3.724 \(\int \frac {x^2}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=115 \[ \frac {x (a d+2 b c)}{3 c \sqrt {c+d x^2} (b c-a d)^2}+\frac {x}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {\sqrt {a} b \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {471, 527, 12, 377, 205} \[ \frac {x (a d+2 b c)}{3 c \sqrt {c+d x^2} (b c-a d)^2}+\frac {x}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {\sqrt {a} b \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 471

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 + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rubi steps

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

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Mathematica [C]  time = 3.08, size = 257, normalized size = 2.23 \[ \frac {12 x^6 \left (c+d x^2\right ) (b c-a d)^3 \, _2F_1\left (2,2;\frac {9}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )-\frac {35 c \left (a+b x^2\right ) \left (5 c+2 d x^2\right ) \left (-3 a^2 \left (c+d x^2\right )^2 \sin ^{-1}\left (\sqrt {\frac {x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )-c \left (a+b x^2\right ) \sqrt {\frac {a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}} \left (-3 a c-4 a d x^2+b c x^2\right )\right )}{\sqrt {\frac {a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}}}}{315 c^3 x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-35*c*(a + b*x^2)*(5*c + 2*d*x^2)*(-(c*(a + b*x^2)*Sqrt[(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)]
*(-3*a*c + b*c*x^2 - 4*a*d*x^2)) - 3*a^2*(c + d*x^2)^2*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]]))/Sqrt[
(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)] + 12*(b*c - a*d)^3*x^6*(c + d*x^2)*Hypergeometric2F1[2, 2
, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(315*c^3*(b*c - a*d)^2*x*(a + b*x^2)^2*(c + d*x^2)^(3/2))

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fricas [B]  time = 1.55, size = 550, normalized size = 4.78 \[ \left [\frac {3 \, {\left (b c d^{2} x^{4} + 2 \, b c^{2} d x^{2} + b c^{3}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (3 \, b c^{2} x + {\left (2 \, b c d + a d^{2}\right )} x^{3}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} x^{2}\right )}}, \frac {3 \, {\left (b c d^{2} x^{4} + 2 \, b c^{2} d x^{2} + b c^{3}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right ) + 2 \, {\left (3 \, b c^{2} x + {\left (2 \, b c d + a d^{2}\right )} x^{3}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(b*c*d^2*x^4 + 2*b*c^2*d*x^2 + b*c^3)*sqrt(-a/(b*c - a*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^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)*x)
*sqrt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(3*b*c^2*x + (2*b*c*d + a*d^2)*x^3)*sq
rt(d*x^2 + c))/(b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^4 + 2*(b^2*c
^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*x^2), 1/6*(3*(b*c*d^2*x^4 + 2*b*c^2*d*x^2 + b*c^3)*sqrt(a/(b*c - a*d))*arc
tan(-1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)) + 2*(3*b*c^2*x + (2*
b*c*d + a*d^2)*x^3)*sqrt(d*x^2 + c))/(b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2
*c*d^4)*x^4 + 2*(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*x^2)]

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giac [B]  time = 0.45, size = 291, normalized size = 2.53 \[ \frac {a b \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {{\left (\frac {{\left (2 \, b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + a^{3} d^{5}\right )} x^{2}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}} + \frac {3 \, {\left (b^{3} c^{4} d - 2 \, a b^{2} c^{3} d^{2} + a^{2} b c^{2} d^{3}\right )}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

a*b*sqrt(d)*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))/((b^2*c^2 -
2*a*b*c*d + a^2*d^2)*sqrt(a*b*c*d - a^2*d^2)) + 1/3*((2*b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 + a^3*d^5)*x^2/(b^4*c^5*
d - 4*a*b^3*c^4*d^2 + 6*a^2*b^2*c^3*d^3 - 4*a^3*b*c^2*d^4 + a^4*c*d^5) + 3*(b^3*c^4*d - 2*a*b^2*c^3*d^2 + a^2*
b*c^2*d^3)/(b^4*c^5*d - 4*a*b^3*c^4*d^2 + 6*a^2*b^2*c^3*d^3 - 4*a^3*b*c^2*d^4 + a^4*c*d^5))*x/(d*x^2 + c)^(3/2
)

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maple [B]  time = 0.02, size = 1134, normalized size = 9.86 \[ \frac {a b \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}\, \left (a d -b c \right )^{2} \sqrt {-\frac {a d -b c}{b}}}-\frac {a b \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}\, \left (a d -b c \right )^{2} \sqrt {-\frac {a d -b c}{b}}}+\frac {a b}{2 \sqrt {-a b}\, \left (a d -b c \right )^{2} \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}}}-\frac {a b}{2 \sqrt {-a b}\, \left (a d -b c \right )^{2} \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}}}+\frac {a d x}{2 \left (a d -b c \right )^{2} \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}}\, c}+\frac {a d x}{2 \left (a d -b c \right )^{2} \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}}\, c}-\frac {a d x}{6 \left (a d -b c \right ) \left (\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 )^{\frac {3}{2}} b c}-\frac {a d x}{6 \left (a d -b c \right ) \left (\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 )^{\frac {3}{2}} b c}-\frac {a}{6 \sqrt {-a b}\, \left (a d -b c \right ) \left (\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 )^{\frac {3}{2}}}+\frac {a}{6 \sqrt {-a b}\, \left (a d -b c \right ) \left (\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 )^{\frac {3}{2}}}-\frac {a d x}{3 \left (a d -b c \right ) \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}}\, b \,c^{2}}-\frac {a d x}{3 \left (a d -b c \right ) \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}}\, b \,c^{2}}+\frac {x}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} b c}+\frac {2 x}{3 \sqrt {d \,x^{2}+c}\, b \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^2/((b*x^2 + a)*(d*x^2 + c)^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a + b*x^2)*(c + d*x^2)^(5/2)),x)

[Out]

int(x^2/((a + b*x^2)*(c + d*x^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x**2+a)/(d*x**2+c)**(5/2),x)

[Out]

Integral(x**2/((a + b*x**2)*(c + d*x**2)**(5/2)), x)

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