3.839 \(\int \frac {(e x)^{5/2} (a+b x^2)^2}{\sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=430 \[ -\frac {c^{5/4} e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{195 d^{15/4} \sqrt {c+d x^2}}+\frac {2 c^{5/4} e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{195 d^{15/4} \sqrt {c+d x^2}}-\frac {2 c e^2 \sqrt {e x} \sqrt {c+d x^2} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right )}{195 d^{7/2} \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {2 e (e x)^{3/2} \sqrt {c+d x^2} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right )}{585 d^3}-\frac {2 b (e x)^{7/2} \sqrt {c+d x^2} (11 b c-26 a d)}{117 d^2 e}+\frac {2 b^2 (e x)^{11/2} \sqrt {c+d x^2}}{13 d e^3} \]

[Out]

2/585*(117*a^2*d^2+7*b*c*(-26*a*d+11*b*c))*e*(e*x)^(3/2)*(d*x^2+c)^(1/2)/d^3-2/117*b*(-26*a*d+11*b*c)*(e*x)^(7
/2)*(d*x^2+c)^(1/2)/d^2/e+2/13*b^2*(e*x)^(11/2)*(d*x^2+c)^(1/2)/d/e^3-2/195*c*(117*a^2*d^2+7*b*c*(-26*a*d+11*b
*c))*e^2*(e*x)^(1/2)*(d*x^2+c)^(1/2)/d^(7/2)/(c^(1/2)+x*d^(1/2))+2/195*c^(5/4)*(117*a^2*d^2+7*b*c*(-26*a*d+11*
b*c))*e^(5/2)*(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)))*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^(15/4)/(d*x^2+c)^(1/2)-1/195*c^(5/4)*(117*a^2*d^2+7*b*c*(-26*a*d+11*
b*c))*e^(5/2)*(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)))*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^(15/4)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 430, 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 = {464, 459, 321, 329, 305, 220, 1196} \[ -\frac {c^{5/4} e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{195 d^{15/4} \sqrt {c+d x^2}}+\frac {2 c^{5/4} e^{5/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{195 d^{15/4} \sqrt {c+d x^2}}-\frac {2 c e^2 \sqrt {e x} \sqrt {c+d x^2} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right )}{195 d^{7/2} \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {2 e (e x)^{3/2} \sqrt {c+d x^2} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right )}{585 d^3}-\frac {2 b (e x)^{7/2} \sqrt {c+d x^2} (11 b c-26 a d)}{117 d^2 e}+\frac {2 b^2 (e x)^{11/2} \sqrt {c+d x^2}}{13 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(5/2)*(a + b*x^2)^2)/Sqrt[c + d*x^2],x]

[Out]

(2*(117*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e*(e*x)^(3/2)*Sqrt[c + d*x^2])/(585*d^3) - (2*b*(11*b*c - 26*a*d)*(
e*x)^(7/2)*Sqrt[c + d*x^2])/(117*d^2*e) + (2*b^2*(e*x)^(11/2)*Sqrt[c + d*x^2])/(13*d*e^3) - (2*c*(117*a^2*d^2
+ 7*b*c*(11*b*c - 26*a*d))*e^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(195*d^(7/2)*(Sqrt[c] + Sqrt[d]*x)) + (2*c^(5/4)*(11
7*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e^(5/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*E
llipticE[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(195*d^(15/4)*Sqrt[c + d*x^2]) - (c^(5/4)*(117
*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e^(5/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*El
lipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(195*d^(15/4)*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 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 321

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

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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(d^2*(e*x)^
(m + n + 1)*(a + b*x^n)^(p + 1))/(b*e^(n + 1)*(m + n*(p + 2) + 1)), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 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 {(e x)^{5/2} \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx &=\frac {2 b^2 (e x)^{11/2} \sqrt {c+d x^2}}{13 d e^3}+\frac {2 \int \frac {(e x)^{5/2} \left (\frac {13 a^2 d}{2}-\frac {1}{2} b (11 b c-26 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{13 d}\\ &=-\frac {2 b (11 b c-26 a d) (e x)^{7/2} \sqrt {c+d x^2}}{117 d^2 e}+\frac {2 b^2 (e x)^{11/2} \sqrt {c+d x^2}}{13 d e^3}-\frac {1}{117} \left (-117 a^2-\frac {7 b c (11 b c-26 a d)}{d^2}\right ) \int \frac {(e x)^{5/2}}{\sqrt {c+d x^2}} \, dx\\ &=\frac {2 \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt {c+d x^2}}{585 d}-\frac {2 b (11 b c-26 a d) (e x)^{7/2} \sqrt {c+d x^2}}{117 d^2 e}+\frac {2 b^2 (e x)^{11/2} \sqrt {c+d x^2}}{13 d e^3}-\frac {\left (c \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e^2\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x^2}} \, dx}{195 d}\\ &=\frac {2 \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt {c+d x^2}}{585 d}-\frac {2 b (11 b c-26 a d) (e x)^{7/2} \sqrt {c+d x^2}}{117 d^2 e}+\frac {2 b^2 (e x)^{11/2} \sqrt {c+d x^2}}{13 d e^3}-\frac {\left (2 c \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{195 d}\\ &=\frac {2 \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt {c+d x^2}}{585 d}-\frac {2 b (11 b c-26 a d) (e x)^{7/2} \sqrt {c+d x^2}}{117 d^2 e}+\frac {2 b^2 (e x)^{11/2} \sqrt {c+d x^2}}{13 d e^3}-\frac {\left (2 c^{3/2} \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{195 d^{3/2}}+\frac {\left (2 c^{3/2} \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e^2\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 )}{195 d^{3/2}}\\ &=\frac {2 \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt {c+d x^2}}{585 d}-\frac {2 b (11 b c-26 a d) (e x)^{7/2} \sqrt {c+d x^2}}{117 d^2 e}+\frac {2 b^2 (e x)^{11/2} \sqrt {c+d x^2}}{13 d e^3}-\frac {2 c \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e^2 \sqrt {e x} \sqrt {c+d x^2}}{195 d^{3/2} \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {2 c^{5/4} \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e^{5/2} \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 )}{195 d^{7/4} \sqrt {c+d x^2}}-\frac {c^{5/4} \left (117 a^2+\frac {7 b c (11 b c-26 a d)}{d^2}\right ) e^{5/2} \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 )}{195 d^{7/4} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 143, normalized size = 0.33 \[ \frac {2 e (e x)^{3/2} \left (\left (c+d x^2\right ) \left (117 a^2 d^2+26 a b d \left (5 d x^2-7 c\right )+b^2 \left (77 c^2-55 c d x^2+45 d^2 x^4\right )\right )-3 c \sqrt {\frac {c}{d x^2}+1} \left (117 a^2 d^2-182 a b c d+77 b^2 c^2\right ) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {c}{d x^2}\right )\right )}{585 d^3 \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(5/2)*(a + b*x^2)^2)/Sqrt[c + d*x^2],x]

[Out]

(2*e*(e*x)^(3/2)*((c + d*x^2)*(117*a^2*d^2 + 26*a*b*d*(-7*c + 5*d*x^2) + b^2*(77*c^2 - 55*c*d*x^2 + 45*d^2*x^4
)) - 3*c*(77*b^2*c^2 - 182*a*b*c*d + 117*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d
*x^2))]))/(585*d^3*Sqrt[c + d*x^2])

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} e^{2} x^{6} + 2 \, a b e^{2} x^{4} + a^{2} e^{2} x^{2}\right )} \sqrt {e x}}{\sqrt {d x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*e^2*x^6 + 2*a*b*e^2*x^4 + a^2*e^2*x^2)*sqrt(e*x)/sqrt(d*x^2 + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{\frac {5}{2}}}{\sqrt {d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*(e*x)^(5/2)/sqrt(d*x^2 + c), x)

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maple [A]  time = 0.04, size = 661, normalized size = 1.54 \[ -\frac {\sqrt {e x}\, \left (-90 b^{2} d^{4} x^{8}-260 a b \,d^{4} x^{6}+20 b^{2} c \,d^{3} x^{6}-234 a^{2} d^{4} x^{4}+104 a b c \,d^{3} x^{4}-44 b^{2} c^{2} d^{2} x^{4}-234 a^{2} c \,d^{3} x^{2}+364 a b \,c^{2} d^{2} x^{2}-154 b^{2} c^{3} d \,x^{2}+702 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c^{2} d^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-351 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c^{2} d^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-1092 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{3} d \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+546 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{3} d \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+462 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{4} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-231 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{4} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )\right ) e^{2}}{585 \sqrt {d \,x^{2}+c}\, d^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x)

[Out]

-1/585*e^2/x*(e*x)^(1/2)/(d*x^2+c)^(1/2)/d^4*(-90*b^2*d^4*x^8-260*a*b*d^4*x^6+20*b^2*c*d^3*x^6+702*((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)*Elli
pticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c^2*d^2-1092*((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^3*d+462*((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^4-351*((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^2*d^2+546*((
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^3*d-231*((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^4-234*a^2*d^4*x^4+104*a*b*c*d^3*x^4-44*b^2*c^2*d^2*x^4-234*a^2
*c*d^3*x^2+364*a*b*c^2*d^2*x^2-154*b^2*c^3*d*x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{\frac {5}{2}}}{\sqrt {d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*(e*x)^(5/2)/sqrt(d*x^2 + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^2}{\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(5/2)*(a + b*x^2)^2)/(c + d*x^2)^(1/2),x)

[Out]

int(((e*x)^(5/2)*(a + b*x^2)^2)/(c + d*x^2)^(1/2), x)

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sympy [C]  time = 49.39, size = 144, normalized size = 0.33 \[ \frac {a^{2} e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} \Gamma \left (\frac {11}{4}\right )} + \frac {a b e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt {c} \Gamma \left (\frac {15}{4}\right )} + \frac {b^{2} e^{\frac {5}{2}} x^{\frac {15}{2}} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} \Gamma \left (\frac {19}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(5/2)*(b*x**2+a)**2/(d*x**2+c)**(1/2),x)

[Out]

a**2*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*gamma(11/4))
+ a*b*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((1/2, 11/4), (15/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*gamma(15/
4)) + b**2*e**(5/2)*x**(15/2)*gamma(15/4)*hyper((1/2, 15/4), (19/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*gam
ma(19/4))

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