\(\int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx\) [42]

Optimal result

Integrand size = 32, antiderivative size = 148 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {a x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {a c \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)} \]

[Out]

1/3*b*(d*x^2+c)^(3/2)*((b*x+a)^2)^(1/2)/d/(b*x+a)+1/2*a*c*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))*((b*x+a)^2)^(1/2)
/(b*x+a)/d^(1/2)+1/2*a*x*((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/(b*x+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {984, 655, 201, 223, 212} \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {a c \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {a x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)} \]

[In]

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

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 655

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

Rule 984

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[(a + b*x + c*x^2)^F
racPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(b + 2*c*x)^(2*p)*(d + f*x^2)^q, x], x] /; Free
Q[{a, b, c, d, 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+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+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {\left (2 a b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \sqrt {c+d x^2} \, dx}{2 a b+2 b^2 x} \\ & = \frac {a x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {\left (a b c \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{2 a b+2 b^2 x} \\ & = \frac {a x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {\left (a b c \sqrt {a^2+2 a b x+b^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 a b+2 b^2 x} \\ & = \frac {a x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {a c \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.57 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (\sqrt {c+d x^2} \left (3 a d x+2 b \left (c+d x^2\right )\right )-3 a c \sqrt {d} \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{6 d (a+b x)} \]

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(Sqrt[c + d*x^2]*(3*a*d*x + 2*b*(c + d*x^2)) - 3*a*c*Sqrt[d]*Log[-(Sqrt[d]*x) + Sqrt[c + d*
x^2]]))/(6*d*(a + b*x))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.51 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.44

method	result	size
default	\(\frac {\operatorname {csgn}\left (b x +a \right ) \left (2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, b +3 a \sqrt {d \,x^{2}+c}\, d^{\frac {3}{2}} x +3 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) a c d \right )}{6 d^{\frac {3}{2}}}\)	\(65\)
risch	\(\frac {\left (2 b d \,x^{2}+3 a d x +2 b c \right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (b x +a \right )^{2}}}{6 d \left (b x +a \right )}+\frac {a c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) \sqrt {\left (b x +a \right )^{2}}}{2 \sqrt {d}\, \left (b x +a \right )}\)	\(88\)

[In]

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

[Out]

1/6*csgn(b*x+a)*(2*(d*x^2+c)^(3/2)*d^(1/2)*b+3*a*(d*x^2+c)^(1/2)*d^(3/2)*x+3*ln(d^(1/2)*x+(d*x^2+c)^(1/2))*a*c
*d)/d^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\left [\frac {3 \, a c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (2 \, b d x^{2} + 3 \, a d x + 2 \, b c\right )} \sqrt {d x^{2} + c}}{12 \, d}, -\frac {3 \, a c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b d x^{2} + 3 \, a d x + 2 \, b c\right )} \sqrt {d x^{2} + c}}{6 \, d}\right ] \]

[In]

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

[Out]

[1/12*(3*a*c*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(2*b*d*x^2 + 3*a*d*x + 2*b*c)*sqrt(d*
x^2 + c))/d, -1/6*(3*a*c*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (2*b*d*x^2 + 3*a*d*x + 2*b*c)*sqrt(d*x^
2 + c))/d]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.53 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=-\frac {a c \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{2 \, \sqrt {d}} + \frac {1}{6} \, \sqrt {d x^{2} + c} {\left ({\left (2 \, b x \mathrm {sgn}\left (b x + a\right ) + 3 \, a \mathrm {sgn}\left (b x + a\right )\right )} x + \frac {2 \, b c \mathrm {sgn}\left (b x + a\right )}{d}\right )} \]

[In]

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

[Out]

-1/2*a*c*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a)/sqrt(d) + 1/6*sqrt(d*x^2 + c)*((2*b*x*sgn(b*x + a
) + 3*a*sgn(b*x + a))*x + 2*b*c*sgn(b*x + a)/d)

Mupad [F(-1)]

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

[In]

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

[Out]

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