3.283 \(\int x^4 (\frac {e (a+b x^2)}{c+d x^2})^{3/2} \, dx\)

Optimal. Leaf size=391 \[ -\frac {\sqrt {c} e \left (a^2 d^2-16 a b c d+16 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{5 b d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {e x \left (-\frac {a^2 d}{b}+16 a c-\frac {16 b c^2}{d}\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 d^2}+\frac {c^{3/2} e (8 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{5 d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {e x \left (c+d x^2\right ) (8 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 d^3}+\frac {6 b e x^3 \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 d^2}-\frac {e x^3 \left (a+b x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d} \]

[Out]

-1/5*(16*a*c-16*b*c^2/d-a^2*d/b)*e*x*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^2-e*x^3*(b*x^2+a)*(e*(b*x^2+a)/(d*x^2+c))
^(1/2)/d-1/5*(-7*a*d+8*b*c)*e*x*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^3+6/5*b*e*x^3*(d*x^2+c)*(e*(b*x^2+a)
/(d*x^2+c))^(1/2)/d^2+1/5*c^(3/2)*(-7*a*d+8*b*c)*e*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)
/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^(7/2)/(c*(b*x^2+a)/a/(d*x^2+c))^
(1/2)-1/5*(a^2*d^2-16*a*b*c*d+16*b^2*c^2)*e*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2
)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b/d^(7/2)/(c*(b*x^2+a)/a/(d*x^2+c
))^(1/2)

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Rubi [A]  time = 0.68, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6719, 467, 581, 582, 531, 418, 492, 411} \[ -\frac {\sqrt {c} e \left (a^2 d^2-16 a b c d+16 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{5 b d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {e x \left (-\frac {a^2 d}{b}+16 a c-\frac {16 b c^2}{d}\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 d^2}+\frac {c^{3/2} e (8 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{5 d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {6 b e x^3 \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 d^2}-\frac {e x \left (c+d x^2\right ) (8 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 d^3}-\frac {e x^3 \left (a+b x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^4*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

-((16*a*c - (16*b*c^2)/d - (a^2*d)/b)*e*x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(5*d^2) - (e*x^3*(a + b*x^2)*Sqrt
[(e*(a + b*x^2))/(c + d*x^2)])/d - ((8*b*c - 7*a*d)*e*x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(5*d^3)
+ (6*b*e*x^3*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(5*d^2) - (Sqrt[c]*(16*b^2*c^2 - 16*a*b*c*d + a^2
*d^2)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(5*b*d^(7/2
)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]) + (c^(3/2)*(8*b*c - 7*a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Ellipt
icF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(5*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))])

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 467

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 581

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*g*(m + n*(p + q + 1) + 1)), x] + Dis
t[1/(b*(m + n*(p + q + 1) + 1)), Int[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*((b*e - a*f)*(m + 1) + b
*e*n*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*n*q*(b*c - a*d) + b*e*d*n*(p + q + 1))*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 582

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

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

\begin {align*} \int x^4 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx &=\frac {\left (e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^4 \left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^{3/2}} \, dx}{\sqrt {a+b x^2}}\\ &=-\frac {e x^3 \left (a+b x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d}+\frac {\left (e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^2 \sqrt {a+b x^2} \left (3 a+6 b x^2\right )}{\sqrt {c+d x^2}} \, dx}{d \sqrt {a+b x^2}}\\ &=-\frac {e x^3 \left (a+b x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d}+\frac {6 b e x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^2}+\frac {\left (e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^2 \left (-3 a (6 b c-5 a d)-3 b (8 b c-7 a d) x^2\right )}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{5 d^2 \sqrt {a+b x^2}}\\ &=-\frac {e x^3 \left (a+b x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d}-\frac {(8 b c-7 a d) e x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^3}+\frac {6 b e x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^2}-\frac {\left (e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {-3 a b c (8 b c-7 a d)-3 b \left (16 b^2 c^2-16 a b c d+a^2 d^2\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b d^3 \sqrt {a+b x^2}}\\ &=-\frac {e x^3 \left (a+b x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d}-\frac {(8 b c-7 a d) e x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^3}+\frac {6 b e x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^2}+\frac {\left (a c (8 b c-7 a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{5 d^3 \sqrt {a+b x^2}}+\frac {\left (\left (16 b^2 c^2-16 a b c d+a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{5 d^3 \sqrt {a+b x^2}}\\ &=\frac {\left (16 b^2 c^2-16 a b c d+a^2 d^2\right ) e x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 b d^3}-\frac {e x^3 \left (a+b x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d}-\frac {(8 b c-7 a d) e x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^3}+\frac {6 b e x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^2}+\frac {c^{3/2} (8 b c-7 a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{5 d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\left (c \left (16 b^2 c^2-16 a b c d+a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{5 b d^3 \sqrt {a+b x^2}}\\ &=\frac {\left (16 b^2 c^2-16 a b c d+a^2 d^2\right ) e x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 b d^3}-\frac {e x^3 \left (a+b x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d}-\frac {(8 b c-7 a d) e x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^3}+\frac {6 b e x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d^2}-\frac {\sqrt {c} \left (16 b^2 c^2-16 a b c d+a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{5 b d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {c^{3/2} (8 b c-7 a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{5 d^{7/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\\ \end {align*}

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Mathematica [C]  time = 0.60, size = 290, normalized size = 0.74 \[ \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (8 i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (a^2 d^2-3 a b c d+2 b^2 c^2\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (a^2 d^2-16 a b c d+16 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+d x \sqrt {\frac {b}{a}} \left (a^2 d \left (7 c+2 d x^2\right )+a b \left (-8 c^2+5 c d x^2+3 d^2 x^4\right )+b^2 x^2 \left (-8 c^2-2 c d x^2+d^2 x^4\right )\right )\right )}{5 d^4 \sqrt {\frac {b}{a}} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[b/a]*d*x*(a^2*d*(7*c + 2*d*x^2) + b^2*x^2*(-8*c^2 - 2*c*d*x^2 + d^2
*x^4) + a*b*(-8*c^2 + 5*c*d*x^2 + 3*d^2*x^4)) - I*c*(16*b^2*c^2 - 16*a*b*c*d + a^2*d^2)*Sqrt[1 + (b*x^2)/a]*Sq
rt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (8*I)*c*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*S
qrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/(5*Sqrt[b/a]*d^4*(a +
b*x^2))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b e x^{6} + a e x^{4}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{d x^{2} + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

integrate(((b*x^2 + a)*e/(d*x^2 + c))^(3/2)*x^4, x)

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maple [B]  time = 0.05, size = 933, normalized size = 2.39 \[ \frac {\left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}} \left (d \,x^{2}+c \right ) \left (\sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, b^{2} d^{3} x^{7}+3 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, a b \,d^{3} x^{5}-2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, b^{2} c \,d^{2} x^{5}+2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, a^{2} d^{3} x^{3}+5 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, a b c \,d^{2} x^{3}-3 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, b^{2} c^{2} d \,x^{3}-5 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, b^{2} c^{2} d \,x^{3}+2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, a^{2} c \,d^{2} x +5 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, a^{2} c \,d^{2} x +\sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c \,d^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-8 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c \,d^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-3 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, a b \,c^{2} d x -5 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {-\frac {b}{a}}\, a b \,c^{2} d x -16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+24 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right )}{5 \left (b \,x^{2}+a \right )^{2} \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*((b*x^2+a)/(d*x^2+c)*e)^(3/2),x)

[Out]

1/5*((b*x^2+a)/(d*x^2+c)*e)^(3/2)*(d*x^2+c)*(((d*x^2+c)*(b*x^2+a))^(1/2)*(-1/a*b)^(1/2)*x^7*b^2*d^3+3*((d*x^2+
c)*(b*x^2+a))^(1/2)*(-1/a*b)^(1/2)*x^5*a*b*d^3-2*((d*x^2+c)*(b*x^2+a))^(1/2)*(-1/a*b)^(1/2)*x^5*b^2*c*d^2+2*((
d*x^2+c)*(b*x^2+a))^(1/2)*(-1/a*b)^(1/2)*x^3*a^2*d^3-3*((d*x^2+c)*(b*x^2+a))^(1/2)*(-1/a*b)^(1/2)*x^3*b^2*c^2*
d+5*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-1/a*b)^(1/2)*x^3*a*b*c*d^2-5*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-1
/a*b)^(1/2)*x^3*b^2*c^2*d-8*((d*x^2+c)*(b*x^2+a))^(1/2)*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/
a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*c*d^2+24*((d*x^2+c)*(b*x^2+a))^(1/2)*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)
*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*b*c^2*d-16*((d*x^2+c)*(b*x^2+a))^(1/2)*((b*x^2+a)/a)^(1/2)*((d*
x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*b^2*c^3+((d*x^2+c)*(b*x^2+a))^(1/2)*((b*x^2+a)/a)^
(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*c*d^2-16*((d*x^2+c)*(b*x^2+a))^(1/2)
*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*b*c^2*d+16*((d*x^2+c)*(
b*x^2+a))^(1/2)*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*b^2*c^3+2*
((d*x^2+c)*(b*x^2+a))^(1/2)*(-1/a*b)^(1/2)*x*a^2*c*d^2-3*((d*x^2+c)*(b*x^2+a))^(1/2)*(-1/a*b)^(1/2)*x*a*b*c^2*
d+5*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-1/a*b)^(1/2)*x*a^2*c*d^2-5*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-1/a
*b)^(1/2)*x*a*b*c^2*d)/d^4/(b*x^2+a)^2/(-1/a*b)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(((b*x^2 + a)*e/(d*x^2 + c))^(3/2)*x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x)

[Out]

int(x^4*((e*(a + b*x^2))/(c + d*x^2))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*(b*x**2+a)/(d*x**2+c))**(3/2),x)

[Out]

Timed out

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