^2 (13 a e+12 b d)-15 b^3 e^3+32 c^3 d^3\right )}{32 c^3}+\frac {1}{2} (d+e x)^3 \sqrt {a+b x+c x^2}+\frac {(d+e x)^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{4 c} \]

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Rubi [A]  time = 0.33, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {832, 779, 621, 206} \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right )-8 c^2 d e (16 a e+5 b d)+4 b c e^2 (13 a e+12 b d)-15 b^3 e^3+32 c^3 d^3\right )}{32 c^3}+\frac {3 e \left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{7/2}}+\frac {1}{2} (d+e x)^3 \sqrt {a+b x+c x^2}+\frac {(d+e x)^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*c*d - b*e)*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(4*c) + ((d + e*x)^3*Sqrt[a + b*x + c*x^2])/2 + ((32*c^3*d^3
- 15*b^3*e^3 + 4*b*c*e^2*(12*b*d + 13*a*e) - 8*c^2*d*e*(5*b*d + 16*a*e) + 2*c*e*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*
e*(2*b*d + 3*a*e))*x)*Sqrt[a + b*x + c*x^2])/(32*c^3) + (3*(b^2 - 4*a*c)*e*(16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*
b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(7/2))

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 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 {(b+2 c x) (d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx &=\frac {1}{2} (d+e x)^3 \sqrt {a+b x+c x^2}+\frac {\int \frac {(d+e x)^2 (3 c (b d-2 a e)+3 c (2 c d-b e) x)}{\sqrt {a+b x+c x^2}} \, dx}{4 c}\\ &=\frac {(2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{4 c}+\frac {1}{2} (d+e x)^3 \sqrt {a+b x+c x^2}+\frac {\int \frac {(d+e x) \left (\frac {3}{2} c \left (b^2 d e-20 a c d e+4 b \left (c d^2+a e^2\right )\right )+\frac {3}{2} c \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{12 c^2}\\ &=\frac {(2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{4 c}+\frac {1}{2} (d+e x)^3 \sqrt {a+b x+c x^2}+\frac {\left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c^3}+\frac {\left (3 \left (b^2-4 a c\right ) e \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{64 c^3}\\ &=\frac {(2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{4 c}+\frac {1}{2} (d+e x)^3 \sqrt {a+b x+c x^2}+\frac {\left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c^3}+\frac {\left (3 \left (b^2-4 a c\right ) e \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\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 )}{32 c^3}\\ &=\frac {(2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{4 c}+\frac {1}{2} (d+e x)^3 \sqrt {a+b x+c x^2}+\frac {\left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c^3}+\frac {3 \left (b^2-4 a c\right ) e \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 302, normalized size = 1.23 \begin {gather*} \frac {-4 a^2 c e^2 (-13 b e+32 c d+6 c e x)+a \left (-15 b^3 e^3+2 b^2 c e^2 (24 d+31 e x)+4 b c^2 e \left (-12 d^2-40 d e x+5 e^2 x^2\right )+8 c^3 \left (8 d^3+12 d^2 e x-8 d e^2 x^2-e^3 x^3\right )\right )+x (b+c x) \left (-15 b^3 e^3+2 b^2 c e^2 (24 d+5 e x)-8 b c^2 e \left (6 d^2+4 d e x+e^2 x^2\right )+16 c^3 \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right )}{32 c^3 \sqrt {a+x (b+c x)}}+\frac {3 e \left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{64 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*c*e^2*(32*c*d - 13*b*e + 6*c*e*x) + a*(-15*b^3*e^3 + 2*b^2*c*e^2*(24*d + 31*e*x) + 4*b*c^2*e*(-12*d^2
- 40*d*e*x + 5*e^2*x^2) + 8*c^3*(8*d^3 + 12*d^2*e*x - 8*d*e^2*x^2 - e^3*x^3)) + x*(b + c*x)*(-15*b^3*e^3 + 2*b
^2*c*e^2*(24*d + 5*e*x) - 8*b*c^2*e*(6*d^2 + 4*d*e*x + e^2*x^2) + 16*c^3*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^
3*x^3)))/(32*c^3*Sqrt[a + x*(b + c*x)]) + (3*(b^2 - 4*a*c)*e*(16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*Ar
cTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(64*c^(7/2))

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IntegrateAlgebraic [A]  time = 0.98, size = 263, normalized size = 1.07 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (52 a b c e^3-128 a c^2 d e^2-24 a c^2 e^3 x-15 b^3 e^3+48 b^2 c d e^2+10 b^2 c e^3 x-48 b c^2 d^2 e-32 b c^2 d e^2 x-8 b c^2 e^3 x^2+64 c^3 d^3+96 c^3 d^2 e x+64 c^3 d e^2 x^2+16 c^3 e^3 x^3\right )}{32 c^3}-\frac {3 \left (16 a^2 c^2 e^3-24 a b^2 c e^3+64 a b c^2 d e^2-64 a c^3 d^2 e+5 b^4 e^3-16 b^3 c d e^2+16 b^2 c^2 d^2 e\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{64 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(64*c^3*d^3 - 48*b*c^2*d^2*e + 48*b^2*c*d*e^2 - 128*a*c^2*d*e^2 - 15*b^3*e^3 + 52*a*b*c
*e^3 + 96*c^3*d^2*e*x - 32*b*c^2*d*e^2*x + 10*b^2*c*e^3*x - 24*a*c^2*e^3*x + 64*c^3*d*e^2*x^2 - 8*b*c^2*e^3*x^
2 + 16*c^3*e^3*x^3))/(32*c^3) - (3*(16*b^2*c^2*d^2*e - 64*a*c^3*d^2*e - 16*b^3*c*d*e^2 + 64*a*b*c^2*d*e^2 + 5*
b^4*e^3 - 24*a*b^2*c*e^3 + 16*a^2*c^2*e^3)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(64*c^(7/2))

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fricas [A]  time = 0.72, size = 545, normalized size = 2.22 \begin {gather*} \left [\frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - 16 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e^{3}\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 (16 \, c^{4} e^{3} x^{3} + 64 \, c^{4} d^{3} - 48 \, b c^{3} d^{2} e + 16 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} d e^{2} - {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} e^{3} + 8 \, {\left (8 \, c^{4} d e^{2} - b c^{3} e^{3}\right )} x^{2} + 2 \, {\left (48 \, c^{4} d^{2} e - 16 \, b c^{3} d e^{2} + {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{128 \, c^{4}}, -\frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - 16 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e^{3}\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 (16 \, c^{4} e^{3} x^{3} + 64 \, c^{4} d^{3} - 48 \, b c^{3} d^{2} e + 16 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} d e^{2} - {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} e^{3} + 8 \, {\left (8 \, c^{4} d e^{2} - b c^{3} e^{3}\right )} x^{2} + 2 \, {\left (48 \, c^{4} d^{2} e - 16 \, b c^{3} d e^{2} + {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{64 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/128*(3*(16*(b^2*c^2 - 4*a*c^3)*d^2*e - 16*(b^3*c - 4*a*b*c^2)*d*e^2 + (5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*e^3
)*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*(16*c^4*e^
3*x^3 + 64*c^4*d^3 - 48*b*c^3*d^2*e + 16*(3*b^2*c^2 - 8*a*c^3)*d*e^2 - (15*b^3*c - 52*a*b*c^2)*e^3 + 8*(8*c^4*
d*e^2 - b*c^3*e^3)*x^2 + 2*(48*c^4*d^2*e - 16*b*c^3*d*e^2 + (5*b^2*c^2 - 12*a*c^3)*e^3)*x)*sqrt(c*x^2 + b*x +
a))/c^4, -1/64*(3*(16*(b^2*c^2 - 4*a*c^3)*d^2*e - 16*(b^3*c - 4*a*b*c^2)*d*e^2 + (5*b^4 - 24*a*b^2*c + 16*a^2*
c^2)*e^3)*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*(16*c^4*
e^3*x^3 + 64*c^4*d^3 - 48*b*c^3*d^2*e + 16*(3*b^2*c^2 - 8*a*c^3)*d*e^2 - (15*b^3*c - 52*a*b*c^2)*e^3 + 8*(8*c^
4*d*e^2 - b*c^3*e^3)*x^2 + 2*(48*c^4*d^2*e - 16*b*c^3*d*e^2 + (5*b^2*c^2 - 12*a*c^3)*e^3)*x)*sqrt(c*x^2 + b*x
+ a))/c^4]

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giac [A]  time = 0.26, size = 252, normalized size = 1.03 \begin {gather*} \frac {1}{32} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, x e^{3} + \frac {8 \, c^{3} d e^{2} - b c^{2} e^{3}}{c^{3}}\right )} x + \frac {48 \, c^{3} d^{2} e - 16 \, b c^{2} d e^{2} + 5 \, b^{2} c e^{3} - 12 \, a c^{2} e^{3}}{c^{3}}\right )} x + \frac {64 \, c^{3} d^{3} - 48 \, b c^{2} d^{2} e + 48 \, b^{2} c d e^{2} - 128 \, a c^{2} d e^{2} - 15 \, b^{3} e^{3} + 52 \, a b c e^{3}}{c^{3}}\right )} - \frac {3 \, {\left (16 \, b^{2} c^{2} d^{2} e - 64 \, a c^{3} d^{2} e - 16 \, b^{3} c d e^{2} + 64 \, a b c^{2} d e^{2} + 5 \, b^{4} e^{3} - 24 \, a b^{2} c e^{3} + 16 \, a^{2} c^{2} e^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{64 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*sqrt(c*x^2 + b*x + a)*(2*(4*(2*x*e^3 + (8*c^3*d*e^2 - b*c^2*e^3)/c^3)*x + (48*c^3*d^2*e - 16*b*c^2*d*e^2
+ 5*b^2*c*e^3 - 12*a*c^2*e^3)/c^3)*x + (64*c^3*d^3 - 48*b*c^2*d^2*e + 48*b^2*c*d*e^2 - 128*a*c^2*d*e^2 - 15*b^
3*e^3 + 52*a*b*c*e^3)/c^3) - 3/64*(16*b^2*c^2*d^2*e - 64*a*c^3*d^2*e - 16*b^3*c*d*e^2 + 64*a*b*c^2*d*e^2 + 5*b
^4*e^3 - 24*a*b^2*c*e^3 + 16*a^2*c^2*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.07, size = 539, normalized size = 2.20 \begin {gather*} \frac {\sqrt {c \,x^{2}+b x +a}\, e^{3} x^{3}}{2}-\frac {\sqrt {c \,x^{2}+b x +a}\, b \,e^{3} x^{2}}{4 c}+2 \sqrt {c \,x^{2}+b x +a}\, d \,e^{2} x^{2}+\frac {3 a^{2} e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}-\frac {9 a \,b^{2} e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}+\frac {3 a b d \,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 {3 a \,d^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+\frac {15 b^{4} e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{64 c^{\frac {7}{2}}}-\frac {3 b^{3} d \,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 {3 b^{2} d^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, a \,e^{3} x}{4 c}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{2} e^{3} x}{16 c^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, b d \,e^{2} x}{c}+3 \sqrt {c \,x^{2}+b x +a}\, d^{2} e x +\frac {13 \sqrt {c \,x^{2}+b x +a}\, a b \,e^{3}}{8 c^{2}}-\frac {4 \sqrt {c \,x^{2}+b x +a}\, a d \,e^{2}}{c}-\frac {15 \sqrt {c \,x^{2}+b x +a}\, b^{3} e^{3}}{32 c^{3}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, b^{2} d \,e^{2}}{2 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b \,d^{2} e}{2 c}+2 \sqrt {c \,x^{2}+b x +a}\, d^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d^2*e+3/4*b^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b
*x+a)^(1/2))*d^2*e+2*(c*x^2+b*x+a)^(1/2)*d^3-3/4*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e^2
-b/c*x*(c*x^2+b*x+a)^(1/2)*d*e^2-9/8/c^(5/2)*e^3*b^2*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-15/32/c^3*e
^3*b^3*(c*x^2+b*x+a)^(1/2)+13/8/c^2*e^3*b*a*(c*x^2+b*x+a)^(1/2)-3/4/c*e^3*a*x*(c*x^2+b*x+a)^(1/2)+3/4/c^(3/2)*
e^3*a^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3/2*b/c*(c*x^2+b*x+a)^(1/2)*d^2*e+2*x^2*(c*x^2+b*x+a)^(1/2
)*d*e^2+3*x*(c*x^2+b*x+a)^(1/2)*d^2*e+5/16/c^2*e^3*b^2*x*(c*x^2+b*x+a)^(1/2)+15/64/c^(7/2)*e^3*b^4*ln((c*x+1/2
*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/4/c*e^3*b*x^2*(c*x^2+b*x+a)^(1/2)+1/2*e^3*x^3*(c*x^2+b*x+a)^(1/2)-4*a/c*(c*
x^2+b*x+a)^(1/2)*d*e^2+3/2*b^2/c^2*(c*x^2+b*x+a)^(1/2)*d*e^2+3*b/c^(3/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a
)^(1/2))*d*e^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((2*c*x+b)*(e*x+d)^3/(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 (b+2\,c\,x\right )\,{\left (d+e\,x\right )}^3}{\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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