3.267 \(\int x^2 \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4} \, dx\)

Optimal. Leaf size=243 \[ \frac{c^2 \sqrt{a^2+2 a b x^2+b^2 x^4} (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{5/2} \left (a+b x^2\right )}-\frac{c x \sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2} (b c-2 a d)}{16 d^2 \left (a+b x^2\right )}+\frac{b x^3 \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{6 d \left (a+b x^2\right )}-\frac{x^3 \sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2} (b c-2 a d)}{8 d \left (a+b x^2\right )} \]

[Out]

-(c*(b*c - 2*a*d)*x*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*d^2*(a + b*x^2)) - ((b*c - 2*a*d)*x^3
*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*d*(a + b*x^2)) + (b*x^3*(c + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*
b*x^2 + b^2*x^4])/(6*d*(a + b*x^2)) + (c^2*(b*c - 2*a*d)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(Sqrt[d]*x)/S
qrt[c + d*x^2]])/(16*d^(5/2)*(a + b*x^2))

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Rubi [A]  time = 0.130078, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {1250, 459, 279, 321, 217, 206} \[ \frac{c^2 \sqrt{a^2+2 a b x^2+b^2 x^4} (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{5/2} \left (a+b x^2\right )}-\frac{c x \sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2} (b c-2 a d)}{16 d^2 \left (a+b x^2\right )}+\frac{b x^3 \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{6 d \left (a+b x^2\right )}-\frac{x^3 \sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2} (b c-2 a d)}{8 d \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

-(c*(b*c - 2*a*d)*x*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*d^2*(a + b*x^2)) - ((b*c - 2*a*d)*x^3
*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*d*(a + b*x^2)) + (b*x^3*(c + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*
b*x^2 + b^2*x^4])/(6*d*(a + b*x^2)) + (c^2*(b*c - 2*a*d)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(Sqrt[d]*x)/S
qrt[c + d*x^2]])/(16*d^(5/2)*(a + b*x^2))

Rule 1250

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dis
t[(a + b*x^2 + c*x^4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(f*x)^m*(d + e*x^2)^q*(b/2
+ c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int x^2 \left (a b+b^2 x^2\right ) \sqrt{c+d x^2} \, dx}{a b+b^2 x^2}\\ &=\frac{b x^3 \left (c+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 d \left (a+b x^2\right )}-\frac{\left (b (b c-2 a d) \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \int x^2 \sqrt{c+d x^2} \, dx}{2 d \left (a b+b^2 x^2\right )}\\ &=-\frac{(b c-2 a d) x^3 \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 d \left (a+b x^2\right )}+\frac{b x^3 \left (c+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 d \left (a+b x^2\right )}-\frac{\left (b c (b c-2 a d) \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \int \frac{x^2}{\sqrt{c+d x^2}} \, dx}{8 d \left (a b+b^2 x^2\right )}\\ &=-\frac{c (b c-2 a d) x \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{16 d^2 \left (a+b x^2\right )}-\frac{(b c-2 a d) x^3 \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 d \left (a+b x^2\right )}+\frac{b x^3 \left (c+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 d \left (a+b x^2\right )}+\frac{\left (b c^2 (b c-2 a d) \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{16 d^2 \left (a b+b^2 x^2\right )}\\ &=-\frac{c (b c-2 a d) x \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{16 d^2 \left (a+b x^2\right )}-\frac{(b c-2 a d) x^3 \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 d \left (a+b x^2\right )}+\frac{b x^3 \left (c+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 d \left (a+b x^2\right )}+\frac{\left (b c^2 (b c-2 a d) \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{16 d^2 \left (a b+b^2 x^2\right )}\\ &=-\frac{c (b c-2 a d) x \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{16 d^2 \left (a+b x^2\right )}-\frac{(b c-2 a d) x^3 \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 d \left (a+b x^2\right )}+\frac{b x^3 \left (c+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 d \left (a+b x^2\right )}+\frac{c^2 (b c-2 a d) \sqrt{a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{5/2} \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.176697, size = 142, normalized size = 0.58 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \sqrt{c+d x^2} \left (\sqrt{d} x \sqrt{\frac{d x^2}{c}+1} \left (6 a d \left (c+2 d x^2\right )+b \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )\right )+3 c^{3/2} (b c-2 a d) \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )\right )}{48 d^{5/2} \left (a+b x^2\right ) \sqrt{\frac{d x^2}{c}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*Sqrt[c + d*x^2]*(Sqrt[d]*x*Sqrt[1 + (d*x^2)/c]*(6*a*d*(c + 2*d*x^2) + b*(-3*c^2 + 2*c*d*x
^2 + 8*d^2*x^4)) + 3*c^(3/2)*(b*c - 2*a*d)*ArcSinh[(Sqrt[d]*x)/Sqrt[c]]))/(48*d^(5/2)*(a + b*x^2)*Sqrt[1 + (d*
x^2)/c])

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Maple [A]  time = 0.012, size = 159, normalized size = 0.7 \begin{align*}{\frac{1}{48\,b{x}^{2}+48\,a}\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( 8\,{d}^{3/2} \left ( d{x}^{2}+c \right ) ^{3/2}{x}^{3}b+12\,{d}^{3/2} \left ( d{x}^{2}+c \right ) ^{3/2}xa-6\,\sqrt{d} \left ( d{x}^{2}+c \right ) ^{3/2}xbc-6\,{d}^{3/2}\sqrt{d{x}^{2}+c}xac+3\,\sqrt{d}\sqrt{d{x}^{2}+c}xb{c}^{2}-6\,\ln \left ( \sqrt{d}x+\sqrt{d{x}^{2}+c} \right ) a{c}^{2}d+3\,\ln \left ( \sqrt{d}x+\sqrt{d{x}^{2}+c} \right ) b{c}^{3} \right ){d}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((b*x^2+a)^2)^(1/2)*(8*d^(3/2)*(d*x^2+c)^(3/2)*x^3*b+12*d^(3/2)*(d*x^2+c)^(3/2)*x*a-6*d^(1/2)*(d*x^2+c)^(
3/2)*x*b*c-6*d^(3/2)*(d*x^2+c)^(1/2)*x*a*c+3*d^(1/2)*(d*x^2+c)^(1/2)*x*b*c^2-6*ln(d^(1/2)*x+(d*x^2+c)^(1/2))*a
*c^2*d+3*ln(d^(1/2)*x+(d*x^2+c)^(1/2))*b*c^3)/(b*x^2+a)/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^{2} + a\right )}^{2}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.94447, size = 475, normalized size = 1.95 \begin{align*} \left [-\frac{3 \,{\left (b c^{3} - 2 \, a c^{2} d\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) - 2 \,{\left (8 \, b d^{3} x^{5} + 2 \,{\left (b c d^{2} + 6 \, a d^{3}\right )} x^{3} - 3 \,{\left (b c^{2} d - 2 \, a c d^{2}\right )} x\right )} \sqrt{d x^{2} + c}}{96 \, d^{3}}, -\frac{3 \,{\left (b c^{3} - 2 \, a c^{2} d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) -{\left (8 \, b d^{3} x^{5} + 2 \,{\left (b c d^{2} + 6 \, a d^{3}\right )} x^{3} - 3 \,{\left (b c^{2} d - 2 \, a c d^{2}\right )} x\right )} \sqrt{d x^{2} + c}}{48 \, d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(b*c^3 - 2*a*c^2*d)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(8*b*d^3*x^5 + 2*(b*
c*d^2 + 6*a*d^3)*x^3 - 3*(b*c^2*d - 2*a*c*d^2)*x)*sqrt(d*x^2 + c))/d^3, -1/48*(3*(b*c^3 - 2*a*c^2*d)*sqrt(-d)*
arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (8*b*d^3*x^5 + 2*(b*c*d^2 + 6*a*d^3)*x^3 - 3*(b*c^2*d - 2*a*c*d^2)*x)*sqr
t(d*x^2 + c))/d^3]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.10844, size = 211, normalized size = 0.87 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (4 \, b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{b c d^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + 6 \, a d^{4} \mathrm{sgn}\left (b x^{2} + a\right )}{d^{4}}\right )} x^{2} - \frac{3 \,{\left (b c^{2} d^{2} \mathrm{sgn}\left (b x^{2} + a\right ) - 2 \, a c d^{3} \mathrm{sgn}\left (b x^{2} + a\right )\right )}}{d^{4}}\right )} \sqrt{d x^{2} + c} x - \frac{{\left (b c^{3} \mathrm{sgn}\left (b x^{2} + a\right ) - 2 \, a c^{2} d \mathrm{sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{16 \, d^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*b*x^2*sgn(b*x^2 + a) + (b*c*d^3*sgn(b*x^2 + a) + 6*a*d^4*sgn(b*x^2 + a))/d^4)*x^2 - 3*(b*c^2*d^2*sg
n(b*x^2 + a) - 2*a*c*d^3*sgn(b*x^2 + a))/d^4)*sqrt(d*x^2 + c)*x - 1/16*(b*c^3*sgn(b*x^2 + a) - 2*a*c^2*d*sgn(b
*x^2 + a))*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)