\(\int \frac {c+d x^3}{(a+b x^3)^{7/3}} \, dx\) [60]

Optimal result

Integrand size = 19, antiderivative size = 47 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx=\frac {3 c x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x \left (c+d x^3\right )}{4 a \left (a+b x^3\right )^{4/3}} \]

[Out]

3/4*c*x/a^2/(b*x^3+a)^(1/3)+1/4*x*(d*x^3+c)/a/(b*x^3+a)^(4/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {386, 197} \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx=\frac {3 c x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x \left (c+d x^3\right )}{4 a \left (a+b x^3\right )^{4/3}} \]

[In]

Int[(c + d*x^3)/(a + b*x^3)^(7/3),x]

[Out]

(3*c*x)/(4*a^2*(a + b*x^3)^(1/3)) + (x*(c + d*x^3))/(4*a*(a + b*x^3)^(4/3))

Rule 197

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 386

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

Rubi steps \begin{align*} \text {integral}& = \frac {x \left (c+d x^3\right )}{4 a \left (a+b x^3\right )^{4/3}}+\frac {(3 c) \int \frac {1}{\left (a+b x^3\right )^{4/3}} \, dx}{4 a} \\ & = \frac {3 c x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x \left (c+d x^3\right )}{4 a \left (a+b x^3\right )^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx=\frac {x \left (4 a c+3 b c x^3+a d x^3\right )}{4 a^2 \left (a+b x^3\right )^{4/3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.90 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72

method	result	size
gosper	\(\frac {x \left (a d \,x^{3}+3 b c \,x^{3}+4 a c \right )}{4 \left (b \,x^{3}+a \right )^{\frac {4}{3}} a^{2}}\)	\(34\)
trager	\(\frac {x \left (a d \,x^{3}+3 b c \,x^{3}+4 a c \right )}{4 \left (b \,x^{3}+a \right )^{\frac {4}{3}} a^{2}}\)	\(34\)
pseudoelliptic	\(\frac {x \left (a d \,x^{3}+3 b c \,x^{3}+4 a c \right )}{4 \left (b \,x^{3}+a \right )^{\frac {4}{3}} a^{2}}\)	\(34\)

[In]

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

[Out]

1/4*x*(a*d*x^3+3*b*c*x^3+4*a*c)/(b*x^3+a)^(4/3)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx=\frac {{\left ({\left (3 \, b c + a d\right )} x^{4} + 4 \, a c x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{4 \, {\left (a^{2} b^{2} x^{6} + 2 \, a^{3} b x^{3} + a^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (41) = 82\).

Time = 20.03 (sec) , antiderivative size = 190, normalized size of antiderivative = 4.04 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx=c \left (\frac {4 a x \Gamma \left (\frac {1}{3}\right )}{9 a^{\frac {10}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {7}{3}\right ) + 9 a^{\frac {7}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {7}{3}\right )} + \frac {3 b x^{4} \Gamma \left (\frac {1}{3}\right )}{9 a^{\frac {10}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {7}{3}\right ) + 9 a^{\frac {7}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {7}{3}\right )}\right ) + \frac {d x^{4} \Gamma \left (\frac {4}{3}\right )}{3 a^{\frac {7}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {7}{3}\right ) + 3 a^{\frac {4}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {7}{3}\right )} \]

[In]

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

[Out]

c*(4*a*x*gamma(1/3)/(9*a**(10/3)*(1 + b*x**3/a)**(1/3)*gamma(7/3) + 9*a**(7/3)*b*x**3*(1 + b*x**3/a)**(1/3)*ga
mma(7/3)) + 3*b*x**4*gamma(1/3)/(9*a**(10/3)*(1 + b*x**3/a)**(1/3)*gamma(7/3) + 9*a**(7/3)*b*x**3*(1 + b*x**3/
a)**(1/3)*gamma(7/3))) + d*x**4*gamma(4/3)/(3*a**(7/3)*(1 + b*x**3/a)**(1/3)*gamma(7/3) + 3*a**(4/3)*b*x**3*(1
+ b*x**3/a)**(1/3)*gamma(7/3))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx=-\frac {{\left (b - \frac {4 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} c x^{4}}{4 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2}} + \frac {d x^{4}}{4 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx=\int { \frac {d x^{3} + c}{{\left (b x^{3} + a\right )}^{\frac {7}{3}}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.46 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.70 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx=\frac {4\,a\,c\,x+a\,d\,x^4+3\,b\,c\,x^4}{4\,a^2\,{\left (b\,x^3+a\right )}^{4/3}} \]

[In]

int((c + d*x^3)/(a + b*x^3)^(7/3),x)

[Out]

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