\(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^{11/2}} \, dx\) [829]

Optimal result

Integrand size = 26, antiderivative size = 386 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^{11/2}} \, dx=-\frac {2 \left (15 b^2 c^2+a d (6 b c-a d)\right ) \sqrt {c+d x^2}}{15 c^2 \sqrt {x}}+\frac {4 \sqrt {d} \left (15 b^2 c^2+a d (6 b c-a d)\right ) \sqrt {x} \sqrt {c+d x^2}}{15 c^2 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac {2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}-\frac {4 \sqrt [4]{d} \left (15 b^2 c^2+a d (6 b c-a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{15 c^{7/4} \sqrt {c+d x^2}}+\frac {2 \sqrt [4]{d} \left (15 b^2 c^2+a d (6 b c-a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{15 c^{7/4} \sqrt {c+d x^2}} \]

[Out]

-2/9*a^2*(d*x^2+c)^(3/2)/c/x^(9/2)-2/15*a*(-a*d+6*b*c)*(d*x^2+c)^(3/2)/c^2/x^(5/2)-2/15*(15*b^2*c^2+a*d*(-a*d+
6*b*c))*(d*x^2+c)^(1/2)/c^2/x^(1/2)+4/15*(15*b^2*c^2+a*d*(-a*d+6*b*c))*d^(1/2)*x^(1/2)*(d*x^2+c)^(1/2)/c^2/(c^
(1/2)+x*d^(1/2))-4/15*d^(1/4)*(15*b^2*c^2+a*d*(-a*d+6*b*c))*(cos(2*arctan(d^(1/4)*x^(1/2)/c^(1/4)))^2)^(1/2)/c
os(2*arctan(d^(1/4)*x^(1/2)/c^(1/4)))*EllipticE(sin(2*arctan(d^(1/4)*x^(1/2)/c^(1/4))),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^(7/4)/(d*x^2+c)^(1/2)+2/15*d^(1/4)*(15*b^2*c^2+a*d*(-a*d+6
*b*c))*(cos(2*arctan(d^(1/4)*x^(1/2)/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^(1/2)/c^(1/4)))*EllipticF(sin(2
*arctan(d^(1/4)*x^(1/2)/c^(1/4))),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^(
7/4)/(d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 383, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {473, 464, 283, 335, 311, 226, 1210} \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^{11/2}} \, dx=-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}+\frac {2 \sqrt [4]{d} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (6 b c-a d)+15 b^2 c^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{15 c^{7/4} \sqrt {c+d x^2}}-\frac {4 \sqrt [4]{d} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (6 b c-a d)+15 b^2 c^2\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{15 c^{7/4} \sqrt {c+d x^2}}-\frac {2 \sqrt {c+d x^2} \left (\frac {a d (6 b c-a d)}{c^2}+15 b^2\right )}{15 \sqrt {x}}+\frac {4 \sqrt {d} \sqrt {x} \sqrt {c+d x^2} \left (a d (6 b c-a d)+15 b^2 c^2\right )}{15 c^2 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a \left (c+d x^2\right )^{3/2} (6 b c-a d)}{15 c^2 x^{5/2}} \]

[In]

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

[Out]

(-2*(15*b^2 + (a*d*(6*b*c - a*d))/c^2)*Sqrt[c + d*x^2])/(15*Sqrt[x]) + (4*Sqrt[d]*(15*b^2*c^2 + a*d*(6*b*c - a
*d))*Sqrt[x]*Sqrt[c + d*x^2])/(15*c^2*(Sqrt[c] + Sqrt[d]*x)) - (2*a^2*(c + d*x^2)^(3/2))/(9*c*x^(9/2)) - (2*a*
(6*b*c - a*d)*(c + d*x^2)^(3/2))/(15*c^2*x^(5/2)) - (4*d^(1/4)*(15*b^2*c^2 + a*d*(6*b*c - a*d))*(Sqrt[c] + Sqr
t[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan[(d^(1/4)*Sqrt[x])/c^(1/4)], 1/2])/(15*c^(
7/4)*Sqrt[c + d*x^2]) + (2*d^(1/4)*(15*b^2*c^2 + a*d*(6*b*c - a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sq
rt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[x])/c^(1/4)], 1/2])/(15*c^(7/4)*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 283

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

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 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 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}+\frac {2 \int \frac {\left (\frac {3}{2} a (6 b c-a d)+\frac {9}{2} b^2 c x^2\right ) \sqrt {c+d x^2}}{x^{7/2}} \, dx}{9 c} \\ & = -\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac {2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}+\frac {1}{15} \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right ) \int \frac {\sqrt {c+d x^2}}{x^{3/2}} \, dx \\ & = -\frac {2 \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right ) \sqrt {c+d x^2}}{15 \sqrt {x}}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac {2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}+\frac {1}{15} \left (2 d \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right )\right ) \int \frac {\sqrt {x}}{\sqrt {c+d x^2}} \, dx \\ & = -\frac {2 \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right ) \sqrt {c+d x^2}}{15 \sqrt {x}}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac {2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}+\frac {1}{15} \left (4 d \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right )\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {c+d x^4}} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {2 \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right ) \sqrt {c+d x^2}}{15 \sqrt {x}}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac {2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}+\frac {1}{15} \left (4 \sqrt {c} \sqrt {d} \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x^4}} \, dx,x,\sqrt {x}\right )-\frac {1}{15} \left (4 \sqrt {c} \sqrt {d} \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right )\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {d} x^2}{\sqrt {c}}}{\sqrt {c+d x^4}} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {2 \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right ) \sqrt {c+d x^2}}{15 \sqrt {x}}+\frac {4 \sqrt {d} \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\right ) \sqrt {x} \sqrt {c+d x^2}}{15 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{9 c x^{9/2}}-\frac {2 a (6 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^{5/2}}-\frac {4 \sqrt [4]{c} \sqrt [4]{d} \left (15 b^2+\frac {a d (6 b c-a d)}{c^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 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{15 \sqrt {c+d x^2}}+\frac {2 \sqrt [4]{c} \sqrt [4]{d} \left (15 b^2+\frac {a d (6 b c-a d)}{c^2}\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 {x}}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{15 \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 = 10.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.38 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^{11/2}} \, dx=\frac {-2 \left (c+d x^2\right ) \left (45 b^2 c^2 x^4+18 a b c x^2 \left (c+2 d x^2\right )+a^2 \left (5 c^2+2 c d x^2-6 d^2 x^4\right )\right )+12 d \left (15 b^2 c^2+6 a b c d-a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x^6 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {c}{d x^2}\right )}{45 c^2 x^{9/2} \sqrt {c+d x^2}} \]

[In]

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

[Out]

(-2*(c + d*x^2)*(45*b^2*c^2*x^4 + 18*a*b*c*x^2*(c + 2*d*x^2) + a^2*(5*c^2 + 2*c*d*x^2 - 6*d^2*x^4)) + 12*d*(15
*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^6*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))])/(45*c
^2*x^(9/2)*Sqrt[c + d*x^2])

Maple [A] (verified)

Time = 3.16 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {2 \sqrt {d \,x^{2}+c}\, \left (-6 a^{2} d^{2} x^{4}+36 x^{4} a b c d +45 b^{2} c^{2} x^{4}+2 a^{2} c d \,x^{2}+18 a b \,c^{2} x^{2}+5 a^{2} c^{2}\right )}{45 x^{\frac {9}{2}} c^{2}}-\frac {2 \left (a^{2} d^{2}-6 a b c d -15 b^{2} c^{2}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right ) \sqrt {x \left (d \,x^{2}+c \right )}}{15 c^{2} \sqrt {d \,x^{3}+c x}\, \sqrt {x}\, \sqrt {d \,x^{2}+c}}\)	\(282\)
elliptic	\(\frac {\sqrt {x \left (d \,x^{2}+c \right )}\, \left (-\frac {2 a^{2} \sqrt {d \,x^{3}+c x}}{9 x^{5}}-\frac {4 a \left (a d +9 b c \right ) \sqrt {d \,x^{3}+c x}}{45 c \,x^{3}}+\frac {2 \left (d \,x^{2}+c \right ) \left (2 a^{2} d^{2}-12 a b c d -15 b^{2} c^{2}\right )}{15 c^{2} \sqrt {x \left (d \,x^{2}+c \right )}}+\frac {\left (b^{2} d -\frac {d \left (2 a^{2} d^{2}-12 a b c d -15 b^{2} c^{2}\right )}{15 c^{2}}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right )}{d \sqrt {d \,x^{3}+c x}}\right )}{\sqrt {x}\, \sqrt {d \,x^{2}+c}}\)	\(311\)
default	\(-\frac {2 \left (6 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2} x^{4}-36 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d \,x^{4}-90 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3} x^{4}-3 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2} x^{4}+18 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d \,x^{4}+45 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3} x^{4}-6 a^{2} d^{3} x^{6}+36 x^{6} d^{2} a b c +45 b^{2} c^{2} d \,x^{6}-4 a^{2} c \,d^{2} x^{4}+54 a b \,c^{2} d \,x^{4}+45 b^{2} c^{3} x^{4}+7 a^{2} c^{2} d \,x^{2}+18 a b \,c^{3} x^{2}+5 a^{2} c^{3}\right )}{45 \sqrt {d \,x^{2}+c}\, x^{\frac {9}{2}} c^{2}}\)	\(659\)

[In]

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

[Out]

-2/45*(d*x^2+c)^(1/2)*(-6*a^2*d^2*x^4+36*a*b*c*d*x^4+45*b^2*c^2*x^4+2*a^2*c*d*x^2+18*a*b*c^2*x^2+5*a^2*c^2)/x^
(9/2)/c^2-2/15*(a^2*d^2-6*a*b*c*d-15*b^2*c^2)/c^2*(-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*x^3+c*x)^(1/2)*(-2*(-c*d)^(1/2)/d*Ellipti
cE(((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)))*(x*(d*x^2+c))^(1/2)/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 = 0.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^{11/2}} \, dx=-\frac {2 \, {\left (6 \, {\left (15 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2}\right )} \sqrt {d} x^{5} {\rm weierstrassZeta}\left (-\frac {4 \, c}{d}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right )\right ) + {\left (3 \, {\left (15 \, b^{2} c^{2} + 12 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{4} + 5 \, a^{2} c^{2} + 2 \, {\left (9 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {x}\right )}}{45 \, c^{2} x^{5}} \]

[In]

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

[Out]

-2/45*(6*(15*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(d)*x^5*weierstrassZeta(-4*c/d, 0, weierstrassPInverse(-4*c/d,
0, x)) + (3*(15*b^2*c^2 + 12*a*b*c*d - 2*a^2*d^2)*x^4 + 5*a^2*c^2 + 2*(9*a*b*c^2 + a^2*c*d)*x^2)*sqrt(d*x^2 +
c)*sqrt(x))/(c^2*x^5)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 29.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.39 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^{11/2}} \, dx=\frac {a^{2} \sqrt {c} \Gamma \left (- \frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {9}{4}, - \frac {1}{2} \\ - \frac {5}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 x^{\frac {9}{2}} \Gamma \left (- \frac {5}{4}\right )} + \frac {a b \sqrt {c} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {b^{2} \sqrt {c} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

a**2*sqrt(c)*gamma(-9/4)*hyper((-9/4, -1/2), (-5/4,), d*x**2*exp_polar(I*pi)/c)/(2*x**(9/2)*gamma(-5/4)) + a*b
*sqrt(c)*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), d*x**2*exp_polar(I*pi)/c)/(x**(5/2)*gamma(-1/4)) + b**2*sqrt
(c)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(x)*gamma(3/4))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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