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

Optimal. Leaf size=198 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) (2 a d-b e) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{16 d^{5/2} (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 d x+e) (2 a d-b e) \sqrt {c+d x^2+e x}}{8 d^2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2}}{3 d (a+b x)} \]

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Rubi [A]  time = 0.10, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {969, 640, 612, 621, 206} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) (2 a d-b e) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{16 d^{5/2} (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 d x+e) (2 a d-b e) \sqrt {c+d x^2+e x}}{8 d^2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2}}{3 d (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*d - b*e)*(e + 2*d*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2])/(8*d^2*(a + b*x)) + (b*Sqrt[a^
2 + 2*a*b*x + b^2*x^2]*(c + e*x + d*x^2)^(3/2))/(3*d*(a + b*x)) + ((2*a*d - b*e)*(4*c*d - e^2)*Sqrt[a^2 + 2*a*
b*x + b^2*x^2]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(16*d^(5/2)*(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 640

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

Rule 969

Int[((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[(b + 2*c*x)^(2*p)*(d + e*x + f*x^2
)^q, x], x] /; FreeQ[{a, b, c, d, e, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \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 \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 {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {\left (b (2 a d-b e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \sqrt {c+e x+d x^2} \, dx}{d \left (2 a b+2 b^2 x\right )}\\ &=\frac {(2 a d-b e) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{8 d^2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {\left (b (2 a d-b e) \left (4 c d-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}{8 d^2 \left (2 a b+2 b^2 x\right )}\\ &=\frac {(2 a d-b e) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{8 d^2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {\left (b (2 a d-b e) \left (4 c d-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 )}{4 d^2 \left (2 a b+2 b^2 x\right )}\\ &=\frac {(2 a d-b e) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{8 d^2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {(2 a d-b e) \left (4 c d-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 )}{16 d^{5/2} (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 134, normalized size = 0.68 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (2 \sqrt {d} \sqrt {c+x (d x+e)} \left (6 a d (2 d x+e)+b \left (8 c d+8 d^2 x^2+2 d e x-3 e^2\right )\right )+3 \left (4 c d-e^2\right ) (2 a d-b e) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )\right )}{48 d^{5/2} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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]*(2*Sqrt[d]*Sqrt[c + x*(e + d*x)]*(6*a*d*(e + 2*d*x) + b*(8*c*d - 3*e^2 + 2*d*e*x + 8*d^2*x^
2)) + 3*(2*a*d - b*e)*(4*c*d - e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])]))/(48*d^(5/2)*(a +
b*x))

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IntegrateAlgebraic [A]  time = 0.60, size = 150, normalized size = 0.76 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (\frac {\sqrt {c+d x^2+e x} \left (12 a d^2 x+6 a d e+8 b c d+8 b d^2 x^2+2 b d e x-3 b e^2\right )}{24 d^2}+\frac {\left (-8 a c d^2+2 a d e^2+4 b c d e-b e^3\right ) \log \left (-2 d^{5/2} \sqrt {c+d x^2+e x}+2 d^3 x+d^2 e\right )}{16 d^{5/2}}\right )}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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]*(8*b*c*d + 6*a*d*e - 3*b*e^2 + 12*a*d^2*x + 2*b*d*e*x + 8*b*d^2*x^2
))/(24*d^2) + ((-8*a*c*d^2 + 4*b*c*d*e + 2*a*d*e^2 - b*e^3)*Log[d^2*e + 2*d^3*x - 2*d^(5/2)*Sqrt[c + e*x + d*x
^2]])/(16*d^(5/2))))/(a + b*x)

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fricas [A]  time = 0.45, size = 287, normalized size = 1.45 \begin {gather*} \left [\frac {3 \, {\left (8 \, a c d^{2} - 4 \, b c d e - 2 \, a d e^{2} + b e^{3}\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 (8 \, b d^{3} x^{2} + 8 \, b c d^{2} + 6 \, a d^{2} e - 3 \, b d e^{2} + 2 \, {\left (6 \, a d^{3} + b d^{2} e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{96 \, d^{3}}, -\frac {3 \, {\left (8 \, a c d^{2} - 4 \, b c d e - 2 \, a d e^{2} + b e^{3}\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 (8 \, b d^{3} x^{2} + 8 \, b c d^{2} + 6 \, a d^{2} e - 3 \, b d e^{2} + 2 \, {\left (6 \, a d^{3} + b d^{2} e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{48 \, d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(8*a*c*d^2 - 4*b*c*d*e - 2*a*d*e^2 + b*e^3)*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*(8*b*d^3*x^2 + 8*b*c*d^2 + 6*a*d^2*e - 3*b*d*e^2 + 2*(6*a*d^3 + b*d^2*
e)*x)*sqrt(d*x^2 + e*x + c))/d^3, -1/48*(3*(8*a*c*d^2 - 4*b*c*d*e - 2*a*d*e^2 + b*e^3)*sqrt(-d)*arctan(1/2*sqr
t(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) - 2*(8*b*d^3*x^2 + 8*b*c*d^2 + 6*a*d^2*e - 3*
b*d*e^2 + 2*(6*a*d^3 + b*d^2*e)*x)*sqrt(d*x^2 + e*x + c))/d^3]

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giac [A]  time = 0.30, size = 185, normalized size = 0.93 \begin {gather*} \frac {1}{24} \, \sqrt {d x^{2} + x e + c} {\left (2 \, {\left (4 \, b x \mathrm {sgn}\left (b x + a\right ) + \frac {6 \, a d^{2} \mathrm {sgn}\left (b x + a\right ) + b d e \mathrm {sgn}\left (b x + a\right )}{d^{2}}\right )} x + \frac {8 \, b c d \mathrm {sgn}\left (b x + a\right ) + 6 \, a d e \mathrm {sgn}\left (b x + a\right ) - 3 \, b e^{2} \mathrm {sgn}\left (b x + a\right )}{d^{2}}\right )} - \frac {{\left (8 \, a c d^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, b c d e \mathrm {sgn}\left (b x + a\right ) - 2 \, a d e^{2} \mathrm {sgn}\left (b x + a\right ) + b e^{3} \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 )}{16 \, d^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(d*x^2 + x*e + c)*(2*(4*b*x*sgn(b*x + a) + (6*a*d^2*sgn(b*x + a) + b*d*e*sgn(b*x + a))/d^2)*x + (8*b*
c*d*sgn(b*x + a) + 6*a*d*e*sgn(b*x + a) - 3*b*e^2*sgn(b*x + a))/d^2) - 1/16*(8*a*c*d^2*sgn(b*x + a) - 4*b*c*d*
e*sgn(b*x + a) - 2*a*d*e^2*sgn(b*x + a) + b*e^3*sgn(b*x + a))*log(abs(-2*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))*s
qrt(d) - e))/d^(5/2)

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maple [C]  time = 0.01, size = 257, normalized size = 1.30 \begin {gather*} \frac {\left (24 a c \,d^{3} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-6 a \,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 )-12 b c \,d^{2} e \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+3 b d \,e^{3} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+24 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {7}{2}} x -12 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {5}{2}} e x +12 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {5}{2}} e -6 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {3}{2}} e^{2}+16 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {5}{2}}\right ) \mathrm {csgn}\left (b x +a \right )}{48 d^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*csgn(b*x+a)*(16*d^(5/2)*(d*x^2+e*x+c)^(3/2)*b+24*d^(7/2)*(d*x^2+e*x+c)^(1/2)*x*a-12*d^(5/2)*(d*x^2+e*x+c)
^(1/2)*x*b*e+12*d^(5/2)*(d*x^2+e*x+c)^(1/2)*a*e-6*d^(3/2)*(d*x^2+e*x+c)^(1/2)*b*e^2+24*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^3-6*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^2-1
2*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+3*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^3)/d^(7/2)

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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}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \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(((a + b*x)^2)^(1/2)*(c + e*x + d*x^2)^(1/2),x)

[Out]

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

[Out]

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

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