\(\int \frac {(a+b x^2)^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx\) [847]

Optimal result

Integrand size = 28, antiderivative size = 438 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx=-\frac {2 a^2 \sqrt {c+d x^2}}{9 c e (e x)^{9/2}}-\frac {2 a (18 b c-7 a d) \sqrt {c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac {2 \left (15 b^2 c^2-18 a b c d+7 a^2 d^2\right ) \sqrt {c+d x^2}}{15 c^3 e^5 \sqrt {e x}}+\frac {2 \sqrt {d} \left (15 b^2 c^2-18 a b c d+7 a^2 d^2\right ) \sqrt {e x} \sqrt {c+d x^2}}{15 c^3 e^6 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 \sqrt [4]{d} \left (15 b^2 c^2-18 a b c d+7 a^2 d^2\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt {c+d x^2}}+\frac {\sqrt [4]{d} \left (15 b^2 c^2-18 a b c d+7 a^2 d^2\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt {c+d x^2}} \]

[Out]

-2/9*a^2*(d*x^2+c)^(1/2)/c/e/(e*x)^(9/2)-2/45*a*(-7*a*d+18*b*c)*(d*x^2+c)^(1/2)/c^2/e^3/(e*x)^(5/2)-2/15*(7*a^
2*d^2-18*a*b*c*d+15*b^2*c^2)*(d*x^2+c)^(1/2)/c^3/e^5/(e*x)^(1/2)+2/15*(7*a^2*d^2-18*a*b*c*d+15*b^2*c^2)*d^(1/2
)*(e*x)^(1/2)*(d*x^2+c)^(1/2)/c^3/e^6/(c^(1/2)+x*d^(1/2))-2/15*d^(1/4)*(7*a^2*d^2-18*a*b*c*d+15*b^2*c^2)*(cos(
2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))*Ell
ipticE(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)
+x*d^(1/2))^2)^(1/2)/c^(11/4)/e^(11/2)/(d*x^2+c)^(1/2)+1/15*d^(1/4)*(7*a^2*d^2-18*a*b*c*d+15*b^2*c^2)*(cos(2*a
rctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))*Ellipt
icF(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*
d^(1/2))^2)^(1/2)/c^(11/4)/e^(11/2)/(d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {473, 464, 331, 335, 311, 226, 1210} \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt [4]{d} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt {c+d x^2}}-\frac {2 \sqrt [4]{d} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt {c+d x^2}}+\frac {2 \sqrt {d} \sqrt {e x} \sqrt {c+d x^2} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right )}{15 c^3 e^6 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 \sqrt {c+d x^2} \left (7 a^2 d^2-18 a b c d+15 b^2 c^2\right )}{15 c^3 e^5 \sqrt {e x}}-\frac {2 a^2 \sqrt {c+d x^2}}{9 c e (e x)^{9/2}}-\frac {2 a \sqrt {c+d x^2} (18 b c-7 a d)}{45 c^2 e^3 (e x)^{5/2}} \]

[In]

Int[(a + b*x^2)^2/((e*x)^(11/2)*Sqrt[c + d*x^2]),x]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(9*c*e*(e*x)^(9/2)) - (2*a*(18*b*c - 7*a*d)*Sqrt[c + d*x^2])/(45*c^2*e^3*(e*x)^(5/2))
- (2*(15*b^2*c^2 - 18*a*b*c*d + 7*a^2*d^2)*Sqrt[c + d*x^2])/(15*c^3*e^5*Sqrt[e*x]) + (2*Sqrt[d]*(15*b^2*c^2 -
18*a*b*c*d + 7*a^2*d^2)*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*c^3*e^6*(Sqrt[c] + Sqrt[d]*x)) - (2*d^(1/4)*(15*b^2*c^
2 - 18*a*b*c*d + 7*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan
[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(15*c^(11/4)*e^(11/2)*Sqrt[c + d*x^2]) + (d^(1/4)*(15*b^2*c^2 -
18*a*b*c*d + 7*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d
^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(15*c^(11/4)*e^(11/2)*Sqrt[c + d*x^2])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

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

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 464

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

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 1210

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2 \sqrt {c+d x^2}}{9 c e (e x)^{9/2}}+\frac {2 \int \frac {\frac {1}{2} a (18 b c-7 a d)+\frac {9}{2} b^2 c x^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx}{9 c e^2} \\ & = -\frac {2 a^2 \sqrt {c+d x^2}}{9 c e (e x)^{9/2}}-\frac {2 a (18 b c-7 a d) \sqrt {c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac {\left (4 \left (-\frac {45}{4} b^2 c^2+\frac {3}{4} a d (18 b c-7 a d)\right )\right ) \int \frac {1}{(e x)^{3/2} \sqrt {c+d x^2}} \, dx}{45 c^2 e^4} \\ & = -\frac {2 a^2 \sqrt {c+d x^2}}{9 c e (e x)^{9/2}}-\frac {2 a (18 b c-7 a d) \sqrt {c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac {2 \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt {c+d x^2}}{15 c^3 e^5 \sqrt {e x}}-\frac {\left (4 d \left (-\frac {45}{4} b^2 c^2+\frac {3}{4} a d (18 b c-7 a d)\right )\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x^2}} \, dx}{45 c^3 e^6} \\ & = -\frac {2 a^2 \sqrt {c+d x^2}}{9 c e (e x)^{9/2}}-\frac {2 a (18 b c-7 a d) \sqrt {c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac {2 \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt {c+d x^2}}{15 c^3 e^5 \sqrt {e x}}-\frac {\left (8 d \left (-\frac {45}{4} b^2 c^2+\frac {3}{4} a d (18 b c-7 a d)\right )\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{45 c^3 e^7} \\ & = -\frac {2 a^2 \sqrt {c+d x^2}}{9 c e (e x)^{9/2}}-\frac {2 a (18 b c-7 a d) \sqrt {c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac {2 \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt {c+d x^2}}{15 c^3 e^5 \sqrt {e x}}-\frac {\left (8 \sqrt {d} \left (-\frac {45}{4} b^2 c^2+\frac {3}{4} a d (18 b c-7 a d)\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{45 c^{5/2} e^6}+\frac {\left (8 \sqrt {d} \left (-\frac {45}{4} b^2 c^2+\frac {3}{4} a d (18 b c-7 a d)\right )\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {d} x^2}{\sqrt {c} e}}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{45 c^{5/2} e^6} \\ & = -\frac {2 a^2 \sqrt {c+d x^2}}{9 c e (e x)^{9/2}}-\frac {2 a (18 b c-7 a d) \sqrt {c+d x^2}}{45 c^2 e^3 (e x)^{5/2}}-\frac {2 \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt {c+d x^2}}{15 c^3 e^5 \sqrt {e x}}+\frac {2 \sqrt {d} \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{15 c^3 e^6 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 \sqrt [4]{d} \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt {c+d x^2}}+\frac {\sqrt [4]{d} \left (15 b^2 c^2-a d (18 b c-7 a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 c^{11/4} e^{11/2} \sqrt {c+d x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 11.15 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.35 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx=\frac {2 \sqrt {e x} \left (-\left (\left (c+d x^2\right ) \left (45 b^2 c^2 x^4+18 a b c x^2 \left (c-3 d x^2\right )+a^2 \left (5 c^2-7 c d x^2+21 d^2 x^4\right )\right )\right )+d \left (15 b^2 c^2-18 a b c d+7 a^2 d^2\right ) x^6 \sqrt {1+\frac {d x^2}{c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {d x^2}{c}\right )\right )}{45 c^3 e^6 x^5 \sqrt {c+d x^2}} \]

[In]

Integrate[(a + b*x^2)^2/((e*x)^(11/2)*Sqrt[c + d*x^2]),x]

[Out]

(2*Sqrt[e*x]*(-((c + d*x^2)*(45*b^2*c^2*x^4 + 18*a*b*c*x^2*(c - 3*d*x^2) + a^2*(5*c^2 - 7*c*d*x^2 + 21*d^2*x^4
))) + d*(15*b^2*c^2 - 18*a*b*c*d + 7*a^2*d^2)*x^6*Sqrt[1 + (d*x^2)/c]*Hypergeometric2F1[1/2, 3/4, 7/4, -((d*x^
2)/c)]))/(45*c^3*e^6*x^5*Sqrt[c + d*x^2])

Maple [A] (verified)

Time = 3.15 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.68

method	result	size
risch	\(-\frac {2 \sqrt {d \,x^{2}+c}\, \left (21 a^{2} d^{2} x^{4}-54 x^{4} a b c d +45 b^{2} c^{2} x^{4}-7 a^{2} c d \,x^{2}+18 a b \,c^{2} x^{2}+5 a^{2} c^{2}\right )}{45 c^{3} x^{4} e^{5} \sqrt {e x}}+\frac {\left (7 a^{2} d^{2}-18 a b c d +15 b^{2} c^{2}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right ) \sqrt {e x \left (d \,x^{2}+c \right )}}{15 c^{3} \sqrt {d e \,x^{3}+c e x}\, e^{5} \sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(299\)
elliptic	\(\frac {\sqrt {e x \left (d \,x^{2}+c \right )}\, \left (-\frac {2 a^{2} \sqrt {d e \,x^{3}+c e x}}{9 e^{6} c \,x^{5}}+\frac {2 a \left (7 a d -18 b c \right ) \sqrt {d e \,x^{3}+c e x}}{45 e^{6} c^{2} x^{3}}-\frac {2 \left (d e \,x^{2}+c e \right ) \left (7 a^{2} d^{2}-18 a b c d +15 b^{2} c^{2}\right )}{15 e^{6} c^{3} \sqrt {x \left (d e \,x^{2}+c e \right )}}+\frac {\left (7 a^{2} d^{2}-18 a b c d +15 b^{2} c^{2}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right )}{15 c^{3} e^{5} \sqrt {d e \,x^{3}+c e x}}\right )}{\sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(331\)
default	\(\frac {42 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2} x^{4}-108 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d \,x^{4}+90 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3} x^{4}-21 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2} x^{4}+54 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d \,x^{4}-45 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3} x^{4}-42 a^{2} d^{3} x^{6}+108 x^{6} d^{2} a b c -90 b^{2} c^{2} d \,x^{6}-28 a^{2} c \,d^{2} x^{4}+72 a b \,c^{2} d \,x^{4}-90 b^{2} c^{3} x^{4}+4 a^{2} c^{2} d \,x^{2}-36 a b \,c^{3} x^{2}-10 a^{2} c^{3}}{45 x^{4} \sqrt {d \,x^{2}+c}\, e^{5} \sqrt {e x}\, c^{3}}\)	\(667\)

[In]

int((b*x^2+a)^2/(e*x)^(11/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/45*(d*x^2+c)^(1/2)*(21*a^2*d^2*x^4-54*a*b*c*d*x^4+45*b^2*c^2*x^4-7*a^2*c*d*x^2+18*a*b*c^2*x^2+5*a^2*c^2)/c^
3/x^4/e^5/(e*x)^(1/2)+1/15*(7*a^2*d^2-18*a*b*c*d+15*b^2*c^2)/c^3*(-c*d)^(1/2)*((x+(-c*d)^(1/2)/d)/(-c*d)^(1/2)
*d)^(1/2)*(-2*(x-(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)/(d*e*x^3+c*e*x)^(1/2)*(-2*(-c
*d)^(1/2)/d*EllipticE(((x+(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2),1/2*2^(1/2))+(-c*d)^(1/2)/d*EllipticF(((x+(-c*
d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2),1/2*2^(1/2)))/e^5*(e*x*(d*x^2+c))^(1/2)/(e*x)^(1/2)/(d*x^2+c)^(1/2)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.30 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx=-\frac {2 \, {\left (3 \, {\left (15 \, b^{2} c^{2} - 18 \, a b c d + 7 \, a^{2} d^{2}\right )} \sqrt {d e} x^{5} {\rm weierstrassZeta}\left (-\frac {4 \, c}{d}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right )\right ) + {\left (3 \, {\left (15 \, b^{2} c^{2} - 18 \, a b c d + 7 \, a^{2} d^{2}\right )} x^{4} + 5 \, a^{2} c^{2} + {\left (18 \, a b c^{2} - 7 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}\right )}}{45 \, c^{3} e^{6} x^{5}} \]

[In]

integrate((b*x^2+a)^2/(e*x)^(11/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

-2/45*(3*(15*b^2*c^2 - 18*a*b*c*d + 7*a^2*d^2)*sqrt(d*e)*x^5*weierstrassZeta(-4*c/d, 0, weierstrassPInverse(-4
*c/d, 0, x)) + (3*(15*b^2*c^2 - 18*a*b*c*d + 7*a^2*d^2)*x^4 + 5*a^2*c^2 + (18*a*b*c^2 - 7*a^2*c*d)*x^2)*sqrt(d
*x^2 + c)*sqrt(e*x))/(c^3*e^6*x^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx=\text {Timed out} \]

[In]

integrate((b*x**2+a)**2/(e*x)**(11/2)/(d*x**2+c)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2}}{\sqrt {d x^{2} + c} \left (e x\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^2/(e*x)^(11/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(11/2)), x)

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2}}{\sqrt {d x^{2} + c} \left (e x\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^2/(e*x)^(11/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(11/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{11/2} \sqrt {c+d x^2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2}{{\left (e\,x\right )}^{11/2}\,\sqrt {d\,x^2+c}} \,d x \]

[In]

int((a + b*x^2)^2/((e*x)^(11/2)*(c + d*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^2/((e*x)^(11/2)*(c + d*x^2)^(1/2)), x)