3.3.68 \(\int x \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [268]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 108, normalized size of antiderivative = 1.00, 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 = {1261, 660, 45} \begin {gather*} \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 )} \end {gather*}

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]

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

Rule 45

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])

Rule 660

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 1261

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]

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} \text {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} \text {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} \text {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*}

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Mathematica [A]
time = 0.03, size = 56, normalized size = 0.52 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (c+d x^2\right )^{3/2} \left (-2 b c+5 a d+3 b d x^2\right )}{15 d^2 \left (a+b x^2\right )} \end {gather*}

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 = 0.12, size = 51, normalized size = 0.47

method	result	size
gosper	\(\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} \left (3 b \,x^{2} d +5 a d -2 b c \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{15 d^{2} \left (b \,x^{2}+a \right )}\)	\(51\)
default	\(\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} \left (3 b \,x^{2} d +5 a d -2 b c \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{15 d^{2} \left (b \,x^{2}+a \right )}\)	\(51\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (3 b \,x^{4} d^{2}+5 a \,d^{2} x^{2}+b c d \,x^{2}+5 a c d -2 b \,c^{2}\right ) \sqrt {d \,x^{2}+c}}{15 \left (b \,x^{2}+a \right ) d^{2}}\)	\(72\)

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,method=_RETURNVERBOSE)

[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 = 0.27, size = 50, normalized size = 0.46 \begin {gather*} \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b x^{2}}{5 \, d} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b c}{15 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a}{3 \, d} \end {gather*}

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]

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

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Fricas [A]
time = 0.34, size = 50, normalized size = 0.46 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {c + d x^{2}} \sqrt {\left (a + b x^{2}\right )^{2}}\, dx \end {gather*}

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]

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

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Giac [A]
time = 3.98, size = 68, normalized size = 0.63 \begin {gather*} \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b \mathrm {sgn}\left (b x^{2} + a\right ) - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b c \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a d \mathrm {sgn}\left (b x^{2} + a\right )}{15 \, d^{2}} \end {gather*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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