3.1.41 \(\int \frac {a+b x^3+c x^6}{(d+e x^3)^{7/2}} \, dx\) [41]

Optimal. Leaf size=349 \[ \frac {2 \left (c d^2-b d e+a e^2\right ) x}{15 d e^2 \left (d+e x^3\right )^{5/2}}-\frac {2 \left (17 c d^2-2 b d e-13 a e^2\right ) x}{135 d^2 e^2 \left (d+e x^3\right )^{3/2}}+\frac {2 \left (16 c d^2+14 b d e+91 a e^2\right ) x}{405 d^3 e^2 \sqrt {d+e x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2+14 b d e+91 a e^2\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right )|-7-4 \sqrt {3}\right )}{405 \sqrt [4]{3} d^3 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \]

[Out]

2/15*(a*e^2-b*d*e+c*d^2)*x/d/e^2/(e*x^3+d)^(5/2)-2/135*(-13*a*e^2-2*b*d*e+17*c*d^2)*x/d^2/e^2/(e*x^3+d)^(3/2)+
2/405*(91*a*e^2+14*b*d*e+16*c*d^2)*x/d^3/e^2/(e*x^3+d)^(1/2)+2/1215*(91*a*e^2+14*b*d*e+16*c*d^2)*(d^(1/3)+e^(1
/3)*x)*EllipticF((e^(1/3)*x+d^(1/3)*(1-3^(1/2)))/(e^(1/3)*x+d^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1
/2*2^(1/2))*((d^(2/3)-d^(1/3)*e^(1/3)*x+e^(2/3)*x^2)/(e^(1/3)*x+d^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/d^3/e^(7
/3)/(e*x^3+d)^(1/2)/(d^(1/3)*(d^(1/3)+e^(1/3)*x)/(e^(1/3)*x+d^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1423, 393, 205, 224} \begin {gather*} \frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (91 a e^2+14 b d e+16 c d^2\right ) F\left (\text {ArcSin}\left (\frac {\sqrt [3]{e} x+\left (1-\sqrt {3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt {3}\right ) \sqrt [3]{d}}\right )|-7-4 \sqrt {3}\right )}{405 \sqrt [4]{3} d^3 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}}-\frac {2 x \left (-13 a e^2-2 b d e+17 c d^2\right )}{135 d^2 e^2 \left (d+e x^3\right )^{3/2}}+\frac {2 x \left (a e^2-b d e+c d^2\right )}{15 d e^2 \left (d+e x^3\right )^{5/2}}+\frac {2 x \left (91 a e^2+14 b d e+16 c d^2\right )}{405 d^3 e^2 \sqrt {d+e x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)/(d + e*x^3)^(7/2),x]

[Out]

(2*(c*d^2 - b*d*e + a*e^2)*x)/(15*d*e^2*(d + e*x^3)^(5/2)) - (2*(17*c*d^2 - 2*b*d*e - 13*a*e^2)*x)/(135*d^2*e^
2*(d + e*x^3)^(3/2)) + (2*(16*c*d^2 + 14*b*d*e + 91*a*e^2)*x)/(405*d^3*e^2*Sqrt[d + e*x^3]) + (2*Sqrt[2 + Sqrt
[3]]*(16*c*d^2 + 14*b*d*e + 91*a*e^2)*(d^(1/3) + e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x + e^(2/3)*x^2)/(
(1 + Sqrt[3])*d^(1/3) + e^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*d^(1/3) + e^(1/3)*x)/((1 + Sqrt[3])*d^(1
/3) + e^(1/3)*x)], -7 - 4*Sqrt[3]])/(405*3^(1/4)*d^3*e^(7/3)*Sqrt[(d^(1/3)*(d^(1/3) + e^(1/3)*x))/((1 + Sqrt[3
])*d^(1/3) + e^(1/3)*x)^2]*Sqrt[d + e*x^3])

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
 + 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[s*((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)
], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 393

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

Rule 1423

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Simp[(-(c*d^2 - b*
d*e + a*e^2))*x*((d + e*x^n)^(q + 1)/(d*e^2*n*(q + 1))), x] + Dist[1/(n*(q + 1)*d*e^2), Int[(d + e*x^n)^(q + 1
)*Simp[c*d^2 - b*d*e + a*e^2*(n*(q + 1) + 1) + c*d*e*n*(q + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, n}, x]
 && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{7/2}} \, dx &=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{15 d e^2 \left (d+e x^3\right )^{5/2}}-\frac {2 \int \frac {\frac {1}{2} \left (2 c d^2-e (2 b d+13 a e)\right )-\frac {15}{2} c d e x^3}{\left (d+e x^3\right )^{5/2}} \, dx}{15 d e^2}\\ &=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{15 d e^2 \left (d+e x^3\right )^{5/2}}-\frac {2 \left (17 c d^2-2 b d e-13 a e^2\right ) x}{135 d^2 e^2 \left (d+e x^3\right )^{3/2}}+\frac {\left (16 c d^2+14 b d e+91 a e^2\right ) \int \frac {1}{\left (d+e x^3\right )^{3/2}} \, dx}{135 d^2 e^2}\\ &=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{15 d e^2 \left (d+e x^3\right )^{5/2}}-\frac {2 \left (17 c d^2-2 b d e-13 a e^2\right ) x}{135 d^2 e^2 \left (d+e x^3\right )^{3/2}}+\frac {2 \left (16 c d^2+14 b d e+91 a e^2\right ) x}{405 d^3 e^2 \sqrt {d+e x^3}}+\frac {\left (16 c d^2+14 b d e+91 a e^2\right ) \int \frac {1}{\sqrt {d+e x^3}} \, dx}{405 d^3 e^2}\\ &=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{15 d e^2 \left (d+e x^3\right )^{5/2}}-\frac {2 \left (17 c d^2-2 b d e-13 a e^2\right ) x}{135 d^2 e^2 \left (d+e x^3\right )^{3/2}}+\frac {2 \left (16 c d^2+14 b d e+91 a e^2\right ) x}{405 d^3 e^2 \sqrt {d+e x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2+14 b d e+91 a e^2\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right )|-7-4 \sqrt {3}\right )}{405 \sqrt [4]{3} d^3 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.14, size = 166, normalized size = 0.48 \begin {gather*} \frac {2 x \left (c d^2 \left (-8 d^2-19 d e x^3+16 e^2 x^6\right )+e \left (b d \left (-7 d^2+34 d e x^3+14 e^2 x^6\right )+a e \left (157 d^2+221 d e x^3+91 e^2 x^6\right )\right )\right )+\left (16 c d^2+7 e (2 b d+13 a e)\right ) x \left (d+e x^3\right )^2 \sqrt {1+\frac {e x^3}{d}} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {e x^3}{d}\right )}{405 d^3 e^2 \left (d+e x^3\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3 + c*x^6)/(d + e*x^3)^(7/2),x]

[Out]

(2*x*(c*d^2*(-8*d^2 - 19*d*e*x^3 + 16*e^2*x^6) + e*(b*d*(-7*d^2 + 34*d*e*x^3 + 14*e^2*x^6) + a*e*(157*d^2 + 22
1*d*e*x^3 + 91*e^2*x^6))) + (16*c*d^2 + 7*e*(2*b*d + 13*a*e))*x*(d + e*x^3)^2*Sqrt[1 + (e*x^3)/d]*Hypergeometr
ic2F1[1/3, 1/2, 4/3, -((e*x^3)/d)])/(405*d^3*e^2*(d + e*x^3)^(5/2))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1094 vs. \(2 (282 ) = 564\).
time = 0.22, size = 1095, normalized size = 3.14

method result size
elliptic \(\frac {2 x \left (a \,e^{2}-d e b +c \,d^{2}\right ) \sqrt {e \,x^{3}+d}}{15 d \,e^{5} \left (x^{3}+\frac {d}{e}\right )^{3}}+\frac {2 x \left (13 a \,e^{2}+2 d e b -17 c \,d^{2}\right ) \sqrt {e \,x^{3}+d}}{135 d^{2} e^{4} \left (x^{3}+\frac {d}{e}\right )^{2}}+\frac {2 x \left (91 a \,e^{2}+14 d e b +16 c \,d^{2}\right )}{405 e^{2} d^{3} \sqrt {\left (x^{3}+\frac {d}{e}\right ) e}}-\frac {2 i \left (91 a \,e^{2}+14 d e b +16 c \,d^{2}\right ) \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{e}}{-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{e \left (-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right )}}\right )}{1215 d^{3} e^{3} \sqrt {e \,x^{3}+d}}\) \(437\)
default \(\text {Expression too large to display}\) \(1095\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^6+b*x^3+a)/(e*x^3+d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

c*(2/15*x*d/e^5*(e*x^3+d)^(1/2)/(x^3+d/e)^3-34/135*x/e^4*(e*x^3+d)^(1/2)/(x^3+d/e)^2+32/405/e^2*x/d/((x^3+d/e)
*e)^(1/2)-32/1215*I/e^3/d*3^(1/2)*(-d*e^2)^(1/3)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^
(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3))/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^
(1/2)*(-I*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)/(e*x^3+d)^(1
/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))
^(1/2),(I*3^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)))+b*(-2/15*x/
e^4*(e*x^3+d)^(1/2)/(x^3+d/e)^3+4/135*x/d/e^3*(e*x^3+d)^(1/2)/(x^3+d/e)^2+28/405/e*x/d^2/((x^3+d/e)*e)^(1/2)-2
8/1215*I/e^2/d^2*3^(1/2)*(-d*e^2)^(1/3)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(
-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3))/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I
*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*Ellip
ticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I
*3^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)))+a*(2/15*x/d/e^3*(e*x
^3+d)^(1/2)/(x^3+d/e)^3+26/135*x/d^2/e^2*(e*x^3+d)^(1/2)/(x^3+d/e)^2+182/405*x/d^3/((x^3+d/e)*e)^(1/2)-182/121
5*I/d^3*3^(1/2)/e*(-d*e^2)^(1/3)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)
^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3))/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2
/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/
3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2
)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^6 + b*x^3 + a)/(x^3*e + d)^(7/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 262, normalized size = 0.75 \begin {gather*} \frac {2 \, {\left ({\left (91 \, a x^{9} e^{5} + 16 \, c d^{5} + 7 \, {\left (2 \, b d x^{9} + 39 \, a d x^{6}\right )} e^{4} + {\left (16 \, c d^{2} x^{9} + 42 \, b d^{2} x^{6} + 273 \, a d^{2} x^{3}\right )} e^{3} + {\left (48 \, c d^{3} x^{6} + 42 \, b d^{3} x^{3} + 91 \, a d^{3}\right )} e^{2} + 2 \, {\left (24 \, c d^{4} x^{3} + 7 \, b d^{4}\right )} e\right )} e^{\frac {1}{2}} {\rm weierstrassPInverse}\left (0, -4 \, d e^{\left (-1\right )}, x\right ) + {\left (91 \, a x^{7} e^{5} - 8 \, c d^{4} x e + {\left (14 \, b d x^{7} + 221 \, a d x^{4}\right )} e^{4} + {\left (16 \, c d^{2} x^{7} + 34 \, b d^{2} x^{4} + 157 \, a d^{2} x\right )} e^{3} - {\left (19 \, c d^{3} x^{4} + 7 \, b d^{3} x\right )} e^{2}\right )} \sqrt {x^{3} e + d}\right )}}{405 \, {\left (d^{3} x^{9} e^{6} + 3 \, d^{4} x^{6} e^{5} + 3 \, d^{5} x^{3} e^{4} + d^{6} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(7/2),x, algorithm="fricas")

[Out]

2/405*((91*a*x^9*e^5 + 16*c*d^5 + 7*(2*b*d*x^9 + 39*a*d*x^6)*e^4 + (16*c*d^2*x^9 + 42*b*d^2*x^6 + 273*a*d^2*x^
3)*e^3 + (48*c*d^3*x^6 + 42*b*d^3*x^3 + 91*a*d^3)*e^2 + 2*(24*c*d^4*x^3 + 7*b*d^4)*e)*e^(1/2)*weierstrassPInve
rse(0, -4*d*e^(-1), x) + (91*a*x^7*e^5 - 8*c*d^4*x*e + (14*b*d*x^7 + 221*a*d*x^4)*e^4 + (16*c*d^2*x^7 + 34*b*d
^2*x^4 + 157*a*d^2*x)*e^3 - (19*c*d^3*x^4 + 7*b*d^3*x)*e^2)*sqrt(x^3*e + d))/(d^3*x^9*e^6 + 3*d^4*x^6*e^5 + 3*
d^5*x^3*e^4 + d^6*e^3)

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Sympy [A]
time = 152.64, size = 119, normalized size = 0.34 \begin {gather*} \frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {7}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {7}{2}} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {7}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {7}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{3}, \frac {7}{2} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {7}{2}} \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**6+b*x**3+a)/(e*x**3+d)**(7/2),x)

[Out]

a*x*gamma(1/3)*hyper((1/3, 7/2), (4/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(7/2)*gamma(4/3)) + b*x**4*gamma(4/3)
*hyper((4/3, 7/2), (7/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(7/2)*gamma(7/3)) + c*x**7*gamma(7/3)*hyper((7/3, 7
/2), (10/3,), e*x**3*exp_polar(I*pi)/d)/(3*d**(7/2)*gamma(10/3))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^(7/2),x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)/(x^3*e + d)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {c\,x^6+b\,x^3+a}{{\left (e\,x^3+d\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3 + c*x^6)/(d + e*x^3)^(7/2),x)

[Out]

int((a + b*x^3 + c*x^6)/(d + e*x^3)^(7/2), x)

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