Integrand size = 25, antiderivative size = 106 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {6 c (c \sin (a+b x))^{3/2}}{77 b d^3 (d \cos (a+b x))^{7/2}}-\frac {8 c (c \sin (a+b x))^{3/2}}{77 b d^5 (d \cos (a+b x))^{3/2}} \]
2/11*c*(c*sin(b*x+a))^(3/2)/b/d/(d*cos(b*x+a))^(11/2)-6/77*c*(c*sin(b*x+a) )^(3/2)/b/d^3/(d*cos(b*x+a))^(7/2)-8/77*c*(c*sin(b*x+a))^(3/2)/b/d^5/(d*co s(b*x+a))^(3/2)
Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.54 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {2 c^4 (9+2 \cos (2 (a+b x))) \tan ^5(a+b x)}{77 b d^6 \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}} \]
(2*c^4*(9 + 2*Cos[2*(a + b*x)])*Tan[a + b*x]^5)/(77*b*d^6*Sqrt[d*Cos[a + b *x]]*(c*Sin[a + b*x])^(3/2))
Time = 0.46 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3046, 3042, 3051, 3042, 3043}
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.
\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}}dx\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {3 c^2 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}}dx}{11 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {3 c^2 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}}dx}{11 d^2}\) |
\(\Big \downarrow \) 3051 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {3 c^2 \left (\frac {4 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}}dx}{7 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {3 c^2 \left (\frac {4 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}}dx}{7 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}\) |
\(\Big \downarrow \) 3043 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {3 c^2 \left (\frac {8 (c \sin (a+b x))^{3/2}}{21 b c d^3 (d \cos (a+b x))^{3/2}}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}\) |
(2*c*(c*Sin[a + b*x])^(3/2))/(11*b*d*(d*Cos[a + b*x])^(11/2)) - (3*c^2*((2 *(c*Sin[a + b*x])^(3/2))/(7*b*c*d*(d*Cos[a + b*x])^(7/2)) + (8*(c*Sin[a + b*x])^(3/2))/(21*b*c*d^3*(d*Cos[a + b*x])^(3/2))))/(11*d^2)
3.3.85.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^( m_.), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Cos[e + f*x])^(n + 1)/ (a*b*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2, 0] & & NeQ[m, -1]
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(a*Sin[e + f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Sin[e + f *x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-(b*Sin[e + f*x])^(n + 1))*((a*Cos[e + f*x])^(m + 1) /(a*b*f*(m + 1))), x] + Simp[(m + n + 2)/(a^2*(m + 1)) Int[(b*Sin[e + f*x ])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m , -1] && IntegersQ[2*m, 2*n]
Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.58
method | result | size |
default | \(\frac {2 c^{2} \sqrt {c \sin \left (b x +a \right )}\, \left (4 \left (\tan ^{3}\left (b x +a \right )\right )+7 \left (\tan ^{3}\left (b x +a \right )\right ) \left (\sec ^{2}\left (b x +a \right )\right )\right )}{77 b \,d^{6} \sqrt {d \cos \left (b x +a \right )}}\) | \(61\) |
2/77/b*c^2*(c*sin(b*x+a))^(1/2)/d^6/(d*cos(b*x+a))^(1/2)*(4*tan(b*x+a)^3+7 *tan(b*x+a)^3*sec(b*x+a)^2)
Time = 0.43 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=-\frac {2 \, {\left (4 \, c^{2} \cos \left (b x + a\right )^{4} + 3 \, c^{2} \cos \left (b x + a\right )^{2} - 7 \, c^{2}\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{77 \, b d^{7} \cos \left (b x + a\right )^{6}} \]
-2/77*(4*c^2*cos(b*x + a)^4 + 3*c^2*cos(b*x + a)^2 - 7*c^2)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*sin(b*x + a)/(b*d^7*cos(b*x + a)^6)
Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {13}{2}}} \,d x } \]
\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {13}{2}}} \,d x } \]
Time = 6.54 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.66 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=-\frac {{\mathrm {e}}^{-a\,5{}\mathrm {i}-b\,x\,5{}\mathrm {i}}\,\sqrt {c\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,\left (\frac {96\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (3\,a+3\,b\,x\right )}{77\,b\,d^6}+\frac {16\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (5\,a+5\,b\,x\right )}{77\,b\,d^6}-\frac {368\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (a+b\,x\right )}{77\,b\,d^6}\right )}{32\,{\cos \left (a+b\,x\right )}^5\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}} \]
-(exp(- a*5i - b*x*5i)*(c*((exp(- a*1i - b*x*1i)*1i)/2 - (exp(a*1i + b*x*1 i)*1i)/2))^(1/2)*((96*c^2*exp(a*5i + b*x*5i)*sin(3*a + 3*b*x))/(77*b*d^6) + (16*c^2*exp(a*5i + b*x*5i)*sin(5*a + 5*b*x))/(77*b*d^6) - (368*c^2*exp(a *5i + b*x*5i)*sin(a + b*x))/(77*b*d^6)))/(32*cos(a + b*x)^5*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2))