3.5.99 \(\int \frac {1}{x^3 (a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\) [499]
Optimal. Leaf size=67 \[ -\frac {\sqrt {1+\frac {d x^3}{c}} F_1\left (-\frac {2}{3};2,\frac {3}{2};\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 c x^2 \sqrt {c+d x^3}} \]
[Out]
-1/2*AppellF1(-2/3,2,3/2,1/3,-b*x^3/a,-d*x^3/c)*(1+d*x^3/c)^(1/2)/a^2/c/x^2/(d*x^3+c)^(1/2)
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Rubi [A]
time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {525, 524}
\begin {gather*} -\frac {\sqrt {\frac {d x^3}{c}+1} F_1\left (-\frac {2}{3};2,\frac {3}{2};\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 c x^2 \sqrt {c+d x^3}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/(x^3*(a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
-1/2*(Sqrt[1 + (d*x^3)/c]*AppellF1[-2/3, 2, 3/2, 1/3, -((b*x^3)/a), -((d*x^3)/c)])/(a^2*c*x^2*Sqrt[c + d*x^3])
Rule 524
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rule 525
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && !(IntegerQ[
p] || GtQ[a, 0])
Rubi steps
\begin {align*} \int \frac {1}{x^3 \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac {\sqrt {1+\frac {d x^3}{c}} \int \frac {1}{x^3 \left (a+b x^3\right )^2 \left (1+\frac {d x^3}{c}\right )^{3/2}} \, dx}{c \sqrt {c+d x^3}}\\ &=-\frac {\sqrt {1+\frac {d x^3}{c}} F_1\left (-\frac {2}{3};2,\frac {3}{2};\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 c x^2 \sqrt {c+d x^3}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(515\) vs. \(2(67)=134\).
time = 10.58, size = 515, normalized size = 7.69 \begin {gather*} \frac {-b d \left (5 b^2 c^2-6 a b c d+7 a^2 d^2\right ) x^6 \sqrt {1+\frac {d x^3}{c}} F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {a \left (32 a c \left (10 b^3 c^2 x^3 \left (3 c+d x^3\right )+3 a^3 d^2 \left (2 c+7 d x^3\right )+3 a b^2 c \left (2 c^2-13 c d x^3-4 d^2 x^6\right )+2 a^2 b d \left (-6 c^2-6 c d x^3+7 d^2 x^6\right )\right ) F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-24 x^3 \left (5 b^3 c^2 x^3 \left (c+d x^3\right )+a^3 d^2 \left (3 c+7 d x^3\right )+3 a b^2 c \left (c^2-c d x^3-2 d^2 x^6\right )+a^2 b d \left (-6 c^2-3 c d x^3+7 d^2 x^6\right )\right ) \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (-8 a c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}}{48 a^3 c^2 (b c-a d)^2 x^2 \sqrt {c+d x^3}} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(x^3*(a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
(-(b*d*(5*b^2*c^2 - 6*a*b*c*d + 7*a^2*d^2)*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -(
(b*x^3)/a)]) + (a*(32*a*c*(10*b^3*c^2*x^3*(3*c + d*x^3) + 3*a^3*d^2*(2*c + 7*d*x^3) + 3*a*b^2*c*(2*c^2 - 13*c*
d*x^3 - 4*d^2*x^6) + 2*a^2*b*d*(-6*c^2 - 6*c*d*x^3 + 7*d^2*x^6))*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b
*x^3)/a)] - 24*x^3*(5*b^3*c^2*x^3*(c + d*x^3) + a^3*d^2*(3*c + 7*d*x^3) + 3*a*b^2*c*(c^2 - c*d*x^3 - 2*d^2*x^6
) + a^2*b*d*(-6*c^2 - 3*c*d*x^3 + 7*d^2*x^6))*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -((b*x^3)/a)] +
a*d*AppellF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)])))/((a + b*x^3)*(-8*a*c*AppellF1[1/3, 1/2, 1, 4/3,
-((d*x^3)/c), -((b*x^3)/a)] + 3*x^3*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + a*d*Appell
F1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)]))))/(48*a^3*c^2*(b*c - a*d)^2*x^2*Sqrt[c + d*x^3])
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
6.
time = 0.48, size = 1919, normalized size = 28.64
| | |
| method |
result |
size |
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| elliptic |
\(\text {Expression too large to display}\) |
\(863\) |
| risch |
\(\text {Expression too large to display}\) |
\(1874\) |
| default |
\(\text {Expression too large to display}\) |
\(1919\) |
| | |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
[Out]
-b/a^2*(2/3*d*x/c/(a*d-b*c)/((x^3+c/d)*d)^(1/2)-2/9*I/c/(a*d-b*c)*3^(1/2)*(-c*d^2)^(1/3)*(I*(x+1/2/d*(-c*d^2)^
(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)*((x-1/d*(-c*d^2)^(1/3))/(-3/2/d*(-c*d^2)
^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1
/2)*d/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(
-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2
)/d*(-c*d^2)^(1/3)))^(1/2))+1/3*I*b/d^2*2^(1/2)*sum(1/(a*d-b*c)^2/_alpha^2*(-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-
I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I
*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^
(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3
)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3
^(1/2)*d/(-c*d^2)^(1/3))^(1/2),1/2*b/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+
I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*
I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))+1/a^2*(-1/2/c^2*(d*x^3+c)^(1/2)/x^2-2/3*d*x/c^2/
((x^3+c/d)*d)^(1/2)+7/18*I/c^2*3^(1/2)*(-c*d^2)^(1/3)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3
))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)*((x-1/d*(-c*d^2)^(1/3))/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/
3)))^(1/2)*(-I*(x+1/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)/(d*x^3+
c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(
1/3))^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)))-b/a*(1
/3*b^2/a/(a*d-b*c)^2*x*(d*x^3+c)^(1/2)/(b*x^3+a)+2/3*d^2*x/c/(a*d-b*c)^2/((x^3+c/d)*d)^(1/2)-2/3*I*(1/6*b*d/a/
(a*d-b*c)^2+1/3*d^2/(a*d-b*c)^2/c)*3^(1/2)/d*(-c*d^2)^(1/3)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2
)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)*((x-1/d*(-c*d^2)^(1/3))/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^
2)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)/(
d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*
d^2)^(1/3))^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2))+1
/18*I/a/d^2*b*2^(1/2)*sum((13*a*d-4*b*c)/(a*d-b*c)^3/_alpha^2*(-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c
*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*
d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+
c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-
c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*
d^2)^(1/3))^(1/2),1/2*b/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d
-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(
-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))
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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(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)*x^3), x)
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (a + b x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x**3/(b*x**3+a)**2/(d*x**3+c)**(3/2),x)
[Out]
Integral(1/(x**3*(a + b*x**3)**2*(c + d*x**3)**(3/2)), x)
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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(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
[Out]
integrate(1/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)*x^3), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\left (b\,x^3+a\right )}^2\,{\left (d\,x^3+c\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(x^3*(a + b*x^3)^2*(c + d*x^3)^(3/2)),x)
[Out]
int(1/(x^3*(a + b*x^3)^2*(c + d*x^3)^(3/2)), x)
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