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

Optimal. Leaf size=227 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) \left (8 a d e+4 b c d-5 b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{128 d^{7/2} (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 d x+e) \sqrt {c+d x^2+e x} \left (8 a d e+4 b c d-5 b e^2\right )}{64 d^3 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2} (8 a d+6 b d x-5 b e)}{24 d^2 (a+b x)} \]

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Rubi [A]  time = 0.13, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {1000, 779, 612, 621, 206} \begin {gather*} -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 d x+e) \sqrt {c+d x^2+e x} \left (8 a d e+4 b c d-5 b e^2\right )}{64 d^3 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) \left (8 a d e+4 b c d-5 b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{128 d^{7/2} (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2} (8 a d+6 b d x-5 b e)}{24 d^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2],x]

[Out]

-((4*b*c*d + 8*a*d*e - 5*b*e^2)*(e + 2*d*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2])/(64*d^3*(a +
b*x)) + ((8*a*d - 5*b*e + 6*b*d*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(c + e*x + d*x^2)^(3/2))/(24*d^2*(a + b*x)) -
((4*c*d - e^2)*(4*b*c*d + 8*a*d*e - 5*b*e^2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqr
t[c + e*x + d*x^2])])/(128*d^(7/2)*(a + b*x))

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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 1000

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_)
, x_Symbol] :> Dist[(a + b*x + c*x^2)^FracPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(g + h*x
)^m*(b + 2*c*x)^(2*p)*(d + e*x + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q}, x] && EqQ[b^2 -
4*a*c, 0]

Rubi steps

\begin {align*} \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x \left (2 a b+2 b^2 x\right ) \sqrt {c+e x+d x^2} \, dx}{2 a b+2 b^2 x}\\ &=\frac {(8 a d-5 b e+6 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac {\left (b \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \sqrt {c+e x+d x^2} \, dx}{8 d^2 \left (2 a b+2 b^2 x\right )}\\ &=-\frac {\left (4 b c d+8 a d e-5 b e^2\right ) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{64 d^3 (a+b x)}+\frac {(8 a d-5 b e+6 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac {\left (b \left (4 c d-e^2\right ) \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{\sqrt {c+e x+d x^2}} \, dx}{64 d^3 \left (2 a b+2 b^2 x\right )}\\ &=-\frac {\left (4 b c d+8 a d e-5 b e^2\right ) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{64 d^3 (a+b x)}+\frac {(8 a d-5 b e+6 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac {\left (b \left (4 c d-e^2\right ) \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{4 d-x^2} \, dx,x,\frac {e+2 d x}{\sqrt {c+e x+d x^2}}\right )}{32 d^3 \left (2 a b+2 b^2 x\right )}\\ &=-\frac {\left (4 b c d+8 a d e-5 b e^2\right ) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{64 d^3 (a+b x)}+\frac {(8 a d-5 b e+6 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac {\left (4 c d-e^2\right ) \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{128 d^{7/2} (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 147, normalized size = 0.65 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left ((c+x (d x+e))^{3/2} (8 a d+6 b d x-5 b e)-\frac {3 \left (8 a d e+4 b c d-5 b e^2\right ) \left (\left (4 c d-e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )+2 \sqrt {d} (2 d x+e) \sqrt {c+x (d x+e)}\right )}{16 d^{3/2}}\right )}{24 d^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2],x]

[Out]

(Sqrt[(a + b*x)^2]*((8*a*d - 5*b*e + 6*b*d*x)*(c + x*(e + d*x))^(3/2) - (3*(4*b*c*d + 8*a*d*e - 5*b*e^2)*(2*Sq
rt[d]*(e + 2*d*x)*Sqrt[c + x*(e + d*x)] + (4*c*d - e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])]
))/(16*d^(3/2))))/(24*d^2*(a + b*x))

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IntegrateAlgebraic [A]  time = 0.70, size = 196, normalized size = 0.86 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (\frac {\left (32 a c d^2 e-8 a d e^3+16 b c^2 d^2-24 b c d e^2+5 b e^4\right ) \log \left (-2 \sqrt {d} \sqrt {c+d x^2+e x}+2 d x+e\right )}{128 d^{7/2}}+\frac {\sqrt {c+d x^2+e x} \left (64 a c d^2+64 a d^3 x^2+16 a d^2 e x-24 a d e^2+24 b c d^2 x-52 b c d e+48 b d^3 x^3+8 b d^2 e x^2-10 b d e^2 x+15 b e^3\right )}{192 d^3}\right )}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2],x]

[Out]

(Sqrt[(a + b*x)^2]*((Sqrt[c + e*x + d*x^2]*(64*a*c*d^2 - 52*b*c*d*e - 24*a*d*e^2 + 15*b*e^3 + 24*b*c*d^2*x + 1
6*a*d^2*e*x - 10*b*d*e^2*x + 64*a*d^3*x^2 + 8*b*d^2*e*x^2 + 48*b*d^3*x^3))/(192*d^3) + ((16*b*c^2*d^2 + 32*a*c
*d^2*e - 24*b*c*d*e^2 - 8*a*d*e^3 + 5*b*e^4)*Log[e + 2*d*x - 2*Sqrt[d]*Sqrt[c + e*x + d*x^2]])/(128*d^(7/2))))
/(a + b*x)

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fricas [A]  time = 0.45, size = 391, normalized size = 1.72 \begin {gather*} \left [\frac {3 \, {\left (16 \, b c^{2} d^{2} + 32 \, a c d^{2} e - 24 \, b c d e^{2} - 8 \, a d e^{3} + 5 \, b e^{4}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x - 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (48 \, b d^{4} x^{3} + 64 \, a c d^{3} - 52 \, b c d^{2} e - 24 \, a d^{2} e^{2} + 15 \, b d e^{3} + 8 \, {\left (8 \, a d^{4} + b d^{3} e\right )} x^{2} + 2 \, {\left (12 \, b c d^{3} + 8 \, a d^{3} e - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{768 \, d^{4}}, \frac {3 \, {\left (16 \, b c^{2} d^{2} + 32 \, a c d^{2} e - 24 \, b c d e^{2} - 8 \, a d e^{3} + 5 \, b e^{4}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \, {\left (48 \, b d^{4} x^{3} + 64 \, a c d^{3} - 52 \, b c d^{2} e - 24 \, a d^{2} e^{2} + 15 \, b d e^{3} + 8 \, {\left (8 \, a d^{4} + b d^{3} e\right )} x^{2} + 2 \, {\left (12 \, b c d^{3} + 8 \, a d^{3} e - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{384 \, d^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(16*b*c^2*d^2 + 32*a*c*d^2*e - 24*b*c*d*e^2 - 8*a*d*e^3 + 5*b*e^4)*sqrt(d)*log(8*d^2*x^2 + 8*d*e*x -
4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(48*b*d^4*x^3 + 64*a*c*d^3 - 52*b*c*d^2*e - 24
*a*d^2*e^2 + 15*b*d*e^3 + 8*(8*a*d^4 + b*d^3*e)*x^2 + 2*(12*b*c*d^3 + 8*a*d^3*e - 5*b*d^2*e^2)*x)*sqrt(d*x^2 +
e*x + c))/d^4, 1/384*(3*(16*b*c^2*d^2 + 32*a*c*d^2*e - 24*b*c*d*e^2 - 8*a*d*e^3 + 5*b*e^4)*sqrt(-d)*arctan(1/
2*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*(48*b*d^4*x^3 + 64*a*c*d^3 - 52*b*c*
d^2*e - 24*a*d^2*e^2 + 15*b*d*e^3 + 8*(8*a*d^4 + b*d^3*e)*x^2 + 2*(12*b*c*d^3 + 8*a*d^3*e - 5*b*d^2*e^2)*x)*sq
rt(d*x^2 + e*x + c))/d^4]

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giac [A]  time = 0.28, size = 268, normalized size = 1.18 \begin {gather*} \frac {1}{192} \, \sqrt {d x^{2} + x e + c} {\left (2 \, {\left (4 \, {\left (6 \, b x \mathrm {sgn}\left (b x + a\right ) + \frac {8 \, a d^{3} \mathrm {sgn}\left (b x + a\right ) + b d^{2} e \mathrm {sgn}\left (b x + a\right )}{d^{3}}\right )} x + \frac {12 \, b c d^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, a d^{2} e \mathrm {sgn}\left (b x + a\right ) - 5 \, b d e^{2} \mathrm {sgn}\left (b x + a\right )}{d^{3}}\right )} x + \frac {64 \, a c d^{2} \mathrm {sgn}\left (b x + a\right ) - 52 \, b c d e \mathrm {sgn}\left (b x + a\right ) - 24 \, a d e^{2} \mathrm {sgn}\left (b x + a\right ) + 15 \, b e^{3} \mathrm {sgn}\left (b x + a\right )}{d^{3}}\right )} + \frac {{\left (16 \, b c^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 32 \, a c d^{2} e \mathrm {sgn}\left (b x + a\right ) - 24 \, b c d e^{2} \mathrm {sgn}\left (b x + a\right ) - 8 \, a d e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | -2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} \sqrt {d} - e \right |}\right )}{128 \, d^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(d*x^2 + x*e + c)*(2*(4*(6*b*x*sgn(b*x + a) + (8*a*d^3*sgn(b*x + a) + b*d^2*e*sgn(b*x + a))/d^3)*x +
(12*b*c*d^2*sgn(b*x + a) + 8*a*d^2*e*sgn(b*x + a) - 5*b*d*e^2*sgn(b*x + a))/d^3)*x + (64*a*c*d^2*sgn(b*x + a)
- 52*b*c*d*e*sgn(b*x + a) - 24*a*d*e^2*sgn(b*x + a) + 15*b*e^3*sgn(b*x + a))/d^3) + 1/128*(16*b*c^2*d^2*sgn(b
*x + a) + 32*a*c*d^2*e*sgn(b*x + a) - 24*b*c*d*e^2*sgn(b*x + a) - 8*a*d*e^3*sgn(b*x + a) + 5*b*e^4*sgn(b*x + a
))*log(abs(-2*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))*sqrt(d) - e))/d^(7/2)

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maple [C]  time = 0.01, size = 381, normalized size = 1.68 \begin {gather*} \frac {\left (-96 a c \,d^{3} e \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+24 a \,d^{2} e^{3} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-48 b \,c^{2} d^{3} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+72 b c \,d^{2} e^{2} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-15 b d \,e^{4} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-96 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {7}{2}} e x -48 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {7}{2}} x +60 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {5}{2}} e^{2} x -48 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {5}{2}} e^{2}-24 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {5}{2}} e +96 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {7}{2}} x +30 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {3}{2}} e^{3}+128 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,d^{\frac {7}{2}}-80 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {5}{2}} e \right ) \mathrm {csgn}\left (b x +a \right )}{384 d^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*csgn(b*x+a)*(96*d^(7/2)*(d*x^2+e*x+c)^(3/2)*x*b+128*d^(7/2)*(d*x^2+e*x+c)^(3/2)*a-80*d^(5/2)*(d*x^2+e*x+
c)^(3/2)*b*e-96*d^(7/2)*(d*x^2+e*x+c)^(1/2)*x*a*e-48*d^(7/2)*(d*x^2+e*x+c)^(1/2)*x*b*c+60*d^(5/2)*(d*x^2+e*x+c
)^(1/2)*x*b*e^2-48*d^(5/2)*(d*x^2+e*x+c)^(1/2)*a*e^2-24*d^(5/2)*(d*x^2+e*x+c)^(1/2)*b*c*e+30*d^(3/2)*(d*x^2+e*
x+c)^(1/2)*b*e^3-96*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^(1/2)*d^(1/2))/d^(1/2))*a*c*d^3*e+24*ln(1/2*(2*d*x+e+2*(d*
x^2+e*x+c)^(1/2)*d^(1/2))/d^(1/2))*a*d^2*e^3-48*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^(1/2)*d^(1/2))/d^(1/2))*b*c^2*
d^3+72*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^(1/2)*d^(1/2))/d^(1/2))*b*c*d^2*e^2-15*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^
(1/2)*d^(1/2))/d^(1/2))*b*d*e^4)/d^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d x^{2} + e x + c} \sqrt {{\left (b x + a\right )}^{2}} x\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x^2 + e*x + c)*sqrt((b*x + a)^2)*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {c + d x^{2} + e x} \sqrt {\left (a + b x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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