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

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

[Out]

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

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Rubi [A]  time = 0.0673629, antiderivative size = 161, normalized size of antiderivative = 1., 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 = {1001, 780, 195, 217, 206} \[ -\frac{b c^2 \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{3/2} (a+b x)}-\frac{b c x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{8 d (a+b x)}+\frac{(4 a+3 b x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{12 d (a+b x)} \]

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]

-(b*c*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/(8*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 1001

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]

Rule 780

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 195

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 217

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 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])

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 ) \operatorname{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*}

Mathematica [A]  time = 0.11876, size = 117, normalized size = 0.73 \[ \frac{\sqrt{(a+b x)^2} \sqrt{c+d x^2} \left (\sqrt{d} \sqrt{\frac{d x^2}{c}+1} \left (8 a \left (c+d x^2\right )+3 b x \left (c+2 d x^2\right )\right )-3 b c^{3/2} \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )\right )}{24 d^{3/2} (a+b x) \sqrt{\frac{d x^2}{c}+1}} \]

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[c + d*x^2]*(Sqrt[d]*Sqrt[1 + (d*x^2)/c]*(8*a*(c + d*x^2) + 3*b*x*(c + 2*d*x^2)) - 3*b*
c^(3/2)*ArcSinh[(Sqrt[d]*x)/Sqrt[c]]))/(24*d^(3/2)*(a + b*x)*Sqrt[1 + (d*x^2)/c])

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Maple [C]  time = 0.223, size = 83, normalized size = 0.5 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) }{24} \left ( 6\,\sqrt{d} \left ( d{x}^{2}+c \right ) ^{3/2}xb+8\,a \left ( d{x}^{2}+c \right ) ^{3/2}\sqrt{d}-3\,\sqrt{d}\sqrt{d{x}^{2}+c}xbc-3\,\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) b{c}^{2} \right ){d}^{-{\frac{3}{2}}}} \end{align*}

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)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d x^{2} + c} \sqrt{{\left (b x + a\right )}^{2}} x\,{d x} \end{align*}

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 = 1.90422, size = 381, normalized size = 2.37 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{c + d x^{2}} \sqrt{\left (a + b x\right )^{2}}\, dx \end{align*}

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 = 1.14935, size = 132, normalized size = 0.82 \begin{align*} \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{align*}

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)