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

Optimal. Leaf size=195 \[ \frac {\sqrt {d} \left (24 a^2 d^2-40 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^4}+\frac {(b c-6 a d) (b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^4}+\frac {d x \sqrt {c+d x^2} (11 b c-12 a d)}{8 b^3}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2} \]

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Rubi [A]  time = 0.24, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {467, 528, 523, 217, 206, 377, 205} \begin {gather*} \frac {\sqrt {d} \left (24 a^2 d^2-40 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^4}+\frac {d x \sqrt {c+d x^2} (11 b c-12 a d)}{8 b^3}+\frac {(b c-6 a d) (b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^4}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(11*b*c - 12*a*d)*x*Sqrt[c + d*x^2])/(8*b^3) + (3*d*x*(c + d*x^2)^(3/2))/(4*b^2) - (x*(c + d*x^2)^(5/2))/(2
*b*(a + b*x^2)) + ((b*c - 6*a*d)*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*S
qrt[a]*b^4) + (Sqrt[d]*(15*b^2*c^2 - 40*a*b*c*d + 24*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*b^4)

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]

Rule 528

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx &=-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {\left (c+d x^2\right )^{3/2} \left (c+6 d x^2\right )}{a+b x^2} \, dx}{2 b}\\ &=\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {\sqrt {c+d x^2} \left (2 c (2 b c-3 a d)+2 d (11 b c-12 a d) x^2\right )}{a+b x^2} \, dx}{8 b^2}\\ &=\frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {2 c \left (4 b^2 c^2-17 a b c d+12 a^2 d^2\right )+2 d \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{16 b^3}\\ &=\frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\left ((b c-6 a d) (b c-a d)^2\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 b^4}+\frac {\left (d \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 b^4}\\ &=\frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\left ((b c-6 a d) (b c-a d)^2\right ) \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^4}+\frac {\left (d \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 b^4}\\ &=\frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {(b c-6 a d) (b c-a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^4}+\frac {\sqrt {d} \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 173, normalized size = 0.89 \begin {gather*} \frac {\sqrt {d} \left (24 a^2 d^2-40 a b c d+15 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )+b x \sqrt {c+d x^2} \left (-\frac {4 (b c-a d)^2}{a+b x^2}+d (9 b c-8 a d)+2 b d^2 x^2\right )+\frac {4 (b c-6 a d) (b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a}}}{8 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*Sqrt[c + d*x^2]*(d*(9*b*c - 8*a*d) + 2*b*d^2*x^2 - (4*(b*c - a*d)^2)/(a + b*x^2)) + (4*(b*c - 6*a*d)*(b*c
- a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/Sqrt[a] + Sqrt[d]*(15*b^2*c^2 - 40*a*b*c*
d + 24*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(8*b^4)

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IntegrateAlgebraic [A]  time = 1.02, size = 245, normalized size = 1.26 \begin {gather*} \frac {\left (-24 a^2 d^{5/2}+40 a b c d^{3/2}-15 b^2 c^2 \sqrt {d}\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{8 b^4}-\frac {\sqrt {b c-a d} \left (6 a^2 d^2-7 a b c d+b^2 c^2\right ) \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^4}+\frac {\sqrt {c+d x^2} \left (-12 a^2 d^2 x+17 a b c d x-6 a b d^2 x^3-4 b^2 c^2 x+9 b^2 c d x^3+2 b^2 d^2 x^5\right )}{8 b^3 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-4*b^2*c^2*x + 17*a*b*c*d*x - 12*a^2*d^2*x + 9*b^2*c*d*x^3 - 6*a*b*d^2*x^3 + 2*b^2*d^2*x^5))
/(8*b^3*(a + b*x^2)) - (Sqrt[b*c - a*d]*(b^2*c^2 - 7*a*b*c*d + 6*a^2*d^2)*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^4) + ((-15*b^2*c^2*Sqrt[d] + 40*a*b*c*d^(3/2) -
24*a^2*d^(5/2))*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(8*b^4)

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fricas [A]  time = 3.14, size = 1379, normalized size = 7.07

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((15*a*b^2*c^2 - 40*a^2*b*c*d + 24*a^3*d^2 + (15*b^3*c^2 - 40*a*b^2*c*d + 24*a^2*b*d^2)*x^2)*sqrt(d)*log
(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(a*b^2*c^2 - 7*a^2*b*c*d + 6*a^3*d^2 + (b^3*c^2 - 7*a*b^2*c*d
+ 6*a^2*b*d^2)*x^2)*sqrt(-(b*c - a*d)/a)*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*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*
x^2 + a^2)) + 2*(2*b^3*d^2*x^5 + 3*(3*b^3*c*d - 2*a*b^2*d^2)*x^3 - (4*b^3*c^2 - 17*a*b^2*c*d + 12*a^2*b*d^2)*x
)*sqrt(d*x^2 + c))/(b^5*x^2 + a*b^4), -1/8*((15*a*b^2*c^2 - 40*a^2*b*c*d + 24*a^3*d^2 + (15*b^3*c^2 - 40*a*b^2
*c*d + 24*a^2*b*d^2)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (a*b^2*c^2 - 7*a^2*b*c*d + 6*a^3*d^2 +
(b^3*c^2 - 7*a*b^2*c*d + 6*a^2*b*d^2)*x^2)*sqrt(-(b*c - a*d)/a)*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*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d
)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - (2*b^3*d^2*x^5 + 3*(3*b^3*c*d - 2*a*b^2*d^2)*x^3 - (4*b^3*c^2 - 17*a*b^2*
c*d + 12*a^2*b*d^2)*x)*sqrt(d*x^2 + c))/(b^5*x^2 + a*b^4), 1/16*(4*(a*b^2*c^2 - 7*a^2*b*c*d + 6*a^3*d^2 + (b^3
*c^2 - 7*a*b^2*c*d + 6*a^2*b*d^2)*x^2)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c
)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)) + (15*a*b^2*c^2 - 40*a^2*b*c*d + 24*a^3*d^2 +
(15*b^3*c^2 - 40*a*b^2*c*d + 24*a^2*b*d^2)*x^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(
2*b^3*d^2*x^5 + 3*(3*b^3*c*d - 2*a*b^2*d^2)*x^3 - (4*b^3*c^2 - 17*a*b^2*c*d + 12*a^2*b*d^2)*x)*sqrt(d*x^2 + c)
)/(b^5*x^2 + a*b^4), -1/8*((15*a*b^2*c^2 - 40*a^2*b*c*d + 24*a^3*d^2 + (15*b^3*c^2 - 40*a*b^2*c*d + 24*a^2*b*d
^2)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 2*(a*b^2*c^2 - 7*a^2*b*c*d + 6*a^3*d^2 + (b^3*c^2 - 7*a
*b^2*c*d + 6*a^2*b*d^2)*x^2)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt((b*
c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)) - (2*b^3*d^2*x^5 + 3*(3*b^3*c*d - 2*a*b^2*d^2)*x^3 - (4
*b^3*c^2 - 17*a*b^2*c*d + 12*a^2*b*d^2)*x)*sqrt(d*x^2 + c))/(b^5*x^2 + a*b^4)]

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giac [B]  time = 0.57, size = 446, normalized size = 2.29 \begin {gather*} \frac {1}{8} \, \sqrt {d x^{2} + c} {\left (\frac {2 \, d^{2} x^{2}}{b^{2}} + \frac {9 \, b^{7} c d^{3} - 8 \, a b^{6} d^{4}}{b^{9} d^{2}}\right )} x - \frac {{\left (15 \, b^{2} c^{2} \sqrt {d} - 40 \, a b c d^{\frac {3}{2}} + 24 \, a^{2} d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{16 \, b^{4}} - \frac {{\left (b^{3} c^{3} \sqrt {d} - 8 \, a b^{2} c^{2} d^{\frac {3}{2}} + 13 \, a^{2} b c d^{\frac {5}{2}} - 6 \, a^{3} d^{\frac {7}{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^{4}} + \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{3} c^{3} \sqrt {d} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{2} c^{2} d^{\frac {3}{2}} + 5 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} b c d^{\frac {5}{2}} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{3} d^{\frac {7}{2}} - b^{3} c^{4} \sqrt {d} + 2 \, a b^{2} c^{3} d^{\frac {3}{2}} - a^{2} b c^{2} d^{\frac {5}{2}}}{{\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^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(d*x^2 + c)*(2*d^2*x^2/b^2 + (9*b^7*c*d^3 - 8*a*b^6*d^4)/(b^9*d^2))*x - 1/16*(15*b^2*c^2*sqrt(d) - 40*
a*b*c*d^(3/2) + 24*a^2*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/b^4 - 1/2*(b^3*c^3*sqrt(d) - 8*a*b^2*c^2*
d^(3/2) + 13*a^2*b*c*d^(5/2) - 6*a^3*d^(7/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqr
t(a*b*c*d - a^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)*b^4) + ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^3*c^3*sqrt(d) - 4*(sq
rt(d)*x - sqrt(d*x^2 + c))^2*a*b^2*c^2*d^(3/2) + 5*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*b*c*d^(5/2) - 2*(sqrt(d
)*x - sqrt(d*x^2 + c))^2*a^3*d^(7/2) - b^3*c^4*sqrt(d) + 2*a*b^2*c^3*d^(3/2) - a^2*b*c^2*d^(5/2))/(((sqrt(d)*x
- sqrt(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^4)

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maple [B]  time = 0.02, size = 7459, normalized size = 38.25 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

integrate((d*x^2 + c)^(5/2)*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\,{\left (d\,x^2+c\right )}^{5/2}}{{\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)^(5/2))/(a + b*x^2)^2,x)

[Out]

int((x^2*(c + d*x^2)^(5/2))/(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} \left (c + d x^{2}\right )^{\frac {5}{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)**(5/2)/(b*x**2+a)**2,x)

[Out]

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

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