Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}+\frac {32 d \sqrt [4]{c+d x}}{45 (b c-a d)^2 (a+b x)^{5/4}}-\frac {128 d^2 \sqrt [4]{c+d x}}{45 (b c-a d)^3 \sqrt [4]{a+b x}} \] Output:
-4/9*(d*x+c)^(1/4)/(-a*d+b*c)/(b*x+a)^(9/4)+32/45*d*(d*x+c)^(1/4)/(-a*d+b* c)^2/(b*x+a)^(5/4)-128/45*d^2*(d*x+c)^(1/4)/(-a*d+b*c)^3/(b*x+a)^(1/4)
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{c+d x} \left (45 a^2 d^2-18 a b d (c-4 d x)+b^2 \left (5 c^2-8 c d x+32 d^2 x^2\right )\right )}{45 (b c-a d)^3 (a+b x)^{9/4}} \] Input:
Integrate[1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x]
Output:
(-4*(c + d*x)^(1/4)*(45*a^2*d^2 - 18*a*b*d*(c - 4*d*x) + b^2*(5*c^2 - 8*c* d*x + 32*d^2*x^2)))/(45*(b*c - a*d)^3*(a + b*x)^(9/4))
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.
\(\displaystyle \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 d \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}}dx}{9 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {8 d \left (-\frac {4 d \int \frac {1}{(a+b x)^{5/4} (c+d x)^{3/4}}dx}{5 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{5 (a+b x)^{5/4} (b c-a d)}\right )}{9 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {8 d \left (\frac {16 d \sqrt [4]{c+d x}}{5 \sqrt [4]{a+b x} (b c-a d)^2}-\frac {4 \sqrt [4]{c+d x}}{5 (a+b x)^{5/4} (b c-a d)}\right )}{9 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)}\) |
Input:
Int[1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x]
Output:
(-4*(c + d*x)^(1/4))/(9*(b*c - a*d)*(a + b*x)^(9/4)) - (8*d*((-4*(c + d*x) ^(1/4))/(5*(b*c - a*d)*(a + b*x)^(5/4)) + (16*d*(c + d*x)^(1/4))/(5*(b*c - a*d)^2*(a + b*x)^(1/4))))/(9*(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 = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {4 \left (x d +c \right )^{\frac {1}{4}} \left (32 d^{2} x^{2} b^{2}+72 x a b \,d^{2}-8 x \,b^{2} c d +45 a^{2} d^{2}-18 a b c d +5 b^{2} c^{2}\right )}{45 \left (b x +a \right )^{\frac {9}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
orering | \(\frac {4 \left (x d +c \right )^{\frac {1}{4}} \left (32 d^{2} x^{2} b^{2}+72 x a b \,d^{2}-8 x \,b^{2} c d +45 a^{2} d^{2}-18 a b c d +5 b^{2} c^{2}\right )}{45 \left (b x +a \right )^{\frac {9}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
Input:
int(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x,method=_RETURNVERBOSE)
Output:
4/45*(d*x+c)^(1/4)*(32*b^2*d^2*x^2+72*a*b*d^2*x-8*b^2*c*d*x+45*a^2*d^2-18* a*b*c*d+5*b^2*c^2)/(b*x+a)^(9/4)/(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. 251 vs. \(2 (83) = 166\).
Time = 0.10 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.49 \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=-\frac {4 \, {\left (32 \, b^{2} d^{2} x^{2} + 5 \, b^{2} c^{2} - 18 \, a b c d + 45 \, a^{2} d^{2} - 8 \, {\left (b^{2} c d - 9 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{45 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \] Input:
integrate(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x, algorithm="fricas")
Output:
-4/45*(32*b^2*d^2*x^2 + 5*b^2*c^2 - 18*a*b*c*d + 45*a^2*d^2 - 8*(b^2*c*d - 9*a*b*d^2)*x)*(b*x + a)^(3/4)*(d*x + c)^(1/4)/(a^3*b^3*c^3 - 3*a^4*b^2*c^ 2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*x^3 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a ^4*b^2*d^3)*x^2 + 3*(a^2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5 *b*d^3)*x)
\[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {13}{4}} \left (c + d x\right )^{\frac {3}{4}}}\, dx \] Input:
integrate(1/(b*x+a)**(13/4)/(d*x+c)**(3/4),x)
Output:
Integral(1/((a + b*x)**(13/4)*(c + d*x)**(3/4)), x)
\[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {13}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(13/4)*(d*x + c)^(3/4)), x)
\[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {13}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(13/4)*(d*x + c)^(3/4)), x)
Time = 0.58 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\frac {{\left (c+d\,x\right )}^{1/4}\,\left (\frac {128\,d^2\,x^2}{45\,{\left (a\,d-b\,c\right )}^3}+\frac {180\,a^2\,d^2-72\,a\,b\,c\,d+20\,b^2\,c^2}{45\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {32\,d\,x\,\left (9\,a\,d-b\,c\right )}{45\,b\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,{\left (a+b\,x\right )}^{1/4}+\frac {a^2\,{\left (a+b\,x\right )}^{1/4}}{b^2}+\frac {2\,a\,x\,{\left (a+b\,x\right )}^{1/4}}{b}} \] Input:
int(1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x)
Output:
((c + d*x)^(1/4)*((128*d^2*x^2)/(45*(a*d - b*c)^3) + (180*a^2*d^2 + 20*b^2 *c^2 - 72*a*b*c*d)/(45*b^2*(a*d - b*c)^3) + (32*d*x*(9*a*d - b*c))/(45*b*( a*d - b*c)^3)))/(x^2*(a + b*x)^(1/4) + (a^2*(a + b*x)^(1/4))/b^2 + (2*a*x* (a + b*x)^(1/4))/b)
\[ \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {3}{4}} \left (b x +a \right )^{\frac {1}{4}} a^{3}+3 \left (d x +c \right )^{\frac {3}{4}} \left (b x +a \right )^{\frac {1}{4}} a^{2} b x +3 \left (d x +c \right )^{\frac {3}{4}} \left (b x +a \right )^{\frac {1}{4}} a \,b^{2} x^{2}+\left (d x +c \right )^{\frac {3}{4}} \left (b x +a \right )^{\frac {1}{4}} b^{3} x^{3}}d x \] Input:
int(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x)
Output:
int(1/((c + d*x)**(3/4)*(a + b*x)**(1/4)*a**3 + 3*(c + d*x)**(3/4)*(a + b* x)**(1/4)*a**2*b*x + 3*(c + d*x)**(3/4)*(a + b*x)**(1/4)*a*b**2*x**2 + (c + d*x)**(3/4)*(a + b*x)**(1/4)*b**3*x**3),x)