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

Optimal. Leaf size=212 \[ -\frac{a c^2 \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{3/2} (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2} (8 b c-15 a d x)}{60 d^2 (a+b x)}-\frac{a c x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{8 d (a+b x)}+\frac{b x^2 \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{5 d (a+b x)} \]

[Out]

-(a*c*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/(8*d*(a + b*x)) + (b*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*
(c + d*x^2)^(3/2))/(5*d*(a + b*x)) - ((8*b*c - 15*a*d*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(c + d*x^2)^(3/2))/(60*
d^2*(a + b*x)) - (a*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 = 0.116095, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {1001, 833, 780, 195, 217, 206} \[ -\frac{a c^2 \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{3/2} (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2} (8 b c-15 a d x)}{60 d^2 (a+b x)}-\frac{a c x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{8 d (a+b x)}+\frac{b x^2 \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{5 d (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/(8*d*(a + b*x)) + (b*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*
(c + d*x^2)^(3/2))/(5*d*(a + b*x)) - ((8*b*c - 15*a*d*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(c + d*x^2)^(3/2))/(60*
d^2*(a + b*x)) - (a*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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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^2 \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^2 \left (2 a b+2 b^2 x\right ) \sqrt{c+d x^2} \, dx}{2 a b+2 b^2 x}\\ &=\frac{b x^2 \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{5 d (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int x \left (-4 b^2 c+10 a b d x\right ) \sqrt{c+d x^2} \, dx}{5 d \left (2 a b+2 b^2 x\right )}\\ &=\frac{b x^2 \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac{(8 b c-15 a d x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{60 d^2 (a+b x)}-\frac{\left (a b 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{a c x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{8 d (a+b x)}+\frac{b x^2 \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac{(8 b c-15 a d x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{60 d^2 (a+b x)}-\frac{\left (a b 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{a c x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{8 d (a+b x)}+\frac{b x^2 \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac{(8 b c-15 a d x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{60 d^2 (a+b x)}-\frac{\left (a b 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{a c x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{8 d (a+b x)}+\frac{b x^2 \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac{(8 b c-15 a d x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{60 d^2 (a+b x)}-\frac{a 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 = 0.156995, size = 129, normalized size = 0.61 \[ \frac{\sqrt{(a+b x)^2} \sqrt{c+d x^2} \left (\sqrt{\frac{d x^2}{c}+1} \left (15 a d x \left (c+2 d x^2\right )+8 b \left (-2 c^2+c d x^2+3 d^2 x^4\right )\right )-15 a c^{3/2} \sqrt{d} \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )\right )}{120 d^2 (a+b x) \sqrt{\frac{d x^2}{c}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*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[1 + (d*x^2)/c]*(15*a*d*x*(c + 2*d*x^2) + 8*b*(-2*c^2 + c*d*x^2 + 3*d^
2*x^4)) - 15*a*c^(3/2)*Sqrt[d]*ArcSinh[(Sqrt[d]*x)/Sqrt[c]]))/(120*d^2*(a + b*x)*Sqrt[1 + (d*x^2)/c])

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Maple [C]  time = 0.211, size = 103, normalized size = 0.5 \begin{align*} -{\frac{{\it csgn} \left ( bx+a \right ) }{120} \left ( -24\,{d}^{3/2} \left ( d{x}^{2}+c \right ) ^{3/2}{x}^{2}b-30\,{d}^{3/2} \left ( d{x}^{2}+c \right ) ^{3/2}xa+16\,\sqrt{d} \left ( d{x}^{2}+c \right ) ^{3/2}bc+15\,{d}^{3/2}\sqrt{d{x}^{2}+c}xac+15\,\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) a{c}^{2}d \right ){d}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*csgn(b*x+a)*(-24*d^(3/2)*(d*x^2+c)^(3/2)*x^2*b-30*d^(3/2)*(d*x^2+c)^(3/2)*x*a+16*d^(1/2)*(d*x^2+c)^(3/2
)*b*c+15*d^(3/2)*(d*x^2+c)^(1/2)*x*a*c+15*ln(x*d^(1/2)+(d*x^2+c)^(1/2))*a*c^2*d)/d^(5/2)

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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^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*((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^2, x)

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Fricas [A]  time = 1.96571, size = 435, normalized size = 2.05 \begin{align*} \left [\frac{15 \, a c^{2} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (24 \, b d^{2} x^{4} + 30 \, a d^{2} x^{3} + 8 \, b c d x^{2} + 15 \, a c d x - 16 \, b c^{2}\right )} \sqrt{d x^{2} + c}}{240 \, d^{2}}, \frac{15 \, a c^{2} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left (24 \, b d^{2} x^{4} + 30 \, a d^{2} x^{3} + 8 \, b c d x^{2} + 15 \, a c d x - 16 \, b c^{2}\right )} \sqrt{d x^{2} + c}}{120 \, d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/240*(15*a*c^2*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(24*b*d^2*x^4 + 30*a*d^2*x^3 + 8*
b*c*d*x^2 + 15*a*c*d*x - 16*b*c^2)*sqrt(d*x^2 + c))/d^2, 1/120*(15*a*c^2*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2
+ c)) + (24*b*d^2*x^4 + 30*a*d^2*x^3 + 8*b*c*d*x^2 + 15*a*c*d*x - 16*b*c^2)*sqrt(d*x^2 + c))/d^2]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \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**2*((b*x+a)**2)**(1/2)*(d*x**2+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.17183, size = 158, normalized size = 0.75 \begin{align*} \frac{a 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}{120} \, \sqrt{d x^{2} + c}{\left ({\left (2 \,{\left (3 \,{\left (4 \, b x \mathrm{sgn}\left (b x + a\right ) + 5 \, a \mathrm{sgn}\left (b x + a\right )\right )} x + \frac{4 \, b c \mathrm{sgn}\left (b x + a\right )}{d}\right )} x + \frac{15 \, a c \mathrm{sgn}\left (b x + a\right )}{d}\right )} x - \frac{16 \, b c^{2} \mathrm{sgn}\left (b x + a\right )}{d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*c^2*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a)/d^(3/2) + 1/120*sqrt(d*x^2 + c)*((2*(3*(4*b*x*sg
n(b*x + a) + 5*a*sgn(b*x + a))*x + 4*b*c*sgn(b*x + a)/d)*x + 15*a*c*sgn(b*x + a)/d)*x - 16*b*c^2*sgn(b*x + a)/
d^2)