3.4.39 \(\int \frac {1}{(a+b x^2)^{9/4} (c+d x^2)^2} \, dx\) [339]

Optimal. Leaf size=371 \[ \frac {b (4 b c+5 a d) x}{10 a c (b c-a d)^2 \left (a+b x^2\right )^{5/4}}-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}+\frac {\sqrt {b} \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{10 a^{3/2} c (b c-a d)^3 \sqrt [4]{a+b x^2}}-\frac {\sqrt [4]{a} d^{3/2} (11 b c-2 a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c (-b c+a d)^{7/2} x}+\frac {\sqrt [4]{a} d^{3/2} (11 b c-2 a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c (-b c+a d)^{7/2} x} \]

[Out]

1/10*b*(5*a*d+4*b*c)*x/a/c/(-a*d+b*c)^2/(b*x^2+a)^(5/4)-1/2*d*x/c/(-a*d+b*c)/(b*x^2+a)^(5/4)/(d*x^2+c)+1/10*(-
5*a^2*d^2-52*a*b*c*d+12*b^2*c^2)*(1+b*x^2/a)^(1/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan
(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)/a^(3/2)/c/(-a*d+b*c)^3/(b*x
^2+a)^(1/4)-1/4*a^(1/4)*d^(3/2)*(-2*a*d+11*b*c)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^
(1/2),I)*(-b*x^2/a)^(1/2)/c/(a*d-b*c)^(7/2)/x+1/4*a^(1/4)*d^(3/2)*(-2*a*d+11*b*c)*EllipticPi((b*x^2+a)^(1/4)/a
^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/c/(a*d-b*c)^(7/2)/x

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Rubi [A]
time = 0.39, antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {425, 541, 544, 235, 233, 202, 408, 504, 1232} \begin {gather*} \frac {\sqrt {b} \sqrt [4]{\frac {b x^2}{a}+1} \left (-5 a^2 d^2-52 a b c d+12 b^2 c^2\right ) E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{10 a^{3/2} c \sqrt [4]{a+b x^2} (b c-a d)^3}-\frac {\sqrt [4]{a} d^{3/2} \sqrt {-\frac {b x^2}{a}} (11 b c-2 a d) \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c x (a d-b c)^{7/2}}+\frac {\sqrt [4]{a} d^{3/2} \sqrt {-\frac {b x^2}{a}} (11 b c-2 a d) \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c x (a d-b c)^{7/2}}-\frac {d x}{2 c \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) (b c-a d)}+\frac {b x (5 a d+4 b c)}{10 a c \left (a+b x^2\right )^{5/4} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(9/4)*(c + d*x^2)^2),x]

[Out]

(b*(4*b*c + 5*a*d)*x)/(10*a*c*(b*c - a*d)^2*(a + b*x^2)^(5/4)) - (d*x)/(2*c*(b*c - a*d)*(a + b*x^2)^(5/4)*(c +
d*x^2)) + (Sqrt[b]*(12*b^2*c^2 - 52*a*b*c*d - 5*a^2*d^2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/S
qrt[a]]/2, 2])/(10*a^(3/2)*c*(b*c - a*d)^3*(a + b*x^2)^(1/4)) - (a^(1/4)*d^(3/2)*(11*b*c - 2*a*d)*Sqrt[-((b*x^
2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(4*c*(-(b*c
) + a*d)^(7/2)*x) + (a^(1/4)*d^(3/2)*(11*b*c - 2*a*d)*Sqrt[-((b*x^2)/a)]*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b
*c) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(4*c*(-(b*c) + a*d)^(7/2)*x)

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 235

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + b*(x^2
/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 408

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[2*(Sqrt[(-b)*(x^2/a)]/x), Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 504

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s
= Denominator[Rt[-a/b, 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), Int[1/
((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 541

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]

Rule 544

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

Rule 1232

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[
a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^2\right )^{9/4} \left (c+d x^2\right )^2} \, dx &=-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}+\frac {\int \frac {2 b c-a d-\frac {7}{2} b d x^2}{\left (a+b x^2\right )^{9/4} \left (c+d x^2\right )} \, dx}{2 c (b c-a d)}\\ &=\frac {b (4 b c+5 a d) x}{10 a c (b c-a d)^2 \left (a+b x^2\right )^{5/4}}-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}-\frac {\int \frac {\frac {1}{2} \left (-6 b^2 c^2+20 a b c d-5 a^2 d^2\right )-\frac {3}{4} b d (4 b c+5 a d) x^2}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx}{5 a c (b c-a d)^2}\\ &=\frac {b (4 b c+5 a d) x}{10 a c (b c-a d)^2 \left (a+b x^2\right )^{5/4}}+\frac {b \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right ) x}{10 a^2 c (b c-a d)^3 \sqrt [4]{a+b x^2}}-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}+\frac {2 \int \frac {\frac {1}{4} \left (-6 b^3 c^3+26 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right )-\frac {1}{8} b d \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right ) x^2}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{5 a^2 c (b c-a d)^3}\\ &=\frac {b (4 b c+5 a d) x}{10 a c (b c-a d)^2 \left (a+b x^2\right )^{5/4}}+\frac {b \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right ) x}{10 a^2 c (b c-a d)^3 \sqrt [4]{a+b x^2}}-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}+\frac {\left (d^2 (11 b c-2 a d)\right ) \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{4 c (b c-a d)^3}-\frac {\left (b \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right )\right ) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{20 a^2 c (b c-a d)^3}\\ &=\frac {b (4 b c+5 a d) x}{10 a c (b c-a d)^2 \left (a+b x^2\right )^{5/4}}+\frac {b \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right ) x}{10 a^2 c (b c-a d)^3 \sqrt [4]{a+b x^2}}-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}+\frac {\left (d^2 (11 b c-2 a d) \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^4}{a}} \left (b c-a d+d x^4\right )} \, dx,x,\sqrt [4]{a+b x^2}\right )}{2 c (b c-a d)^3 x}-\frac {\left (b \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right ) \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{20 a^2 c (b c-a d)^3 \sqrt [4]{a+b x^2}}\\ &=\frac {b (4 b c+5 a d) x}{10 a c (b c-a d)^2 \left (a+b x^2\right )^{5/4}}-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}-\frac {\left (d^{3/2} (11 b c-2 a d) \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {-b c+a d}-\sqrt {d} x^2\right ) \sqrt {1-\frac {x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{4 c (b c-a d)^3 x}+\frac {\left (d^{3/2} (11 b c-2 a d) \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {-b c+a d}+\sqrt {d} x^2\right ) \sqrt {1-\frac {x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{4 c (b c-a d)^3 x}+\frac {\left (b \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right ) \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{20 a^2 c (b c-a d)^3 \sqrt [4]{a+b x^2}}\\ &=\frac {b (4 b c+5 a d) x}{10 a c (b c-a d)^2 \left (a+b x^2\right )^{5/4}}-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}+\frac {\sqrt {b} \left (12 b^2 c^2-52 a b c d-5 a^2 d^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{10 a^{3/2} c (b c-a d)^3 \sqrt [4]{a+b x^2}}-\frac {\sqrt [4]{a} d^{3/2} (11 b c-2 a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c (-b c+a d)^{7/2} x}+\frac {\sqrt [4]{a} d^{3/2} (11 b c-2 a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c (-b c+a d)^{7/2} x}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.76, size = 536, normalized size = 1.44 \begin {gather*} \frac {b d \left (-12 b^2 c^2+52 a b c d+5 a^2 d^2\right ) x^3 \sqrt [4]{1+\frac {b x^2}{a}} F_1\left (\frac {3}{2};\frac {1}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {6 c \left (-6 a c x \left (10 a^4 d^3+15 a^3 b d^2 \left (-2 c+d x^2\right )-6 b^4 c^2 x^2 \left (c+2 d x^2\right )+a^2 b^2 d \left (30 c^2+26 c d x^2+5 d^2 x^4\right )+2 a b^3 c \left (-5 c^2+5 c d x^2+26 d^2 x^4\right )\right ) F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^3 \left (5 a^4 d^3+10 a^3 b d^3 x^2-12 b^4 c^2 x^2 \left (c+d x^2\right )+a^2 b^2 d \left (56 c^2+56 c d x^2+5 d^2 x^4\right )+4 a b^3 c \left (-4 c^2+9 c d x^2+13 d^2 x^4\right )\right ) \left (4 a d F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{\left (a+b x^2\right ) \left (c+d x^2\right ) \left (6 a c F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-x^2 \left (4 a d F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}}{60 a^2 c^2 (b c-a d)^3 \sqrt [4]{a+b x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^2)^(9/4)*(c + d*x^2)^2),x]

[Out]

(b*d*(-12*b^2*c^2 + 52*a*b*c*d + 5*a^2*d^2)*x^3*(1 + (b*x^2)/a)^(1/4)*AppellF1[3/2, 1/4, 1, 5/2, -((b*x^2)/a),
-((d*x^2)/c)] + (6*c*(-6*a*c*x*(10*a^4*d^3 + 15*a^3*b*d^2*(-2*c + d*x^2) - 6*b^4*c^2*x^2*(c + 2*d*x^2) + a^2*
b^2*d*(30*c^2 + 26*c*d*x^2 + 5*d^2*x^4) + 2*a*b^3*c*(-5*c^2 + 5*c*d*x^2 + 26*d^2*x^4))*AppellF1[1/2, 1/4, 1, 3
/2, -((b*x^2)/a), -((d*x^2)/c)] + x^3*(5*a^4*d^3 + 10*a^3*b*d^3*x^2 - 12*b^4*c^2*x^2*(c + d*x^2) + a^2*b^2*d*(
56*c^2 + 56*c*d*x^2 + 5*d^2*x^4) + 4*a*b^3*c*(-4*c^2 + 9*c*d*x^2 + 13*d^2*x^4))*(4*a*d*AppellF1[3/2, 1/4, 2, 5
/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))/((a + b*x^2)*(
c + d*x^2)*(6*a*c*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] - x^2*(4*a*d*AppellF1[3/2, 1/4, 2, 5/
2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)]))))/(60*a^2*c^2*(b
*c - a*d)^3*(a + b*x^2)^(1/4))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {9}{4}} \left (d \,x^{2}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(9/4)/(d*x^2+c)^2,x)

[Out]

int(1/(b*x^2+a)^(9/4)/(d*x^2+c)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(9/4)*(d*x^2 + c)^2), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x^{2}\right )^{\frac {9}{4}} \left (c + d x^{2}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(9/4)/(d*x**2+c)**2,x)

[Out]

Integral(1/((a + b*x**2)**(9/4)*(c + d*x**2)**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4)/(d*x^2+c)^2,x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(9/4)*(d*x^2 + c)^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b\,x^2+a\right )}^{9/4}\,{\left (d\,x^2+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^2)^(9/4)*(c + d*x^2)^2),x)

[Out]

int(1/((a + b*x^2)^(9/4)*(c + d*x^2)^2), x)

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