Integrand size = 19, antiderivative size = 101 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=-\frac {3 (c+d x)^{4/3}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {9 d (c+d x)^{4/3}}{35 (b c-a d)^2 (a+b x)^{7/3}}-\frac {27 d^2 (c+d x)^{4/3}}{140 (b c-a d)^3 (a+b x)^{4/3}} \] Output:
-3/10*(d*x+c)^(4/3)/(-a*d+b*c)/(b*x+a)^(10/3)+9/35*d*(d*x+c)^(4/3)/(-a*d+b *c)^2/(b*x+a)^(7/3)-27/140*d^2*(d*x+c)^(4/3)/(-a*d+b*c)^3/(b*x+a)^(4/3)
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=-\frac {3 (c+d x)^{4/3} \left (35 a^2 d^2+10 a b d (-4 c+3 d x)+b^2 \left (14 c^2-12 c d x+9 d^2 x^2\right )\right )}{140 (b c-a d)^3 (a+b x)^{10/3}} \] Input:
Integrate[(c + d*x)^(1/3)/(a + b*x)^(13/3),x]
Output:
(-3*(c + d*x)^(4/3)*(35*a^2*d^2 + 10*a*b*d*(-4*c + 3*d*x) + b^2*(14*c^2 - 12*c*d*x + 9*d^2*x^2)))/(140*(b*c - a*d)^3*(a + b*x)^(10/3))
Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {55, 55, 48}
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.
\(\displaystyle \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {3 d \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}}dx}{5 (b c-a d)}-\frac {3 (c+d x)^{4/3}}{10 (a+b x)^{10/3} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {3 d \left (-\frac {3 d \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}}dx}{7 (b c-a d)}-\frac {3 (c+d x)^{4/3}}{7 (a+b x)^{7/3} (b c-a d)}\right )}{5 (b c-a d)}-\frac {3 (c+d x)^{4/3}}{10 (a+b x)^{10/3} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {3 (c+d x)^{4/3}}{10 (a+b x)^{10/3} (b c-a d)}-\frac {3 d \left (\frac {9 d (c+d x)^{4/3}}{28 (a+b x)^{4/3} (b c-a d)^2}-\frac {3 (c+d x)^{4/3}}{7 (a+b x)^{7/3} (b c-a d)}\right )}{5 (b c-a d)}\) |
Input:
Int[(c + d*x)^(1/3)/(a + b*x)^(13/3),x]
Output:
(-3*(c + d*x)^(4/3))/(10*(b*c - a*d)*(a + b*x)^(10/3)) - (3*d*((-3*(c + d* x)^(4/3))/(7*(b*c - a*d)*(a + b*x)^(7/3)) + (9*d*(c + d*x)^(4/3))/(28*(b*c - a*d)^2*(a + b*x)^(4/3))))/(5*(b*c - a*d))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {3 \left (x d +c \right )^{\frac {4}{3}} \left (9 d^{2} x^{2} b^{2}+30 x a b \,d^{2}-12 x \,b^{2} c d +35 a^{2} d^{2}-40 a b c d +14 b^{2} c^{2}\right )}{140 \left (b x +a \right )^{\frac {10}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
orering | \(\frac {3 \left (x d +c \right )^{\frac {4}{3}} \left (9 d^{2} x^{2} b^{2}+30 x a b \,d^{2}-12 x \,b^{2} c d +35 a^{2} d^{2}-40 a b c d +14 b^{2} c^{2}\right )}{140 \left (b x +a \right )^{\frac {10}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
Input:
int((d*x+c)^(1/3)/(b*x+a)^(13/3),x,method=_RETURNVERBOSE)
Output:
3/140*(d*x+c)^(4/3)*(9*b^2*d^2*x^2+30*a*b*d^2*x-12*b^2*c*d*x+35*a^2*d^2-40 *a*b*c*d+14*b^2*c^2)/(b*x+a)^(10/3)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b ^3*c^3)
Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (83) = 166\).
Time = 0.08 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.34 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=-\frac {3 \, {\left (9 \, b^{2} d^{3} x^{3} + 14 \, b^{2} c^{3} - 40 \, a b c^{2} d + 35 \, a^{2} c d^{2} - 3 \, {\left (b^{2} c d^{2} - 10 \, a b d^{3}\right )} x^{2} + {\left (2 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{140 \, {\left (a^{4} b^{3} c^{3} - 3 \, a^{5} b^{2} c^{2} d + 3 \, a^{6} b c d^{2} - a^{7} d^{3} + {\left (b^{7} c^{3} - 3 \, a b^{6} c^{2} d + 3 \, a^{2} b^{5} c d^{2} - a^{3} b^{4} d^{3}\right )} x^{4} + 4 \, {\left (a b^{6} c^{3} - 3 \, a^{2} b^{5} c^{2} d + 3 \, a^{3} b^{4} c d^{2} - a^{4} b^{3} d^{3}\right )} x^{3} + 6 \, {\left (a^{2} b^{5} c^{3} - 3 \, a^{3} b^{4} c^{2} d + 3 \, a^{4} b^{3} c d^{2} - a^{5} b^{2} d^{3}\right )} x^{2} + 4 \, {\left (a^{3} b^{4} c^{3} - 3 \, a^{4} b^{3} c^{2} d + 3 \, a^{5} b^{2} c d^{2} - a^{6} b d^{3}\right )} x\right )}} \] Input:
integrate((d*x+c)^(1/3)/(b*x+a)^(13/3),x, algorithm="fricas")
Output:
-3/140*(9*b^2*d^3*x^3 + 14*b^2*c^3 - 40*a*b*c^2*d + 35*a^2*c*d^2 - 3*(b^2* c*d^2 - 10*a*b*d^3)*x^2 + (2*b^2*c^2*d - 10*a*b*c*d^2 + 35*a^2*d^3)*x)*(b* x + a)^(2/3)*(d*x + c)^(1/3)/(a^4*b^3*c^3 - 3*a^5*b^2*c^2*d + 3*a^6*b*c*d^ 2 - a^7*d^3 + (b^7*c^3 - 3*a*b^6*c^2*d + 3*a^2*b^5*c*d^2 - a^3*b^4*d^3)*x^ 4 + 4*(a*b^6*c^3 - 3*a^2*b^5*c^2*d + 3*a^3*b^4*c*d^2 - a^4*b^3*d^3)*x^3 + 6*(a^2*b^5*c^3 - 3*a^3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x^2 + 4* (a^3*b^4*c^3 - 3*a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*b*d^3)*x)
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\int \frac {\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac {13}{3}}}\, dx \] Input:
integrate((d*x+c)**(1/3)/(b*x+a)**(13/3),x)
Output:
Integral((c + d*x)**(1/3)/(a + b*x)**(13/3), x)
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {13}{3}}} \,d x } \] Input:
integrate((d*x+c)^(1/3)/(b*x+a)^(13/3),x, algorithm="maxima")
Output:
integrate((d*x + c)^(1/3)/(b*x + a)^(13/3), x)
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {13}{3}}} \,d x } \] Input:
integrate((d*x+c)^(1/3)/(b*x+a)^(13/3),x, algorithm="giac")
Output:
integrate((d*x + c)^(1/3)/(b*x + a)^(13/3), x)
Time = 0.58 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.01 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {105\,a^2\,c\,d^2-120\,a\,b\,c^2\,d+42\,b^2\,c^3}{140\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {x\,\left (105\,a^2\,d^3-30\,a\,b\,c\,d^2+6\,b^2\,c^2\,d\right )}{140\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {27\,d^3\,x^3}{140\,b\,{\left (a\,d-b\,c\right )}^3}+\frac {9\,d^2\,x^2\,\left (10\,a\,d-b\,c\right )}{140\,b^2\,{\left (a\,d-b\,c\right )}^3}\right )}{x^3\,{\left (a+b\,x\right )}^{1/3}+\frac {a^3\,{\left (a+b\,x\right )}^{1/3}}{b^3}+\frac {3\,a\,x^2\,{\left (a+b\,x\right )}^{1/3}}{b}+\frac {3\,a^2\,x\,{\left (a+b\,x\right )}^{1/3}}{b^2}} \] Input:
int((c + d*x)^(1/3)/(a + b*x)^(13/3),x)
Output:
((c + d*x)^(1/3)*((42*b^2*c^3 + 105*a^2*c*d^2 - 120*a*b*c^2*d)/(140*b^3*(a *d - b*c)^3) + (x*(105*a^2*d^3 + 6*b^2*c^2*d - 30*a*b*c*d^2))/(140*b^3*(a* d - b*c)^3) + (27*d^3*x^3)/(140*b*(a*d - b*c)^3) + (9*d^2*x^2*(10*a*d - b* c))/(140*b^2*(a*d - b*c)^3)))/(x^3*(a + b*x)^(1/3) + (a^3*(a + b*x)^(1/3)) /b^3 + (3*a*x^2*(a + b*x)^(1/3))/b + (3*a^2*x*(a + b*x)^(1/3))/b^2)
Time = 0.28 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.99 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\frac {3 \left (d x +c \right )^{\frac {1}{3}} \left (9 b^{2} d^{3} x^{3}+30 a b \,d^{3} x^{2}-3 b^{2} c \,d^{2} x^{2}+35 a^{2} d^{3} x -10 a b c \,d^{2} x +2 b^{2} c^{2} d x +35 a^{2} c \,d^{2}-40 a b \,c^{2} d +14 b^{2} c^{3}\right )}{140 \left (b x +a \right )^{\frac {1}{3}} \left (a^{3} b^{3} d^{3} x^{3}-3 a^{2} b^{4} c \,d^{2} x^{3}+3 a \,b^{5} c^{2} d \,x^{3}-b^{6} c^{3} x^{3}+3 a^{4} b^{2} d^{3} x^{2}-9 a^{3} b^{3} c \,d^{2} x^{2}+9 a^{2} b^{4} c^{2} d \,x^{2}-3 a \,b^{5} c^{3} x^{2}+3 a^{5} b \,d^{3} x -9 a^{4} b^{2} c \,d^{2} x +9 a^{3} b^{3} c^{2} d x -3 a^{2} b^{4} c^{3} x +a^{6} d^{3}-3 a^{5} b c \,d^{2}+3 a^{4} b^{2} c^{2} d -a^{3} b^{3} c^{3}\right )} \] Input:
int((d*x+c)^(1/3)/(b*x+a)^(13/3),x)
Output:
(3*(c + d*x)**(1/3)*(35*a**2*c*d**2 + 35*a**2*d**3*x - 40*a*b*c**2*d - 10* a*b*c*d**2*x + 30*a*b*d**3*x**2 + 14*b**2*c**3 + 2*b**2*c**2*d*x - 3*b**2* c*d**2*x**2 + 9*b**2*d**3*x**3))/(140*(a + b*x)**(1/3)*(a**6*d**3 - 3*a**5 *b*c*d**2 + 3*a**5*b*d**3*x + 3*a**4*b**2*c**2*d - 9*a**4*b**2*c*d**2*x + 3*a**4*b**2*d**3*x**2 - a**3*b**3*c**3 + 9*a**3*b**3*c**2*d*x - 9*a**3*b** 3*c*d**2*x**2 + a**3*b**3*d**3*x**3 - 3*a**2*b**4*c**3*x + 9*a**2*b**4*c** 2*d*x**2 - 3*a**2*b**4*c*d**2*x**3 - 3*a*b**5*c**3*x**2 + 3*a*b**5*c**2*d* x**3 - b**6*c**3*x**3))