3.1.41 \(\int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx\) [41]

Optimal. Leaf size=161 \[ -\frac {b c x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{8 d (a+b x)}+\frac {(4 a+3 b x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{12 d (a+b x)}-\frac {b c^2 \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2} (a+b x)} \]

[Out]

1/12*(3*b*x+4*a)*(d*x^2+c)^(3/2)*((b*x+a)^2)^(1/2)/d/(b*x+a)-1/8*b*c^2*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))*((b*
x+a)^2)^(1/2)/d^(3/2)/(b*x+a)-1/8*b*c*x*((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/d/(b*x+a)

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Rubi [A]
time = 0.04, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {1015, 794, 201, 223, 212} \begin {gather*} -\frac {b c^2 \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2} (a+b x)}-\frac {b c x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{8 d (a+b x)}+\frac {(4 a+3 b x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{12 d (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 + d*x^2],x]

[Out]

-1/8*(b*c*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/(d*(a + b*x)) + ((4*a + 3*b*x)*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]*(c + d*x^2)^(3/2))/(12*d*(a + b*x)) - (b*c^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(Sqrt[d]*x)/Sqrt[
c + d*x^2]])/(8*d^(3/2)*(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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 1015

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (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 + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, 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+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+d x^2} \, dx}{2 a b+2 b^2 x}\\ &=\frac {(4 a+3 b x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{12 d (a+b x)}-\frac {\left (b^2 c \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \sqrt {c+d x^2} \, dx}{2 d \left (2 a b+2 b^2 x\right )}\\ &=-\frac {b c x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{8 d (a+b x)}+\frac {(4 a+3 b x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{12 d (a+b x)}-\frac {\left (b^2 c^2 \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{4 d \left (2 a b+2 b^2 x\right )}\\ &=-\frac {b c x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{8 d (a+b x)}+\frac {(4 a+3 b x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{12 d (a+b x)}-\frac {\left (b^2 c^2 \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 )}{4 d \left (2 a b+2 b^2 x\right )}\\ &=-\frac {b c x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{8 d (a+b x)}+\frac {(4 a+3 b x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{12 d (a+b x)}-\frac {b c^2 \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2} (a+b x)}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 95, normalized size = 0.59 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (\sqrt {d} \sqrt {c+d x^2} \left (8 a \left (c+d x^2\right )+3 b x \left (c+2 d x^2\right )\right )+3 b c^2 \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{24 d^{3/2} (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 + d*x^2],x]

[Out]

(Sqrt[(a + b*x)^2]*(Sqrt[d]*Sqrt[c + d*x^2]*(8*a*(c + d*x^2) + 3*b*x*(c + 2*d*x^2)) + 3*b*c^2*Log[-(Sqrt[d]*x)
+ Sqrt[c + d*x^2]]))/(24*d^(3/2)*(a + b*x))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.08, size = 83, normalized size = 0.52

method	result	size
default	\(\frac {\mathrm {csgn}\left (b x +a \right ) \left (6 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, b x +8 a \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}-3 \sqrt {d \,x^{2}+c}\, \sqrt {d}\, b c x -3 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b \,c^{2}\right )}{24 d^{\frac {3}{2}}}\)	\(83\)
risch	\(\frac {\left (6 b d \,x^{3}+8 a d \,x^{2}+3 b c x +8 a c \right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (b x +a \right )^{2}}}{24 d \left (b x +a \right )}-\frac {c^{2} b \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) \sqrt {\left (b x +a \right )^{2}}}{8 d^{\frac {3}{2}} \left (b x +a \right )}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*csgn(b*x+a)*(6*(d*x^2+c)^(3/2)*d^(1/2)*b*x+8*a*(d*x^2+c)^(3/2)*d^(1/2)-3*(d*x^2+c)^(1/2)*d^(1/2)*b*c*x-3*
ln(d^(1/2)*x+(d*x^2+c)^(1/2))*b*c^2)/d^(3/2)

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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(x*((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, x)

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Fricas [A]
time = 0.40, size = 157, normalized size = 0.98 \begin {gather*} \left [\frac {3 \, b c^{2} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (6 \, b d^{2} x^{3} + 8 \, a d^{2} x^{2} + 3 \, b c d x + 8 \, a c d\right )} \sqrt {d x^{2} + c}}{48 \, d^{2}}, \frac {3 \, b c^{2} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (6 \, b d^{2} x^{3} + 8 \, a d^{2} x^{2} + 3 \, b c d x + 8 \, a c d\right )} \sqrt {d x^{2} + c}}{24 \, d^{2}}\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+c)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*b*c^2*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(6*b*d^2*x^3 + 8*a*d^2*x^2 + 3*b*c*
d*x + 8*a*c*d)*sqrt(d*x^2 + c))/d^2, 1/24*(3*b*c^2*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (6*b*d^2*x^3
+ 8*a*d^2*x^2 + 3*b*c*d*x + 8*a*c*d)*sqrt(d*x^2 + c))/d^2]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {c + d x^{2}} \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+c)**(1/2),x)

[Out]

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

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Giac [A]
time = 3.30, size = 98, normalized size = 0.61 \begin {gather*} \frac {b c^{2} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{8 \, d^{\frac {3}{2}}} + \frac {1}{24} \, \sqrt {d x^{2} + c} {\left ({\left (2 \, {\left (3 \, b x \mathrm {sgn}\left (b x + a\right ) + 4 \, a \mathrm {sgn}\left (b x + a\right )\right )} x + \frac {3 \, b c \mathrm {sgn}\left (b x + a\right )}{d}\right )} x + \frac {8 \, a c \mathrm {sgn}\left (b x + a\right )}{d}\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+c)^(1/2),x, algorithm="giac")

[Out]

1/8*b*c^2*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a)/d^(3/2) + 1/24*sqrt(d*x^2 + c)*((2*(3*b*x*sgn(b*
x + a) + 4*a*sgn(b*x + a))*x + 3*b*c*sgn(b*x + a)/d)*x + 8*a*c*sgn(b*x + a)/d)

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

[Out]

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

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