### 3.41 $$\int \frac{a+b x^3+c x^6}{(d+e x^3)^{7/2}} \, dx$$

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

[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])

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Rubi [A]  time = 0.324033, antiderivative size = 349, normalized size of antiderivative = 1., 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 = {1409, 385, 199, 218} $\frac{2 x \left (91 a e^2+14 b d e+16 c d^2\right )}{405 d^3 e^2 \sqrt{d+e x^3}}-\frac{2 x \left (-13 a e^2-2 b d e+17 c d^2\right )}{135 d^2 e^2 \left (d+e x^3\right )^{3/2}}+\frac{2 x \left (a e^2-b d e+c d^2\right )}{15 d e^2 \left (d+e x^3\right )^{5/2}}+\frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (91 a e^2+14 b d e+16 c d^2\right ) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{e} x+\left (1-\sqrt{3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt{3}\right ) \sqrt [3]{d}}\right )|-7-4 \sqrt{3}\right )}{405 \sqrt [4]{3} d^3 e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}$

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 1409

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]

Rule 385

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 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 218

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

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*}

Mathematica [C]  time = 0.191137, size = 166, normalized size = 0.48 $\frac{2 x \left (e \left (a e \left (157 d^2+221 d e x^3+91 e^2 x^6\right )+b d \left (-7 d^2+34 d e x^3+14 e^2 x^6\right )\right )+c d^2 \left (-8 d^2-19 d e x^3+16 e^2 x^6\right )\right )+x \sqrt{\frac{e x^3}{d}+1} \left (d+e x^3\right )^2 \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{e x^3}{d}\right ) \left (7 e (13 a e+2 b d)+16 c d^2\right )}{405 d^3 e^2 \left (d+e x^3\right )^{5/2}}$

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]  time = 0.043, size = 1095, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

c*(2/15*d*x/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/d*x/((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/d*x/e^3*(e*x^3+d)^(1/2)/(x^3+d/e)^2+28/405/e/d^2*x/((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/d*x/e^3*(e*x
^3+d)^(1/2)/(x^3+d/e)^3+26/135/d^2*x/e^2*(e*x^3+d)^(1/2)/(x^3+d/e)^2+182/405/d^3*x/((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., size = 0, normalized size = 0. \begin{align*} \int \frac{c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

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)/(e*x^3 + d)^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{6} + b x^{3} + a\right )} \sqrt{e x^{3} + d}}{e^{4} x^{12} + 4 \, d e^{3} x^{9} + 6 \, d^{2} e^{2} x^{6} + 4 \, d^{3} e x^{3} + d^{4}}, x\right ) \end{align*}

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]

integral((c*x^6 + b*x^3 + a)*sqrt(e*x^3 + d)/(e^4*x^12 + 4*d*e^3*x^9 + 6*d^2*e^2*x^6 + 4*d^3*e*x^3 + d^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

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)/(e*x^3 + d)^(7/2), x)