\(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx\) [827]

Optimal result

Integrand size = 28, antiderivative size = 421 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx=\frac {2 \left (b^2 c^2+a d (10 b c+a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{5 c^2 e^5}+\frac {4 \left (b^2 c^2+a d (10 b c+a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{5 c \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c+a d) \left (c+d x^2\right )^{3/2}}{5 c^2 e^3 \sqrt {e x}}-\frac {4 \left (b^2 c^2+a d (10 b c+a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}+\frac {2 \left (b^2 c^2+a d (10 b c+a d)\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}} \]

[Out]

-2/5*a^2*(d*x^2+c)^(3/2)/c/e/(e*x)^(5/2)-2/5*a*(a*d+10*b*c)*(d*x^2+c)^(3/2)/c^2/e^3/(e*x)^(1/2)+2/5*(b^2*c^2+a
*d*(a*d+10*b*c))*(e*x)^(3/2)*(d*x^2+c)^(1/2)/c^2/e^5+4/5*(b^2*c^2+a*d*(a*d+10*b*c))*(e*x)^(1/2)*(d*x^2+c)^(1/2
)/c/e^4/d^(1/2)/(c^(1/2)+x*d^(1/2))-4/5*(b^2*c^2+a*d*(a*d+10*b*c))*(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)))*EllipticE(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^(3/4)/d^(3/4)/e
^(7/2)/(d*x^2+c)^(1/2)+2/5*(b^2*c^2+a*d*(a*d+10*b*c))*(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)))*EllipticF(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^(3/4)/d^(3/4)/e^(7/2)/(d*x^2
+c)^(1/2)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 421, 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, 285, 335, 311, 226, 1210} \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx=-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}+\frac {2 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (a d+10 b c)+b^2 c^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}-\frac {4 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (a d+10 b c)+b^2 c^2\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}+\frac {2 (e x)^{3/2} \sqrt {c+d x^2} \left (a d (a d+10 b c)+b^2 c^2\right )}{5 c^2 e^5}+\frac {4 \sqrt {e x} \sqrt {c+d x^2} \left (a d (a d+10 b c)+b^2 c^2\right )}{5 c \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a \left (c+d x^2\right )^{3/2} (a d+10 b c)}{5 c^2 e^3 \sqrt {e x}} \]

[In]

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/(e*x)^(7/2),x]

[Out]

(2*(b^2*c^2 + a*d*(10*b*c + a*d))*(e*x)^(3/2)*Sqrt[c + d*x^2])/(5*c^2*e^5) + (4*(b^2*c^2 + a*d*(10*b*c + a*d))
*Sqrt[e*x]*Sqrt[c + d*x^2])/(5*c*Sqrt[d]*e^4*(Sqrt[c] + Sqrt[d]*x)) - (2*a^2*(c + d*x^2)^(3/2))/(5*c*e*(e*x)^(
5/2)) - (2*a*(10*b*c + a*d)*(c + d*x^2)^(3/2))/(5*c^2*e^3*Sqrt[e*x]) - (4*(b^2*c^2 + a*d*(10*b*c + a*d))*(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)*Sqr
t[e])], 1/2])/(5*c^(3/4)*d^(3/4)*e^(7/2)*Sqrt[c + d*x^2]) + (2*(b^2*c^2 + a*d*(10*b*c + a*d))*(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
])/(5*c^(3/4)*d^(3/4)*e^(7/2)*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 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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}}{5 c e (e x)^{5/2}}+\frac {2 \int \frac {\left (\frac {1}{2} a (10 b c+a d)+\frac {5}{2} b^2 c x^2\right ) \sqrt {c+d x^2}}{(e x)^{3/2}} \, dx}{5 c e^2} \\ & = -\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c+a d) \left (c+d x^2\right )^{3/2}}{5 c^2 e^3 \sqrt {e x}}+\frac {\left (b^2 c^2+a d (10 b c+a d)\right ) \int \sqrt {e x} \sqrt {c+d x^2} \, dx}{c^2 e^4} \\ & = \frac {2 \left (b^2 c^2+a d (10 b c+a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{5 c^2 e^5}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c+a d) \left (c+d x^2\right )^{3/2}}{5 c^2 e^3 \sqrt {e x}}+\frac {\left (2 \left (b^2 c^2+a d (10 b c+a d)\right )\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x^2}} \, dx}{5 c e^4} \\ & = \frac {2 \left (b^2 c^2+a d (10 b c+a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{5 c^2 e^5}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c+a d) \left (c+d x^2\right )^{3/2}}{5 c^2 e^3 \sqrt {e x}}+\frac {\left (4 \left (b^2 c^2+a d (10 b c+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 )}{5 c e^5} \\ & = \frac {2 \left (b^2 c^2+a d (10 b c+a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{5 c^2 e^5}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c+a d) \left (c+d x^2\right )^{3/2}}{5 c^2 e^3 \sqrt {e x}}+\frac {\left (4 \left (b^2 c^2+a d (10 b c+a d)\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 \sqrt {c} \sqrt {d} e^4}-\frac {\left (4 \left (b^2 c^2+a d (10 b c+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 )}{5 \sqrt {c} \sqrt {d} e^4} \\ & = \frac {2 \left (b^2 c^2+a d (10 b c+a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{5 c^2 e^5}+\frac {4 \left (b^2 c^2+a d (10 b c+a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{5 c \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c+a d) \left (c+d x^2\right )^{3/2}}{5 c^2 e^3 \sqrt {e x}}-\frac {4 \left (b^2 c^2+a d (10 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 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}+\frac {2 \left (b^2 c^2+a d (10 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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \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 = 11.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.30 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx=\frac {x \left (-2 \left (c+d x^2\right ) \left (10 a b c x^2-b^2 c x^4+a^2 \left (c+2 d x^2\right )\right )+4 \left (b^2 c^2+10 a b c d+a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {c}{d x^2}\right )\right )}{5 c (e x)^{7/2} \sqrt {c+d x^2}} \]

[In]

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/(e*x)^(7/2),x]

[Out]

(x*(-2*(c + d*x^2)*(10*a*b*c*x^2 - b^2*c*x^4 + a^2*(c + 2*d*x^2)) + 4*(b^2*c^2 + 10*a*b*c*d + a^2*d^2)*Sqrt[1
+ c/(d*x^2)]*x^4*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))]))/(5*c*(e*x)^(7/2)*Sqrt[c + d*x^2])

Maple [A] (verified)

Time = 3.13 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.65

method	result	size
risch	\(-\frac {2 \sqrt {d \,x^{2}+c}\, \left (-b^{2} c \,x^{4}+2 a^{2} d \,x^{2}+10 a b c \,x^{2}+a^{2} c \right )}{5 x^{2} c \,e^{3} \sqrt {e x}}+\frac {2 \left (a^{2} d^{2}+10 a b c d +b^{2} c^{2}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right ) \sqrt {e x \left (d \,x^{2}+c \right )}}{5 c d \sqrt {d e \,x^{3}+c e x}\, e^{3} \sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(272\)
elliptic	\(\frac {\sqrt {e x \left (d \,x^{2}+c \right )}\, \left (-\frac {2 a^{2} \sqrt {d e \,x^{3}+c e x}}{5 e^{4} x^{3}}-\frac {4 \left (d e \,x^{2}+c e \right ) a \left (a d +5 b c \right )}{5 e^{4} c \sqrt {x \left (d e \,x^{2}+c e \right )}}+\frac {2 b^{2} x \sqrt {d e \,x^{3}+c e x}}{5 e^{4}}+\frac {\left (\frac {b \left (2 a d +b c \right )}{e^{3}}+\frac {2 d a \left (a d +5 b c \right )}{5 c \,e^{3}}-\frac {3 b^{2} c}{5 e^{3}}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right )}{d \sqrt {d e \,x^{3}+c e x}}\right )}{\sqrt {e x}\, \sqrt {d \,x^{2}+c}}\)	\(316\)
default	\(\frac {\frac {4 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2} x^{2}}{5}+8 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d \,x^{2}+\frac {4 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3} x^{2}}{5}-\frac {2 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2} x^{2}}{5}-4 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d \,x^{2}-\frac {2 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3} x^{2}}{5}+\frac {2 b^{2} c \,d^{2} x^{6}}{5}-\frac {4 a^{2} d^{3} x^{4}}{5}-4 a b c \,d^{2} x^{4}+\frac {2 b^{2} c^{2} d \,x^{4}}{5}-\frac {6 a^{2} c \,d^{2} x^{2}}{5}-4 a b \,c^{2} d \,x^{2}-\frac {2 a^{2} c^{2} d}{5}}{x^{2} \sqrt {d \,x^{2}+c}\, d \,e^{3} \sqrt {e x}\, c}\)	\(648\)

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/5*(d*x^2+c)^(1/2)*(-b^2*c*x^4+2*a^2*d*x^2+10*a*b*c*x^2+a^2*c)/x^2/c/e^3/(e*x)^(1/2)+2/5*(a^2*d^2+10*a*b*c*d
+b^2*c^2)/c*(-c*d)^(1/2)/d*((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^3
*(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 = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx=-\frac {2 \, {\left (2 \, {\left (b^{2} c^{2} + 10 \, a b c d + a^{2} d^{2}\right )} \sqrt {d e} x^{3} {\rm weierstrassZeta}\left (-\frac {4 \, c}{d}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right )\right ) - {\left (b^{2} c d x^{4} - a^{2} c d - 2 \, {\left (5 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}\right )}}{5 \, c d e^{4} x^{3}} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(7/2),x, algorithm="fricas")

[Out]

-2/5*(2*(b^2*c^2 + 10*a*b*c*d + a^2*d^2)*sqrt(d*e)*x^3*weierstrassZeta(-4*c/d, 0, weierstrassPInverse(-4*c/d,
0, x)) - (b^2*c*d*x^4 - a^2*c*d - 2*(5*a*b*c*d + a^2*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(e*x))/(c*d*e^4*x^3)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 16.89 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.38 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx=\frac {a^{2} \sqrt {c} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {a b \sqrt {c} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {b^{2} \sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/(e*x)**(7/2),x)

[Out]

a**2*sqrt(c)*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(7/2)*x**(5/2)*gamma(-1/
4)) + a*b*sqrt(c)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), d*x**2*exp_polar(I*pi)/c)/(e**(7/2)*sqrt(x)*gamma(3/
4)) + b**2*sqrt(c)*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(7/2)*gamma(
7/4))

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c}}{\left (e x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(7/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/(e*x)^(7/2), x)

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c}}{\left (e x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(7/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/(e*x)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{7/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{{\left (e\,x\right )}^{7/2}} \,d x \]

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/(e*x)^(7/2),x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/(e*x)^(7/2), x)