Integrand size = 19, antiderivative size = 98 \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx=-\frac {6}{(b c-a d) \sqrt [6]{a+b x} (c+d x)^{11/6}}-\frac {72 d (a+b x)^{5/6}}{11 (b c-a d)^2 (c+d x)^{11/6}}-\frac {432 b d (a+b x)^{5/6}}{55 (b c-a d)^3 (c+d x)^{5/6}} \] Output:
-6/(-a*d+b*c)/(b*x+a)^(1/6)/(d*x+c)^(11/6)-72/11*d*(b*x+a)^(5/6)/(-a*d+b*c )^2/(d*x+c)^(11/6)-432/55*b*d*(b*x+a)^(5/6)/(-a*d+b*c)^3/(d*x+c)^(5/6)
Time = 0.35 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx=-\frac {6 \left (-5 a^2 d^2+2 a b d (11 c+6 d x)+b^2 \left (55 c^2+132 c d x+72 d^2 x^2\right )\right )}{55 (b c-a d)^3 \sqrt [6]{a+b x} (c+d x)^{11/6}} \] Input:
Integrate[1/((a + b*x)^(7/6)*(c + d*x)^(17/6)),x]
Output:
(-6*(-5*a^2*d^2 + 2*a*b*d*(11*c + 6*d*x) + b^2*(55*c^2 + 132*c*d*x + 72*d^ 2*x^2)))/(55*(b*c - a*d)^3*(a + b*x)^(1/6)*(c + d*x)^(11/6))
Time = 0.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.12, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {12 d \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{17/6}}dx}{b c-a d}-\frac {6}{\sqrt [6]{a+b x} (c+d x)^{11/6} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {12 d \left (\frac {6 b \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{11/6}}dx}{11 (b c-a d)}+\frac {6 (a+b x)^{5/6}}{11 (c+d x)^{11/6} (b c-a d)}\right )}{b c-a d}-\frac {6}{\sqrt [6]{a+b x} (c+d x)^{11/6} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {12 d \left (\frac {36 b (a+b x)^{5/6}}{55 (c+d x)^{5/6} (b c-a d)^2}+\frac {6 (a+b x)^{5/6}}{11 (c+d x)^{11/6} (b c-a d)}\right )}{b c-a d}-\frac {6}{\sqrt [6]{a+b x} (c+d x)^{11/6} (b c-a d)}\) |
Input:
Int[1/((a + b*x)^(7/6)*(c + d*x)^(17/6)),x]
Output:
-6/((b*c - a*d)*(a + b*x)^(1/6)*(c + d*x)^(11/6)) - (12*d*((6*(a + b*x)^(5 /6))/(11*(b*c - a*d)*(c + d*x)^(11/6)) + (36*b*(a + b*x)^(5/6))/(55*(b*c - a*d)^2*(c + d*x)^(5/6))))/(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 = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07
method | result | size |
gosper | \(-\frac {6 \left (-72 d^{2} x^{2} b^{2}-12 x a b \,d^{2}-132 x \,b^{2} c d +5 a^{2} d^{2}-22 a b c d -55 b^{2} c^{2}\right )}{55 \left (b x +a \right )^{\frac {1}{6}} \left (x d +c \right )^{\frac {11}{6}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
orering | \(-\frac {6 \left (-72 d^{2} x^{2} b^{2}-12 x a b \,d^{2}-132 x \,b^{2} c d +5 a^{2} d^{2}-22 a b c d -55 b^{2} c^{2}\right )}{55 \left (b x +a \right )^{\frac {1}{6}} \left (x d +c \right )^{\frac {11}{6}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
Input:
int(1/(b*x+a)^(7/6)/(d*x+c)^(17/6),x,method=_RETURNVERBOSE)
Output:
-6/55*(-72*b^2*d^2*x^2-12*a*b*d^2*x-132*b^2*c*d*x+5*a^2*d^2-22*a*b*c*d-55* b^2*c^2)/(b*x+a)^(1/6)/(d*x+c)^(11/6)/(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. 273 vs. \(2 (82) = 164\).
Time = 0.10 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.79 \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx=-\frac {6 \, {\left (72 \, b^{2} d^{2} x^{2} + 55 \, b^{2} c^{2} + 22 \, a b c d - 5 \, a^{2} d^{2} + 12 \, {\left (11 \, b^{2} c d + a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {1}{6}}}{55 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}} \] Input:
integrate(1/(b*x+a)^(7/6)/(d*x+c)^(17/6),x, algorithm="fricas")
Output:
-6/55*(72*b^2*d^2*x^2 + 55*b^2*c^2 + 22*a*b*c*d - 5*a^2*d^2 + 12*(11*b^2*c *d + a*b*d^2)*x)*(b*x + a)^(5/6)*(d*x + c)^(1/6)/(a*b^3*c^5 - 3*a^2*b^2*c^ 4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a ^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2 *c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2 *c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)
Timed out. \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x+a)**(7/6)/(d*x+c)**(17/6),x)
Output:
Timed out
\[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {7}{6}} {\left (d x + c\right )}^{\frac {17}{6}}} \,d x } \] Input:
integrate(1/(b*x+a)^(7/6)/(d*x+c)^(17/6),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(7/6)*(d*x + c)^(17/6)), x)
\[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {7}{6}} {\left (d x + c\right )}^{\frac {17}{6}}} \,d x } \] Input:
integrate(1/(b*x+a)^(7/6)/(d*x+c)^(17/6),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(7/6)*(d*x + c)^(17/6)), x)
Time = 0.60 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx=\frac {{\left (c+d\,x\right )}^{1/6}\,\left (\frac {432\,b^2\,x^2}{55\,{\left (a\,d-b\,c\right )}^3}+\frac {-30\,a^2\,d^2+132\,a\,b\,c\,d+330\,b^2\,c^2}{55\,d^2\,{\left (a\,d-b\,c\right )}^3}+\frac {72\,b\,x\,\left (a\,d+11\,b\,c\right )}{55\,d\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,{\left (a+b\,x\right )}^{1/6}+\frac {c^2\,{\left (a+b\,x\right )}^{1/6}}{d^2}+\frac {2\,c\,x\,{\left (a+b\,x\right )}^{1/6}}{d}} \] Input:
int(1/((a + b*x)^(7/6)*(c + d*x)^(17/6)),x)
Output:
((c + d*x)^(1/6)*((432*b^2*x^2)/(55*(a*d - b*c)^3) + (330*b^2*c^2 - 30*a^2 *d^2 + 132*a*b*c*d)/(55*d^2*(a*d - b*c)^3) + (72*b*x*(a*d + 11*b*c))/(55*d *(a*d - b*c)^3)))/(x^2*(a + b*x)^(1/6) + (c^2*(a + b*x)^(1/6))/d^2 + (2*c* x*(a + b*x)^(1/6))/d)
\[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx=\int \frac {\left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}}}{\sqrt {d x +c}\, \sqrt {b x +a}\, a \,c^{3}+3 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,c^{2} d x +3 \sqrt {d x +c}\, \sqrt {b x +a}\, a c \,d^{2} x^{2}+\sqrt {d x +c}\, \sqrt {b x +a}\, a \,d^{3} x^{3}+\sqrt {d x +c}\, \sqrt {b x +a}\, b \,c^{3} x +3 \sqrt {d x +c}\, \sqrt {b x +a}\, b \,c^{2} d \,x^{2}+3 \sqrt {d x +c}\, \sqrt {b x +a}\, b c \,d^{2} x^{3}+\sqrt {d x +c}\, \sqrt {b x +a}\, b \,d^{3} x^{4}}d x \] Input:
int(1/(b*x+a)^(7/6)/(d*x+c)^(17/6),x)
Output:
int(((c + d*x)**(2/3)*(a + b*x)**(1/3))/(sqrt(c + d*x)*sqrt(a + b*x)*a*c** 3 + 3*sqrt(c + d*x)*sqrt(a + b*x)*a*c**2*d*x + 3*sqrt(c + d*x)*sqrt(a + b* x)*a*c*d**2*x**2 + sqrt(c + d*x)*sqrt(a + b*x)*a*d**3*x**3 + sqrt(c + d*x) *sqrt(a + b*x)*b*c**3*x + 3*sqrt(c + d*x)*sqrt(a + b*x)*b*c**2*d*x**2 + 3* sqrt(c + d*x)*sqrt(a + b*x)*b*c*d**2*x**3 + sqrt(c + d*x)*sqrt(a + b*x)*b* d**3*x**4),x)