3.61 \(\int \frac {c+d x^3}{(a+b x^3)^{10/3}} \, dx\)

Optimal. Leaf size=91 \[ \frac {3 x (a d+6 b c)}{28 a^3 b \sqrt [3]{a+b x^3}}+\frac {x (a d+6 b c)}{28 a^2 b \left (a+b x^3\right )^{4/3}}+\frac {x (b c-a d)}{7 a b \left (a+b x^3\right )^{7/3}} \]

[Out]

1/7*(-a*d+b*c)*x/a/b/(b*x^3+a)^(7/3)+1/28*(a*d+6*b*c)*x/a^2/b/(b*x^3+a)^(4/3)+3/28*(a*d+6*b*c)*x/a^3/b/(b*x^3+
a)^(1/3)

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Rubi [A]  time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {385, 192, 191} \[ \frac {3 x (a d+6 b c)}{28 a^3 b \sqrt [3]{a+b x^3}}+\frac {x (a d+6 b c)}{28 a^2 b \left (a+b x^3\right )^{4/3}}+\frac {x (b c-a d)}{7 a b \left (a+b x^3\right )^{7/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)/(a + b*x^3)^(10/3),x]

[Out]

((b*c - a*d)*x)/(7*a*b*(a + b*x^3)^(7/3)) + ((6*b*c + a*d)*x)/(28*a^2*b*(a + b*x^3)^(4/3)) + (3*(6*b*c + a*d)*
x)/(28*a^3*b*(a + b*x^3)^(1/3))

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

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, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -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])

Rubi steps

\begin {align*} \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx &=\frac {(b c-a d) x}{7 a b \left (a+b x^3\right )^{7/3}}+\frac {(6 b c+a d) \int \frac {1}{\left (a+b x^3\right )^{7/3}} \, dx}{7 a b}\\ &=\frac {(b c-a d) x}{7 a b \left (a+b x^3\right )^{7/3}}+\frac {(6 b c+a d) x}{28 a^2 b \left (a+b x^3\right )^{4/3}}+\frac {(3 (6 b c+a d)) \int \frac {1}{\left (a+b x^3\right )^{4/3}} \, dx}{28 a^2 b}\\ &=\frac {(b c-a d) x}{7 a b \left (a+b x^3\right )^{7/3}}+\frac {(6 b c+a d) x}{28 a^2 b \left (a+b x^3\right )^{4/3}}+\frac {3 (6 b c+a d) x}{28 a^3 b \sqrt [3]{a+b x^3}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 59, normalized size = 0.65 \[ \frac {7 a^2 \left (4 c x+d x^4\right )+3 a b x^4 \left (14 c+d x^3\right )+18 b^2 c x^7}{28 a^3 \left (a+b x^3\right )^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)/(a + b*x^3)^(10/3),x]

[Out]

(18*b^2*c*x^7 + 3*a*b*x^4*(14*c + d*x^3) + 7*a^2*(4*c*x + d*x^4))/(28*a^3*(a + b*x^3)^(7/3))

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fricas [A]  time = 0.72, size = 87, normalized size = 0.96 \[ \frac {{\left (3 \, {\left (6 \, b^{2} c + a b d\right )} x^{7} + 7 \, {\left (6 \, a b c + a^{2} d\right )} x^{4} + 28 \, a^{2} c x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{28 \, {\left (a^{3} b^{3} x^{9} + 3 \, a^{4} b^{2} x^{6} + 3 \, a^{5} b x^{3} + a^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(10/3),x, algorithm="fricas")

[Out]

1/28*(3*(6*b^2*c + a*b*d)*x^7 + 7*(6*a*b*c + a^2*d)*x^4 + 28*a^2*c*x)*(b*x^3 + a)^(2/3)/(a^3*b^3*x^9 + 3*a^4*b
^2*x^6 + 3*a^5*b*x^3 + a^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{3} + c}{{\left (b x^{3} + a\right )}^{\frac {10}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(10/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)/(b*x^3 + a)^(10/3), x)

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maple [A]  time = 0.04, size = 57, normalized size = 0.63 \[ \frac {\left (3 a b d \,x^{6}+18 b^{2} c \,x^{6}+7 a^{2} d \,x^{3}+42 a b c \,x^{3}+28 a^{2} c \right ) x}{28 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)/(b*x^3+a)^(10/3),x)

[Out]

1/28*x*(3*a*b*d*x^6+18*b^2*c*x^6+7*a^2*d*x^3+42*a*b*c*x^3+28*a^2*c)/(b*x^3+a)^(7/3)/a^3

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maxima [A]  time = 0.61, size = 86, normalized size = 0.95 \[ -\frac {{\left (4 \, b - \frac {7 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} d x^{7}}{28 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{2}} + \frac {{\left (2 \, b^{2} - \frac {7 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {14 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} c x^{7}}{14 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(10/3),x, algorithm="maxima")

[Out]

-1/28*(4*b - 7*(b*x^3 + a)/x^3)*d*x^7/((b*x^3 + a)^(7/3)*a^2) + 1/14*(2*b^2 - 7*(b*x^3 + a)*b/x^3 + 14*(b*x^3
+ a)^2/x^6)*c*x^7/((b*x^3 + a)^(7/3)*a^3)

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mupad [B]  time = 1.42, size = 87, normalized size = 0.96 \[ \frac {3\,a\,d\,x\,{\left (b\,x^3+a\right )}^2-4\,a^3\,d\,x+18\,b\,c\,x\,{\left (b\,x^3+a\right )}^2+a^2\,d\,x\,\left (b\,x^3+a\right )+4\,a^2\,b\,c\,x+6\,a\,b\,c\,x\,\left (b\,x^3+a\right )}{28\,a^3\,b\,{\left (b\,x^3+a\right )}^{7/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)/(a + b*x^3)^(10/3),x)

[Out]

(3*a*d*x*(a + b*x^3)^2 - 4*a^3*d*x + 18*b*c*x*(a + b*x^3)^2 + a^2*d*x*(a + b*x^3) + 4*a^2*b*c*x + 6*a*b*c*x*(a
+ b*x^3))/(28*a^3*b*(a + b*x^3)^(7/3))

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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((d*x**3+c)/(b*x**3+a)**(10/3),x)

[Out]

Timed out

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