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

Optimal. Leaf size=120 \[ \frac {(b c-2 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^2 \sqrt {b c-a d}}-\frac {x \sqrt {c+d x^2}}{2 b \left (a+b x^2\right )}+\frac {\sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2} \]

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Rubi [A]  time = 0.08, antiderivative size = 120, 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 = {467, 523, 217, 206, 377, 205} \begin {gather*} \frac {(b c-2 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^2 \sqrt {b c-a d}}-\frac {x \sqrt {c+d x^2}}{2 b \left (a+b x^2\right )}+\frac {\sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 467

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*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && 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}}{\left (a+b x^2\right )^2} \, dx &=-\frac {x \sqrt {c+d x^2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {c+2 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 \left (a+b x^2\right )}+\frac {d \int \frac {1}{\sqrt {c+d x^2}} \, dx}{b^2}+\frac {(b c-2 a d) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 b^2}\\ &=-\frac {x \sqrt {c+d x^2}}{2 b \left (a+b x^2\right )}+\frac {d \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{b^2}+\frac {(b c-2 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 )}{2 b^2}\\ &=-\frac {x \sqrt {c+d x^2}}{2 b \left (a+b x^2\right )}+\frac {(b c-2 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^2 \sqrt {b c-a d}}+\frac {\sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 118, normalized size = 0.98 \begin {gather*} \frac {-\frac {b x \sqrt {c+d x^2}}{a+b x^2}+\frac {(b c-2 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} \sqrt {b c-a d}}+2 \sqrt {d} \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.60, size = 145, normalized size = 1.21 \begin {gather*} \frac {(2 a d-b c) \tan ^{-1}\left (\frac {a \sqrt {d}-b x \sqrt {c+d x^2}+b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 \sqrt {a} b^2 \sqrt {b c-a d}}-\frac {x \sqrt {c+d x^2}}{2 b \left (a+b x^2\right )}-\frac {\sqrt {d} \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(x*Sqrt[c + d*x^2])/(b*(a + b*x^2)) + ((-(b*c) + 2*a*d)*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^2 - b*x*Sqrt[c +
d*x^2])/(Sqrt[a]*Sqrt[b*c - a*d])])/(2*Sqrt[a]*b^2*Sqrt[b*c - a*d]) - (Sqrt[d]*Log[-(Sqrt[d]*x) + Sqrt[c + d*x
^2]])/b^2

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fricas [B]  time = 1.11, size = 1069, normalized size = 8.91 \begin {gather*} \left [-\frac {4 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{2} + c} x - 4 \, {\left (a^{2} b c - a^{3} d + {\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (a b c - 2 \, a^{2} d + {\left (b^{2} c - 2 \, a b d\right )} x^{2}\right )} \sqrt {-a b c + a^{2} 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 )}{8 \, {\left (a^{2} b^{3} c - a^{3} b^{2} d + {\left (a b^{4} c - a^{2} b^{3} d\right )} x^{2}\right )}}, -\frac {4 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{2} + c} x + 8 \, {\left (a^{2} b c - a^{3} d + {\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (a b c - 2 \, a^{2} d + {\left (b^{2} c - 2 \, a b d\right )} x^{2}\right )} \sqrt {-a b c + a^{2} 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 )}{8 \, {\left (a^{2} b^{3} c - a^{3} b^{2} d + {\left (a b^{4} c - a^{2} b^{3} d\right )} x^{2}\right )}}, -\frac {2 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{2} + c} x - \sqrt {a b c - a^{2} d} {\left (a b c - 2 \, a^{2} d + {\left (b^{2} c - 2 \, a b d\right )} x^{2}\right )} \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 ) - 2 \, {\left (a^{2} b c - a^{3} d + {\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right )}{4 \, {\left (a^{2} b^{3} c - a^{3} b^{2} d + {\left (a b^{4} c - a^{2} b^{3} d\right )} x^{2}\right )}}, -\frac {2 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{2} + c} x - \sqrt {a b c - a^{2} d} {\left (a b c - 2 \, a^{2} d + {\left (b^{2} c - 2 \, a b d\right )} x^{2}\right )} \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 ) + 4 \, {\left (a^{2} b c - a^{3} d + {\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right )}{4 \, {\left (a^{2} b^{3} c - a^{3} b^{2} d + {\left (a b^{4} c - a^{2} b^{3} d\right )} x^{2}\right )}}\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,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.49, size = 251, normalized size = 2.09 \begin {gather*} -\frac {{\left (b c \sqrt {d} - 2 \, a d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} b^{2}} - \frac {\sqrt {d} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{2 \, b^{2}} + \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d^{\frac {3}{2}} - b c^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} b^{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,x, algorithm="giac")

[Out]

-1/2*(b*c*sqrt(d) - 2*a*d^(3/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a
^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)*b^2) - 1/2*sqrt(d)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/b^2 + ((sqrt(d)*x -
sqrt(d*x^2 + c))^2*b*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d^(3/2) - b*c^2*sqrt(d))/(((sqrt(d)*x - s
qrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)*b
^2)

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maple [B]  time = 0.01, size = 2547, normalized size = 21.22

result too large to display

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,x)

[Out]

-1/4/(-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)+1/4/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))-1/4/(-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))*a*d+1/4/(-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+1/4/b/(a*d-b*c)/(x+(-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)^(3/2)+1/4/b^2*(-a*b)^(1/2)*d/(a*d-b*c)*((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)+1/4/b^2*a*d^(3/2)/(a*d-b*c)*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))+1/4/b^3
*(-a*b)^(1/2)*d^2/(a*d-b*c)/(-(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))*a-1/4/b^2*(-a*b)^(1/2)*d/(a*d-b*c)/(-(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-1/4/b*d/(a*d-b*c)*((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-1/4/b*d^(1/2)/(a*d-b*c)*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))*c+1/4/(-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)+1/4/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))+1/4/(-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))*a*d-1/4
/(-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+
1/4/b/(a*d-b*c)/(x-(-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)^
(3/2)-1/4/b^2*(-a*b)^(1/2)*d/(a*d-b*c)*((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)+1/4/b^2*a*d^(3/2)/(a*d-b*c)*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))-1/4/b^3*(-a*b)^(1/2)*d^2/(a*d-b*c)/(-(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))*a+1/4/b^2*(-a*b)^(1/2)*d/(a*d-b*
c)/(-(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-1/4/b*d/(a*d-
b*c)*((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-1/4/b*d^(1/2)/(a*d-b*c
)*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))*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}}{{\left (b x^{2} + a\right )}^{2}}\,{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)^2,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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

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