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

Optimal result

Integrand size = 35, antiderivative size = 198 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\frac {(2 a d-b e) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{8 d^2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {(2 a d-b e) \left (4 c d-e^2\right ) \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{16 d^{5/2} (a+b x)} \]

[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] (verified)

Time = 0.07 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {983, 654, 626, 635, 212} \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) (2 a d-b e) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{16 d^{5/2} (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 d x+e) (2 a d-b e) \sqrt {c+d x^2+e x}}{8 d^2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2}}{3 d (a+b x)} \]

[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*} \text {integral}& = \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] (verified)

Time = 0.67 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.87 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (\sqrt {d} \sqrt {c+x (e+d x)} \left (6 a d (e+2 d x)+b \left (8 c d-3 e^2+2 d e x+8 d^2 x^2\right )\right )+6 d e (2 b c+a e) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+x (e+d x)}}\right )+3 \left (8 a c d^2+b e^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+x (e+d x)}}\right )\right )}{24 d^{5/2} (a+b x)} \]

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

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\left (8 b \,x^{2} d^{2}+12 a \,d^{2} x +2 b d e x +6 a d e +8 b c d -3 b \,e^{2}\right ) \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b x +a \right )^{2}}}{24 d^{2} \left (b x +a \right )}+\frac {\left (8 c \,d^{2} a -2 a d \,e^{2}-4 b c d e +b \,e^{3}\right ) \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) \sqrt {\left (b x +a \right )^{2}}}{16 d^{\frac {5}{2}} \left (b x +a \right )}\)	\(146\)
default	\(\frac {\operatorname {csgn}\left (b x +a \right ) \left (16 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {5}{2}} b +24 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} a x -12 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b e x +12 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a e -6 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b \,e^{2}+24 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{3}-6 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a \,d^{2} e^{2}-12 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2} e +3 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{3}\right )}{48 d^{\frac {7}{2}}}\)	\(257\)

[In]

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

[Out]

1/24*(8*b*d^2*x^2+12*a*d^2*x+2*b*d*e*x+6*a*d*e+8*b*c*d-3*b*e^2)*(d*x^2+e*x+c)^(1/2)/d^2*((b*x+a)^2)^(1/2)/(b*x
+a)+1/16*(8*a*c*d^2-2*a*d*e^2-4*b*c*d*e+b*e^3)/d^(5/2)*ln((1/2*e+d*x)/d^(1/2)+(d*x^2+e*x+c)^(1/2))*((b*x+a)^2)
^(1/2)/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.45 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\left [\frac {3 \, {\left (8 \, a c d^{2} - 4 \, b c d e - 2 \, a d e^{2} + b e^{3}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (8 \, b d^{3} x^{2} + 8 \, b c d^{2} + 6 \, a d^{2} e - 3 \, b d e^{2} + 2 \, {\left (6 \, a d^{3} + b d^{2} e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{96 \, d^{3}}, -\frac {3 \, {\left (8 \, a c d^{2} - 4 \, b c d e - 2 \, a d e^{2} + b e^{3}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) - 2 \, {\left (8 \, b d^{3} x^{2} + 8 \, b c d^{2} + 6 \, a d^{2} e - 3 \, b d e^{2} + 2 \, {\left (6 \, a d^{3} + b d^{2} e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{48 \, d^{3}}\right ] \]

[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]

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.91 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\frac {1}{24} \, \sqrt {d x^{2} + e x + c} {\left (2 \, {\left (4 \, b x \mathrm {sgn}\left (b x + a\right ) + \frac {6 \, a d^{2} \mathrm {sgn}\left (b x + a\right ) + b d e \mathrm {sgn}\left (b x + a\right )}{d^{2}}\right )} x + \frac {8 \, b c d \mathrm {sgn}\left (b x + a\right ) + 6 \, a d e \mathrm {sgn}\left (b x + a\right ) - 3 \, b e^{2} \mathrm {sgn}\left (b x + a\right )}{d^{2}}\right )} - \frac {{\left (8 \, a c d^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, b c d e \mathrm {sgn}\left (b x + a\right ) - 2 \, a d e^{2} \mathrm {sgn}\left (b x + a\right ) + b e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} \sqrt {d} + e \right |}\right )}{16 \, d^{\frac {5}{2}}} \]

[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 + e*x + 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 + e*x + c))*sq
rt(d) + e))/d^(5/2)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx=\int \sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+e\,x+c} \,d x \]

[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)