Integrand size = 25, antiderivative size = 106 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\frac {2 c \sqrt {c \sin (a+b x)}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {2 c \sqrt {c \sin (a+b x)}}{45 b d^3 (d \cos (a+b x))^{5/2}}-\frac {8 c \sqrt {c \sin (a+b x)}}{45 b d^5 \sqrt {d \cos (a+b x)}} \]
2/9*c*(c*sin(b*x+a))^(1/2)/b/d/(d*cos(b*x+a))^(9/2)-2/45*c*(c*sin(b*x+a))^ (1/2)/b/d^3/(d*cos(b*x+a))^(5/2)-8/45*c*(c*sin(b*x+a))^(1/2)/b/d^5/(d*cos( b*x+a))^(1/2)
Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.54 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\frac {2 \sqrt {d \cos (a+b x)} (7+2 \cos (2 (a+b x))) \sec ^5(a+b x) (c \sin (a+b x))^{5/2}}{45 b c d^6} \]
(2*Sqrt[d*Cos[a + b*x]]*(7 + 2*Cos[2*(a + b*x)])*Sec[a + b*x]^5*(c*Sin[a + b*x])^(5/2))/(45*b*c*d^6)
Time = 0.49 (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.
\(\displaystyle \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}}dx\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {c^2 \int \frac {1}{(d \cos (a+b x))^{7/2} \sqrt {c \sin (a+b x)}}dx}{9 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {c^2 \int \frac {1}{(d \cos (a+b x))^{7/2} \sqrt {c \sin (a+b x)}}dx}{9 d^2}\) |
\(\Big \downarrow \) 3051 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {c^2 \left (\frac {4 \int \frac {1}{(d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}dx}{5 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{5 b c d (d \cos (a+b x))^{5/2}}\right )}{9 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {c^2 \left (\frac {4 \int \frac {1}{(d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}dx}{5 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{5 b c d (d \cos (a+b x))^{5/2}}\right )}{9 d^2}\) |
\(\Big \downarrow \) 3043 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {c^2 \left (\frac {8 \sqrt {c \sin (a+b x)}}{5 b c d^3 \sqrt {d \cos (a+b x)}}+\frac {2 \sqrt {c \sin (a+b x)}}{5 b c d (d \cos (a+b x))^{5/2}}\right )}{9 d^2}\) |
(2*c*Sqrt[c*Sin[a + b*x]])/(9*b*d*(d*Cos[a + b*x])^(9/2)) - (c^2*((2*Sqrt[ c*Sin[a + b*x]])/(5*b*c*d*(d*Cos[a + b*x])^(5/2)) + (8*Sqrt[c*Sin[a + b*x] ])/(5*b*c*d^3*Sqrt[d*Cos[a + b*x]])))/(9*d^2)
3.3.74.3.1 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 = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.56
method | result | size |
default | \(\frac {2 c \sqrt {c \sin \left (b x +a \right )}\, \left (4 \left (\tan ^{2}\left (b x +a \right )\right )+5 \left (\tan ^{2}\left (b x +a \right )\right ) \left (\sec ^{2}\left (b x +a \right )\right )\right )}{45 b \,d^{5} \sqrt {d \cos \left (b x +a \right )}}\) | \(59\) |
2/45/b*c*(c*sin(b*x+a))^(1/2)/d^5/(d*cos(b*x+a))^(1/2)*(4*tan(b*x+a)^2+5*t an(b*x+a)^2*sec(b*x+a)^2)
Time = 0.38 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.58 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=-\frac {2 \, {\left (4 \, c \cos \left (b x + a\right )^{4} + c \cos \left (b x + a\right )^{2} - 5 \, c\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )}}{45 \, b d^{6} \cos \left (b x + a\right )^{5}} \]
-2/45*(4*c*cos(b*x + a)^4 + c*cos(b*x + a)^2 - 5*c)*sqrt(d*cos(b*x + a))*s qrt(c*sin(b*x + a))/(b*d^6*cos(b*x + a)^5)
Timed out. \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}}} \,d x } \]
\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}}} \,d x } \]
Time = 6.18 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.95 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=-\frac {\sqrt {c\,\sin \left (a+b\,x\right )}\,\left (2\,{\sin \left (2\,a+2\,b\,x\right )}^2+\sin \left (4\,a+4\,b\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {32\,c\,\left (-2\,{\sin \left (2\,a+2\,b\,x\right )}^2+\sin \left (4\,a+4\,b\,x\right )\,1{}\mathrm {i}+1\right )}{15\,b\,d^5}+\frac {16\,c\,\left (2\,{\sin \left (2\,a+2\,b\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,a+2\,b\,x\right )}^2+\sin \left (4\,a+4\,b\,x\right )\,1{}\mathrm {i}+1\right )}{45\,b\,d^5}+\frac {16\,c\,\left (2\,{\sin \left (a+b\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,a+2\,b\,x\right )}^2+\sin \left (4\,a+4\,b\,x\right )\,1{}\mathrm {i}+1\right )}{9\,b\,d^5}\right )}{16\,{\left ({\sin \left (a+b\,x\right )}^2-1\right )}^2\,\sqrt {-d\,\left (2\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )}} \]
-((c*sin(a + b*x))^(1/2)*(sin(4*a + 4*b*x)*1i + 2*sin(2*a + 2*b*x)^2 - 1)* ((32*c*(sin(4*a + 4*b*x)*1i - 2*sin(2*a + 2*b*x)^2 + 1))/(15*b*d^5) + (16* c*(2*sin(2*a + 2*b*x)^2 - 1)*(sin(4*a + 4*b*x)*1i - 2*sin(2*a + 2*b*x)^2 + 1))/(45*b*d^5) + (16*c*(2*sin(a + b*x)^2 - 1)*(sin(4*a + 4*b*x)*1i - 2*si n(2*a + 2*b*x)^2 + 1))/(9*b*d^5)))/(16*(sin(a + b*x)^2 - 1)^2*(-d*(2*sin(a /2 + (b*x)/2)^2 - 1))^(1/2))