Integrand size = 20, antiderivative size = 67 \[ \int \frac {\left (a+b x^3\right )^p \left (c+d x^3\right )}{x} \, dx=\frac {d \left (a+b x^3\right )^{1+p}}{3 b (1+p)}-\frac {c \left (a+b x^3\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^3}{a}\right )}{3 a (1+p)} \] Output:
1/3*d*(b*x^3+a)^(p+1)/b/(p+1)-1/3*c*(b*x^3+a)^(p+1)*hypergeom([1, p+1],[2+ p],1+b*x^3/a)/a/(p+1)
Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^3\right )^p \left (c+d x^3\right )}{x} \, dx=\frac {\left (a+b x^3\right )^{1+p} \left (a d-b c \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^3}{a}\right )\right )}{3 a b (1+p)} \] Input:
Integrate[((a + b*x^3)^p*(c + d*x^3))/x,x]
Output:
((a + b*x^3)^(1 + p)*(a*d - b*c*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b* x^3)/a]))/(3*a*b*(1 + p))
Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {948, 90, 75}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (c+d x^3\right ) \left (a+b x^3\right )^p}{x} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {\left (b x^3+a\right )^p \left (d x^3+c\right )}{x^3}dx^3\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{3} \left (c \int \frac {\left (b x^3+a\right )^p}{x^3}dx^3+\frac {d \left (a+b x^3\right )^{p+1}}{b (p+1)}\right )\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \frac {1}{3} \left (\frac {d \left (a+b x^3\right )^{p+1}}{b (p+1)}-\frac {c \left (a+b x^3\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^3}{a}+1\right )}{a (p+1)}\right )\) |
Input:
Int[((a + b*x^3)^p*(c + d*x^3))/x,x]
Output:
((d*(a + b*x^3)^(1 + p))/(b*(1 + p)) - (c*(a + b*x^3)^(1 + p)*Hypergeometr ic2F1[1, 1 + p, 2 + p, 1 + (b*x^3)/a])/(a*(1 + p)))/3
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
\[\int \frac {\left (b \,x^{3}+a \right )^{p} \left (d \,x^{3}+c \right )}{x}d x\]
Input:
int((b*x^3+a)^p*(d*x^3+c)/x,x)
Output:
int((b*x^3+a)^p*(d*x^3+c)/x,x)
\[ \int \frac {\left (a+b x^3\right )^p \left (c+d x^3\right )}{x} \, dx=\int { \frac {{\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{p}}{x} \,d x } \] Input:
integrate((b*x^3+a)^p*(d*x^3+c)/x,x, algorithm="fricas")
Output:
integral((d*x^3 + c)*(b*x^3 + a)^p/x, x)
Time = 13.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^3\right )^p \left (c+d x^3\right )}{x} \, dx=- \frac {b^{p} c x^{3 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 \Gamma \left (1 - p\right )} + d \left (\begin {cases} \frac {a^{p} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{3}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{3} \right )} & \text {otherwise} \end {cases}}{3 b} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x**3+a)**p*(d*x**3+c)/x,x)
Output:
-b**p*c*x**(3*p)*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b* x**3))/(3*gamma(1 - p)) + d*Piecewise((a**p*x**3/3, Eq(b, 0)), (Piecewise( ((a + b*x**3)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + b*x**3), True))/(3*b) , True))
\[ \int \frac {\left (a+b x^3\right )^p \left (c+d x^3\right )}{x} \, dx=\int { \frac {{\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{p}}{x} \,d x } \] Input:
integrate((b*x^3+a)^p*(d*x^3+c)/x,x, algorithm="maxima")
Output:
integrate((d*x^3 + c)*(b*x^3 + a)^p/x, x)
\[ \int \frac {\left (a+b x^3\right )^p \left (c+d x^3\right )}{x} \, dx=\int { \frac {{\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{p}}{x} \,d x } \] Input:
integrate((b*x^3+a)^p*(d*x^3+c)/x,x, algorithm="giac")
Output:
integrate((d*x^3 + c)*(b*x^3 + a)^p/x, x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^p \left (c+d x^3\right )}{x} \, dx=\int \frac {{\left (b\,x^3+a\right )}^p\,\left (d\,x^3+c\right )}{x} \,d x \] Input:
int(((a + b*x^3)^p*(c + d*x^3))/x,x)
Output:
int(((a + b*x^3)^p*(c + d*x^3))/x, x)
\[ \int \frac {\left (a+b x^3\right )^p \left (c+d x^3\right )}{x} \, dx=\frac {\left (b \,x^{3}+a \right )^{p} a d p +\left (b \,x^{3}+a \right )^{p} b c p +\left (b \,x^{3}+a \right )^{p} b c +\left (b \,x^{3}+a \right )^{p} b d p \,x^{3}+3 \left (\int \frac {\left (b \,x^{3}+a \right )^{p}}{b \,x^{4}+a x}d x \right ) a b c \,p^{2}+3 \left (\int \frac {\left (b \,x^{3}+a \right )^{p}}{b \,x^{4}+a x}d x \right ) a b c p}{3 b p \left (p +1\right )} \] Input:
int((b*x^3+a)^p*(d*x^3+c)/x,x)
Output:
((a + b*x**3)**p*a*d*p + (a + b*x**3)**p*b*c*p + (a + b*x**3)**p*b*c + (a + b*x**3)**p*b*d*p*x**3 + 3*int((a + b*x**3)**p/(a*x + b*x**4),x)*a*b*c*p* *2 + 3*int((a + b*x**3)**p/(a*x + b*x**4),x)*a*b*c*p)/(3*b*p*(p + 1))