3.825 \(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{(e x)^{3/2}} \, dx\)

Optimal. Leaf size=421 \[ -\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{c e \sqrt {e x}}-\frac {2 \sqrt [4]{c} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (b^2 c^2-3 a d (5 a d+2 b c)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 d^{7/4} e^{3/2} \sqrt {c+d x^2}}+\frac {4 \sqrt [4]{c} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (b^2 c^2-3 a d (5 a d+2 b c)\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 d^{7/4} e^{3/2} \sqrt {c+d x^2}}-\frac {4 \sqrt {e x} \sqrt {c+d x^2} \left (b^2 c^2-3 a d (5 a d+2 b c)\right )}{15 d^{3/2} e^2 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 (e x)^{3/2} \sqrt {c+d x^2} \left (b^2 c^2-3 a d (5 a d+2 b c)\right )}{15 c d e^3}+\frac {2 b^2 (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 d e^3} \]

[Out]

2/9*b^2*(e*x)^(3/2)*(d*x^2+c)^(3/2)/d/e^3-2*a^2*(d*x^2+c)^(3/2)/c/e/(e*x)^(1/2)-2/15*(b^2*c^2-3*a*d*(5*a*d+2*b
*c))*(e*x)^(3/2)*(d*x^2+c)^(1/2)/c/d/e^3-4/15*(b^2*c^2-3*a*d*(5*a*d+2*b*c))*(e*x)^(1/2)*(d*x^2+c)^(1/2)/d^(3/2
)/e^2/(c^(1/2)+x*d^(1/2))+4/15*c^(1/4)*(b^2*c^2-3*a*d*(5*a*d+2*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)/d^(7/4)/e^(3/2)
/(d*x^2+c)^(1/2)-2/15*c^(1/4)*(b^2*c^2-3*a*d*(5*a*d+2*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)/d^(7/4)/e^(3/2)/(d*x^2+c
)^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 421, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {462, 459, 279, 329, 305, 220, 1196} \[ -\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{c e \sqrt {e x}}-\frac {4 \sqrt {e x} \sqrt {c+d x^2} \left (b^2 c^2-3 a d (5 a d+2 b c)\right )}{15 d^{3/2} e^2 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 \sqrt [4]{c} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (b^2 c^2-3 a d (5 a d+2 b c)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 d^{7/4} e^{3/2} \sqrt {c+d x^2}}+\frac {4 \sqrt [4]{c} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (b^2 c^2-3 a d (5 a d+2 b c)\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 d^{7/4} e^{3/2} \sqrt {c+d x^2}}-\frac {2 (e x)^{3/2} \sqrt {c+d x^2} \left (b^2 c^2-3 a d (5 a d+2 b c)\right )}{15 c d e^3}+\frac {2 b^2 (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/(e*x)^(3/2),x]

[Out]

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

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 279

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 305

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 329

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 459

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

Rule 462

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 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{(e x)^{3/2}} \, dx &=-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{c e \sqrt {e x}}+\frac {2 \int \sqrt {e x} \left (\frac {1}{2} a (2 b c+5 a d)+\frac {1}{2} b^2 c x^2\right ) \sqrt {c+d x^2} \, dx}{c e^2}\\ &=-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{c e \sqrt {e x}}+\frac {2 b^2 (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 d e^3}-\frac {\left (b^2 c^2-3 a d (2 b c+5 a d)\right ) \int \sqrt {e x} \sqrt {c+d x^2} \, dx}{3 c d e^2}\\ &=-\frac {2 \left (b^2 c^2-3 a d (2 b c+5 a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{15 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{c e \sqrt {e x}}+\frac {2 b^2 (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 d e^3}-\frac {\left (2 \left (b^2 c^2-3 a d (2 b c+5 a d)\right )\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x^2}} \, dx}{15 d e^2}\\ &=-\frac {2 \left (b^2 c^2-3 a d (2 b c+5 a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{15 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{c e \sqrt {e x}}+\frac {2 b^2 (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 d e^3}-\frac {\left (4 \left (b^2 c^2-3 a d (2 b c+5 a d)\right )\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{15 d e^3}\\ &=-\frac {2 \left (b^2 c^2-3 a d (2 b c+5 a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{15 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{c e \sqrt {e x}}+\frac {2 b^2 (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 d e^3}-\frac {\left (4 \sqrt {c} \left (b^2 c^2-3 a d (2 b c+5 a d)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{15 d^{3/2} e^2}+\frac {\left (4 \sqrt {c} \left (b^2 c^2-3 a d (2 b c+5 a d)\right )\right ) \operatorname {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 )}{15 d^{3/2} e^2}\\ &=-\frac {2 \left (b^2 c^2-3 a d (2 b c+5 a d)\right ) (e x)^{3/2} \sqrt {c+d x^2}}{15 c d e^3}-\frac {4 \left (b^2 c^2-3 a d (2 b c+5 a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{15 d^{3/2} e^2 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{c e \sqrt {e x}}+\frac {2 b^2 (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 d e^3}+\frac {4 \sqrt [4]{c} \left (b^2 c^2-3 a d (2 b c+5 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 d^{7/4} e^{3/2} \sqrt {c+d x^2}}-\frac {2 \sqrt [4]{c} \left (b^2 c^2-3 a d (2 b c+5 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 d^{7/4} e^{3/2} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 129, normalized size = 0.31 \[ \frac {x \left (12 x^2 \sqrt {\frac {c}{d x^2}+1} \left (15 a^2 d^2+6 a b c d-b^2 c^2\right ) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {c}{d x^2}\right )+2 \left (c+d x^2\right ) \left (-45 a^2 d+18 a b d x^2+b^2 x^2 \left (2 c+5 d x^2\right )\right )\right )}{45 d (e x)^{3/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/(e*x)^(3/2),x]

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{e^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(3/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x^2 + c)*sqrt(e*x)/(e^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c}}{\left (e x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/(e*x)^(3/2), x)

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maple [A]  time = 0.05, size = 624, normalized size = 1.48 \[ \frac {\frac {2 b^{2} d^{3} x^{6}}{9}+\frac {4 a b \,d^{3} x^{4}}{5}+\frac {14 b^{2} c \,d^{2} x^{4}}{45}-2 a^{2} d^{3} x^{2}+\frac {4 a b c \,d^{2} x^{2}}{5}+\frac {4 b^{2} c^{2} d \,x^{2}}{45}+4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c \,d^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c \,d^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+\frac {8 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{2} d \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{2} d \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{3} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{15}+\frac {2 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{3} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{15}-2 a^{2} c \,d^{2}}{\sqrt {d \,x^{2}+c}\, \sqrt {e x}\, d^{2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(3/2),x)

[Out]

2/45*(5*x^6*b^2*d^3+90*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2
)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c*d^2+36*((d*
x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/
2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^2*d-6*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2)
)^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticE(((d*x+(-c*d)^(1
/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c^3-45*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)
^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*
2^(1/2))*a^2*c*d^2-18*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)
*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^2*d+3*((d*x+
(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)
*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c^3+18*x^4*a*b*d^3+7*x^4*b^2*c*d^2-45*a^2*
d^3*x^2+18*a*b*c*d^2*x^2+2*x^2*b^2*c^2*d-45*a^2*c*d^2)/(d*x^2+c)^(1/2)/d^2/e/(e*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c}}{\left (e x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/(e*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*sqrt(d*x^2 + c)/(e*x)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{{\left (e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/(e*x)^(3/2),x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/(e*x)^(3/2), x)

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sympy [C]  time = 6.65, size = 153, normalized size = 0.36 \[ \frac {a^{2} \sqrt {c} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {a b \sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {b^{2} \sqrt {c} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/(e*x)**(3/2),x)

[Out]

a**2*sqrt(c)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(3/2)*sqrt(x)*gamma(3/4))
+ a*b*sqrt(c)*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), d*x**2*exp_polar(I*pi)/c)/(e**(3/2)*gamma(7/4))
+ b**2*sqrt(c)*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(3/2)*gamma(11/
4))

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