Integrand size = 19, antiderivative size = 136 \[ \int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{c+d x}}{13 (b c-a d) (a+b x)^{13/4}}+\frac {16 d \sqrt [4]{c+d x}}{39 (b c-a d)^2 (a+b x)^{9/4}}-\frac {128 d^2 \sqrt [4]{c+d x}}{195 (b c-a d)^3 (a+b x)^{5/4}}+\frac {512 d^3 \sqrt [4]{c+d x}}{195 (b c-a d)^4 \sqrt [4]{a+b x}} \]
-4/13*(d*x+c)^(1/4)/(-a*d+b*c)/(b*x+a)^(13/4)+16/39*d*(d*x+c)^(1/4)/(-a*d+ b*c)^2/(b*x+a)^(9/4)-128/195*d^2*(d*x+c)^(1/4)/(-a*d+b*c)^3/(b*x+a)^(5/4)+ 512/195*d^3*(d*x+c)^(1/4)/(-a*d+b*c)^4/(b*x+a)^(1/4)
Time = 0.76 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx=\frac {4 \sqrt [4]{c+d x} \left (195 a^3 d^3-117 a^2 b d^2 (c-4 d x)+13 a b^2 d \left (5 c^2-8 c d x+32 d^2 x^2\right )+b^3 \left (-15 c^3+20 c^2 d x-32 c d^2 x^2+128 d^3 x^3\right )\right )}{195 (b c-a d)^4 (a+b x)^{13/4}} \]
(4*(c + d*x)^(1/4)*(195*a^3*d^3 - 117*a^2*b*d^2*(c - 4*d*x) + 13*a*b^2*d*( 5*c^2 - 8*c*d*x + 32*d^2*x^2) + b^3*(-15*c^3 + 20*c^2*d*x - 32*c*d^2*x^2 + 128*d^3*x^3)))/(195*(b*c - a*d)^4*(a + b*x)^(13/4))
Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {55, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {12 d \int \frac {1}{(a+b x)^{13/4} (c+d x)^{3/4}}dx}{13 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{13 (a+b x)^{13/4} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {12 d \left (-\frac {8 d \int \frac {1}{(a+b x)^{9/4} (c+d x)^{3/4}}dx}{9 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)}\right )}{13 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{13 (a+b x)^{13/4} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {12 d \left (-\frac {8 d \left (-\frac {4 d \int \frac {1}{(a+b x)^{5/4} (c+d x)^{3/4}}dx}{5 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{5 (a+b x)^{5/4} (b c-a d)}\right )}{9 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)}\right )}{13 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{13 (a+b x)^{13/4} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {12 d \left (-\frac {8 d \left (\frac {16 d \sqrt [4]{c+d x}}{5 \sqrt [4]{a+b x} (b c-a d)^2}-\frac {4 \sqrt [4]{c+d x}}{5 (a+b x)^{5/4} (b c-a d)}\right )}{9 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)}\right )}{13 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{13 (a+b x)^{13/4} (b c-a d)}\) |
(-4*(c + d*x)^(1/4))/(13*(b*c - a*d)*(a + b*x)^(13/4)) - (12*d*((-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))))/(13*(b*c - a*d))
3.18.11.3.1 Defintions of rubi rules used
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 = 0.34 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.26
method | result | size |
gosper | \(\frac {4 \left (d x +c \right )^{\frac {1}{4}} \left (128 d^{3} x^{3} b^{3}+416 x^{2} a \,b^{2} d^{3}-32 x^{2} b^{3} c \,d^{2}+468 x \,a^{2} b \,d^{3}-104 x a \,b^{2} c \,d^{2}+20 x \,b^{3} c^{2} d +195 a^{3} d^{3}-117 a^{2} b c \,d^{2}+65 a \,b^{2} c^{2} d -15 b^{3} c^{3}\right )}{195 \left (b x +a \right )^{\frac {13}{4}} \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}\) | \(171\) |
4/195*(d*x+c)^(1/4)*(128*b^3*d^3*x^3+416*a*b^2*d^3*x^2-32*b^3*c*d^2*x^2+46 8*a^2*b*d^3*x-104*a*b^2*c*d^2*x+20*b^3*c^2*d*x+195*a^3*d^3-117*a^2*b*c*d^2 +65*a*b^2*c^2*d-15*b^3*c^3)/(b*x+a)^(13/4)/(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^ 2*c^2*d^2-4*a*b^3*c^3*d+b^4*c^4)
Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (112) = 224\).
Time = 0.24 (sec) , antiderivative size = 419, normalized size of antiderivative = 3.08 \[ \int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx=\frac {4 \, {\left (128 \, b^{3} d^{3} x^{3} - 15 \, b^{3} c^{3} + 65 \, a b^{2} c^{2} d - 117 \, a^{2} b c d^{2} + 195 \, a^{3} d^{3} - 32 \, {\left (b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2} + 4 \, {\left (5 \, b^{3} c^{2} d - 26 \, a b^{2} c d^{2} + 117 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{195 \, {\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \]
4/195*(128*b^3*d^3*x^3 - 15*b^3*c^3 + 65*a*b^2*c^2*d - 117*a^2*b*c*d^2 + 1 95*a^3*d^3 - 32*(b^3*c*d^2 - 13*a*b^2*d^3)*x^2 + 4*(5*b^3*c^2*d - 26*a*b^2 *c*d^2 + 117*a^2*b*d^3)*x)*(b*x + a)^(3/4)*(d*x + c)^(1/4)/(a^4*b^4*c^4 - 4*a^5*b^3*c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 + a^8*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d + 6*a^2*b^6*c^2*d^2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^4 + 4*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a^4*b^4*c*d^3 + a^5 *b^3*d^4)*x^3 + 6*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a ^5*b^3*c*d^3 + a^6*b^2*d^4)*x^2 + 4*(a^3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^5 *b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4)*x)
Timed out. \[ \int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {17}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {17}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]
Time = 1.20 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.54 \[ \int \frac {1}{(a+b x)^{17/4} (c+d x)^{3/4}} \, dx=\frac {{\left (c+d\,x\right )}^{1/4}\,\left (\frac {512\,d^3\,x^3}{195\,{\left (a\,d-b\,c\right )}^4}+\frac {780\,a^3\,d^3-468\,a^2\,b\,c\,d^2+260\,a\,b^2\,c^2\,d-60\,b^3\,c^3}{195\,b^3\,{\left (a\,d-b\,c\right )}^4}+\frac {16\,d\,x\,\left (117\,a^2\,d^2-26\,a\,b\,c\,d+5\,b^2\,c^2\right )}{195\,b^2\,{\left (a\,d-b\,c\right )}^4}+\frac {128\,d^2\,x^2\,\left (13\,a\,d-b\,c\right )}{195\,b\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,{\left (a+b\,x\right )}^{1/4}+\frac {a^3\,{\left (a+b\,x\right )}^{1/4}}{b^3}+\frac {3\,a\,x^2\,{\left (a+b\,x\right )}^{1/4}}{b}+\frac {3\,a^2\,x\,{\left (a+b\,x\right )}^{1/4}}{b^2}} \]
((c + d*x)^(1/4)*((512*d^3*x^3)/(195*(a*d - b*c)^4) + (780*a^3*d^3 - 60*b^ 3*c^3 + 260*a*b^2*c^2*d - 468*a^2*b*c*d^2)/(195*b^3*(a*d - b*c)^4) + (16*d *x*(117*a^2*d^2 + 5*b^2*c^2 - 26*a*b*c*d))/(195*b^2*(a*d - b*c)^4) + (128* d^2*x^2*(13*a*d - b*c))/(195*b*(a*d - b*c)^4)))/(x^3*(a + b*x)^(1/4) + (a^ 3*(a + b*x)^(1/4))/b^3 + (3*a*x^2*(a + b*x)^(1/4))/b + (3*a^2*x*(a + b*x)^ (1/4))/b^2)