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

Optimal. Leaf size=108 \[ \frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{5/2}}{5 d^2 \left (a+b x^2\right )}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2} (b c-a d)}{3 d^2 \left (a+b x^2\right )} \]

[Out]

-((b*c - a*d)*(c + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^2*(a + b*x^2)) + (b*(c + d*x^2)^(5/2)*Sq
rt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d^2*(a + b*x^2))

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Rubi [A]  time = 0.101093, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {1247, 646, 43} \[ \frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{5/2}}{5 d^2 \left (a+b x^2\right )}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2} (b c-a d)}{3 d^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c - a*d)*(c + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^2*(a + b*x^2)) + (b*(c + d*x^2)^(5/2)*Sq
rt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d^2*(a + b*x^2))

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 646

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0303362, size = 56, normalized size = 0.52 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \left (c+d x^2\right )^{3/2} \left (5 a d-2 b c+3 b d x^2\right )}{15 d^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x*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]*(c + d*x^2)^(3/2)*(-2*b*c + 5*a*d + 3*b*d*x^2))/(15*d^2*(a + b*x^2))

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Maple [A]  time = 0.004, size = 51, normalized size = 0.5 \begin{align*}{\frac{3\,b{x}^{2}d+5\,ad-2\,bc}{15\,{d}^{2} \left ( b{x}^{2}+a \right ) } \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(d*x^2+c)^(3/2)*(3*b*d*x^2+5*a*d-2*b*c)*((b*x^2+a)^2)^(1/2)/d^2/(b*x^2+a)

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Maxima [A]  time = 0.96501, size = 68, normalized size = 0.63 \begin{align*} \frac{{\left (3 \, b d^{2} x^{4} - 2 \, b c^{2} + 5 \, a c d +{\left (b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{15 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b*d^2*x^4 - 2*b*c^2 + 5*a*c*d + (b*c*d + 5*a*d^2)*x^2)*sqrt(d*x^2 + c)/d^2

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Fricas [A]  time = 1.80305, size = 113, normalized size = 1.05 \begin{align*} \frac{{\left (3 \, b d^{2} x^{4} - 2 \, b c^{2} + 5 \, a c d +{\left (b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{15 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b*d^2*x^4 - 2*b*c^2 + 5*a*c*d + (b*c*d + 5*a*d^2)*x^2)*sqrt(d*x^2 + c)/d^2

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

[Out]

Timed out

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Giac [A]  time = 1.09989, size = 85, normalized size = 0.79 \begin{align*} \frac{5 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{{\left (3 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} - 5 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} c\right )} b \mathrm{sgn}\left (b x^{2} + a\right )}{d}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(5*(d*x^2 + c)^(3/2)*a*sgn(b*x^2 + a) + (3*(d*x^2 + c)^(5/2) - 5*(d*x^2 + c)^(3/2)*c)*b*sgn(b*x^2 + a)/d)
/d