3.3.67 \(\int x^2 \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [267]

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

[Out]

1/6*b*x^3*(d*x^2+c)^(3/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)+1/16*c^2*(-2*a*d+b*c)*arctanh(x*d^(1/2)/(d*x^2+c)^(1
/2))*((b*x^2+a)^2)^(1/2)/d^(5/2)/(b*x^2+a)-1/16*c*(-2*a*d+b*c)*x*(d*x^2+c)^(1/2)*((b*x^2+a)^2)^(1/2)/d^2/(b*x^
2+a)-1/8*(-2*a*d+b*c)*x^3*(d*x^2+c)^(1/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)

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Rubi [A]
time = 0.09, antiderivative size = 243, normalized size of antiderivative = 1.00, 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 = {1264, 470, 285, 327, 223, 212} \begin {gather*} \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 )} \end {gather*}

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]

-1/16*(c*(b*c - 2*a*d)*x*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(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)
/Sqrt[c + d*x^2]])/(16*d^(5/2)*(a + b*x^2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 285

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 327

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 470

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 1264

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.12, size = 120, normalized size = 0.49 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (\sqrt {d} x \sqrt {c+d x^2} \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^2 (b c-2 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{48 d^{5/2} \left (a+b x^2\right )} \end {gather*}

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

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Maple [A]
time = 0.12, size = 159, normalized size = 0.65

method	result	size
risch	\(\frac {x \left (8 b \,x^{4} d^{2}+12 a \,d^{2} x^{2}+2 b c d \,x^{2}+6 a c d -3 b \,c^{2}\right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (b \,x^{2}+a \right )^{2}}}{48 d^{2} \left (b \,x^{2}+a \right )}+\frac {\left (-\frac {c^{2} a \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {3}{2}}}+\frac {c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b}{16 d^{\frac {5}{2}}}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(147\)
default	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (8 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{\frac {3}{2}} b \,x^{3}+12 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{\frac {3}{2}} a x -6 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, b c x -6 \sqrt {d \,x^{2}+c}\, d^{\frac {3}{2}} a c x +3 \sqrt {d \,x^{2}+c}\, \sqrt {d}\, b \,c^{2} x -6 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) a \,c^{2} d +3 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b \,c^{3}\right )}{48 \left (b \,x^{2}+a \right ) d^{\frac {5}{2}}}\)	\(159\)

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,method=_RETURNVERBOSE)

[Out]

1/48*((b*x^2+a)^2)^(1/2)*(8*(d*x^2+c)^(3/2)*d^(3/2)*b*x^3+12*(d*x^2+c)^(3/2)*d^(3/2)*a*x-6*(d*x^2+c)^(3/2)*d^(
1/2)*b*c*x-6*(d*x^2+c)^(1/2)*d^(3/2)*a*c*x+3*(d*x^2+c)^(1/2)*d^(1/2)*b*c^2*x-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 [A]
time = 0.29, size = 124, normalized size = 0.51 \begin {gather*} \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b x^{3}}{6 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b c x}{8 \, d^{2}} + \frac {\sqrt {d x^{2} + c} b c^{2} x}{16 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a x}{4 \, d} - \frac {\sqrt {d x^{2} + c} a c x}{8 \, d} + \frac {b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {5}{2}}} - \frac {a c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {3}{2}}} \end {gather*}

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]

1/6*(d*x^2 + c)^(3/2)*b*x^3/d - 1/8*(d*x^2 + c)^(3/2)*b*c*x/d^2 + 1/16*sqrt(d*x^2 + c)*b*c^2*x/d^2 + 1/4*(d*x^
2 + c)^(3/2)*a*x/d - 1/8*sqrt(d*x^2 + c)*a*c*x/d + 1/16*b*c^3*arcsinh(d*x/sqrt(c*d))/d^(5/2) - 1/8*a*c^2*arcsi
nh(d*x/sqrt(c*d))/d^(3/2)

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Fricas [A]
time = 0.38, size = 206, normalized size = 0.85 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \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**2*(d*x**2+c)**(1/2)*((b*x**2+a)**2)**(1/2),x)

[Out]

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

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Giac [A]
time = 4.42, size = 156, normalized size = 0.64 \begin {gather*} \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 {gather*}

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)

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

[Out]

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

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