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

Optimal. Leaf size=112 \[ -\frac {\sqrt {a} \sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^2}+\frac {(b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 b^2 \sqrt {d}}+\frac {x \sqrt {c+d x^2}}{2 b} \]

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Rubi [A]  time = 0.10, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {478, 523, 217, 206, 377, 205} \begin {gather*} -\frac {\sqrt {a} \sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^2}+\frac {(b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 b^2 \sqrt {d}}+\frac {x \sqrt {c+d x^2}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[c + d*x^2])/(2*b) - (Sqrt[a]*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/b^
2 + ((b*c - 2*a*d)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*b^2*Sqrt[d])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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])

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 377

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

Rule 478

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 108, normalized size = 0.96 \begin {gather*} \frac {\frac {(b c-2 a d) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{\sqrt {d}}-2 \sqrt {a} \sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )+b x \sqrt {c+d x^2}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[c + d*x^2])/(a + b*x^2),x]

[Out]

(b*x*Sqrt[c + d*x^2] - 2*Sqrt[a]*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])] + ((b*c
- 2*a*d)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/(2*b^2)

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IntegrateAlgebraic [A]  time = 0.19, size = 167, normalized size = 1.49 \begin {gather*} \frac {(2 a d-b c) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{2 b^2 \sqrt {d}}+\frac {\sqrt {a} \sqrt {b c-a d} \tan ^{-1}\left (\frac {b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}-\frac {b x \sqrt {c+d x^2}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{b^2}+\frac {x \sqrt {c+d x^2}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*Sqrt[c + d*x^2])/(a + b*x^2),x]

[Out]

(x*Sqrt[c + d*x^2])/(2*b) + (Sqrt[a]*Sqrt[b*c - a*d]*ArcTan[(Sqrt[a]*Sqrt[d])/Sqrt[b*c - a*d] + (b*Sqrt[d]*x^2
)/(Sqrt[a]*Sqrt[b*c - a*d]) - (b*x*Sqrt[c + d*x^2])/(Sqrt[a]*Sqrt[b*c - a*d])])/b^2 + ((-(b*c) + 2*a*d)*Log[-(
Sqrt[d]*x) + Sqrt[c + d*x^2]])/(2*b^2*Sqrt[d])

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fricas [A]  time = 1.18, size = 690, normalized size = 6.16 \begin {gather*} \left [\frac {2 \, \sqrt {d x^{2} + c} b d x - {\left (b c - 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + \sqrt {-a b c + a^{2} d} d \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, b^{2} d}, \frac {2 \, \sqrt {d x^{2} + c} b d x - 2 \, {\left (b c - 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + \sqrt {-a b c + a^{2} d} d \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, b^{2} d}, \frac {2 \, \sqrt {d x^{2} + c} b d x - 2 \, \sqrt {a b c - a^{2} d} d \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - {\left (b c - 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right )}{4 \, b^{2} d}, \frac {\sqrt {d x^{2} + c} b d x - \sqrt {a b c - a^{2} d} d \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - {\left (b c - 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right )}{2 \, b^{2} d}\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),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(d*x^2 + c)*b*d*x - (b*c - 2*a*d)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + sqrt(-
a*b*c + a^2*d)*d*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b*
c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(b^2*d), 1/4*(2*sq
rt(d*x^2 + c)*b*d*x - 2*(b*c - 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + sqrt(-a*b*c + a^2*d)*d*log
(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b*c - 2*a*d)*x^3 - a*c
*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(b^2*d), 1/4*(2*sqrt(d*x^2 + c)*b*d*x
- 2*sqrt(a*b*c - a^2*d)*d*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d -
a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - (b*c - 2*a*d)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c
))/(b^2*d), 1/2*(sqrt(d*x^2 + c)*b*d*x - sqrt(a*b*c - a^2*d)*d*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x
^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - (b*c - 2*a*d)*sqrt(-d)*arctan(s
qrt(-d)*x/sqrt(d*x^2 + c)))/(b^2*d)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:

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maple [B]  time = 0.01, size = 1010, normalized size = 9.02 \begin {gather*} -\frac {a^{2} d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {a^{2} d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {a c \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b}-\frac {a c \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b}-\frac {a \sqrt {d}\, \ln \left (\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) d -\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 b^{2}}-\frac {a \sqrt {d}\, \ln \left (\frac {\left (x -\frac {\sqrt {-a b}}{b}\right ) d +\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 b^{2}}+\frac {c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 b \sqrt {d}}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a}{2 \sqrt {-a b}\, b}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a}{2 \sqrt {-a b}\, b}+\frac {\sqrt {d \,x^{2}+c}\, x}{2 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d x^{2} + c} x^{2}}{b x^{2} + a}\,{d x} \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),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt {c + d x^{2}}}{a + b x^{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),x)

[Out]

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

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