3.1.48 \(\int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx\) [48]

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

[Out]

1/3*b*(d*x^2+e*x+c)^(3/2)*((b*x+a)^2)^(1/2)/d/(b*x+a)+1/16*(2*a*d-b*e)*(4*c*d-e^2)*arctanh(1/2*(2*d*x+e)/d^(1/
2)/(d*x^2+e*x+c)^(1/2))*((b*x+a)^2)^(1/2)/d^(5/2)/(b*x+a)+1/8*(2*a*d-b*e)*(2*d*x+e)*((b*x+a)^2)^(1/2)*(d*x^2+e
*x+c)^(1/2)/d^2/(b*x+a)

________________________________________________________________________________________

Rubi [A]
time = 0.06, 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 = {983, 654, 626, 635, 212} \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 212

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

Rule 626

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 635

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 654

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 983

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 ) \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.40, size = 136, normalized size = 0.69 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (2 \sqrt {d} \sqrt {c+x (e+d x)} \left (6 a d (e+2 d x)+b \left (8 c d-3 e^2+2 d e x+8 d^2 x^2\right )\right )-3 (2 a d-b e) \left (4 c d-e^2\right ) \log \left (d^2 \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\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)*Log[d^2*(e + 2*d*x - 2*Sqrt[d]*Sqrt[c + x*(e + d*x)])]))/(48*d^(5/2)*(a +
b*x))

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.12, size = 257, normalized size = 1.30

method	result	size
risch	\(\frac {\left (8 b \,x^{2} d^{2}+12 a \,d^{2} x +2 b d e x +6 a d e +8 b c d -3 e^{2} b \right ) \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b x +a \right )^{2}}}{24 d^{2} \left (b x +a \right )}+\frac {\left (\frac {\ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) a c}{2 \sqrt {d}}-\frac {\ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) e^{2} a}{8 d^{\frac {3}{2}}}-\frac {\ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) b c e}{4 d^{\frac {3}{2}}}+\frac {\ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) e^{3} b}{16 d^{\frac {5}{2}}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(225\)
default	\(\frac {\mathrm {csgn}\left (b x +a \right ) \left (16 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {5}{2}} b +24 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} a x -12 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b e x +12 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a e -6 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b \,e^{2}+24 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{3}-6 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a \,d^{2} e^{2}-12 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2} e +3 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{3}\right )}{48 d^{\frac {7}{2}}}\)	\(257\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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 + x*e + c)*sqrt((b*x + a)^2), x)

________________________________________________________________________________________

Fricas [A]
time = 0.41, size = 292, normalized size = 1.47 \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 x e + 4 \, \sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (8 \, b d^{3} x^{2} + 12 \, a d^{3} x + 8 \, b c d^{2} - 3 \, b d e^{2} + 2 \, {\left (b d^{2} x + 3 \, a d^{2}\right )} e\right )} \sqrt {d x^{2} + x e + 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} + x e + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d x e + c d\right )}}\right ) - 2 \, {\left (8 \, b d^{3} x^{2} + 12 \, a d^{3} x + 8 \, b c d^{2} - 3 \, b d e^{2} + 2 \, {\left (b d^{2} x + 3 \, a d^{2}\right )} e\right )} \sqrt {d x^{2} + x e + 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*x*e + 4*sqrt(d*x^2 + x*e + c)
*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(8*b*d^3*x^2 + 12*a*d^3*x + 8*b*c*d^2 - 3*b*d*e^2 + 2*(b*d^2*x + 3*a*d
^2)*e)*sqrt(d*x^2 + x*e + 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*sq
rt(d*x^2 + x*e + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*x*e + c*d)) - 2*(8*b*d^3*x^2 + 12*a*d^3*x + 8*b*c*d^2 -
3*b*d*e^2 + 2*(b*d^2*x + 3*a*d^2)*e)*sqrt(d*x^2 + x*e + c))/d^3]

________________________________________________________________________________________

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)

________________________________________________________________________________________

Giac [A]
time = 2.06, 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)

________________________________________________________________________________________

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)

________________________________________________________________________________________