Integrand size = 21, antiderivative size = 45 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx=\frac {2}{9 b d (d \cos (a+b x))^{9/2}}-\frac {2}{5 b d^3 (d \cos (a+b x))^{5/2}} \] Output:
2/9/b/d/(d*cos(b*x+a))^(9/2)-2/5/b/d^3/(d*cos(b*x+a))^(5/2)
Leaf count is larger than twice the leaf count of optimal. \(94\) vs. \(2(45)=90\).
Time = 0.60 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.09 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx=\frac {2 \left (4 \sqrt [4]{\cos ^2(a+b x)}+\left (9-8 \sqrt [4]{\cos ^2(a+b x)}\right ) \csc ^2(a+b x)+4 \left (-1+\sqrt [4]{\cos ^2(a+b x)}\right ) \csc ^4(a+b x)\right ) \tan ^4(a+b x)}{45 b d^5 \sqrt {d \cos (a+b x)}} \] Input:
Integrate[Sin[a + b*x]^3/(d*Cos[a + b*x])^(11/2),x]
Output:
(2*(4*(Cos[a + b*x]^2)^(1/4) + (9 - 8*(Cos[a + b*x]^2)^(1/4))*Csc[a + b*x] ^2 + 4*(-1 + (Cos[a + b*x]^2)^(1/4))*Csc[a + b*x]^4)*Tan[a + b*x]^4)/(45*b *d^5*Sqrt[d*Cos[a + b*x]])
Time = 0.41 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3045, 27, 244, 2009}
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.
\(\displaystyle \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)^3}{(d \cos (a+b x))^{11/2}}dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {d^2-d^2 \cos ^2(a+b x)}{d^2 (d \cos (a+b x))^{11/2}}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {d^2-d^2 \cos ^2(a+b x)}{(d \cos (a+b x))^{11/2}}d(d \cos (a+b x))}{b d^3}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (\frac {d^2}{(d \cos (a+b x))^{11/2}}-\frac {1}{(d \cos (a+b x))^{7/2}}\right )d(d \cos (a+b x))}{b d^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {2}{5 (d \cos (a+b x))^{5/2}}-\frac {2 d^2}{9 (d \cos (a+b x))^{9/2}}}{b d^3}\) |
Input:
Int[Sin[a + b*x]^3/(d*Cos[a + b*x])^(11/2),x]
Output:
-(((-2*d^2)/(9*(d*Cos[a + b*x])^(9/2)) + 2/(5*(d*Cos[a + b*x])^(5/2)))/(b* d^3))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Time = 3.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {-\frac {2}{5 \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}+\frac {2 d^{2}}{9 \left (d \cos \left (b x +a \right )\right )^{\frac {9}{2}}}}{b \,d^{3}}\) | \(37\) |
default | \(\frac {-\frac {2}{5 \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}+\frac {2 d^{2}}{9 \left (d \cos \left (b x +a \right )\right )^{\frac {9}{2}}}}{b \,d^{3}}\) | \(37\) |
Input:
int(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x,method=_RETURNVERBOSE)
Output:
2/b/d^3*(-1/5/(d*cos(b*x+a))^(5/2)+1/9*d^2/(d*cos(b*x+a))^(9/2))
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx=-\frac {2 \, \sqrt {d \cos \left (b x + a\right )} {\left (9 \, \cos \left (b x + a\right )^{2} - 5\right )}}{45 \, b d^{6} \cos \left (b x + a\right )^{5}} \] Input:
integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x, algorithm="fricas")
Output:
-2/45*sqrt(d*cos(b*x + a))*(9*cos(b*x + a)^2 - 5)/(b*d^6*cos(b*x + a)^5)
Timed out. \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)**3/(d*cos(b*x+a))**(11/2),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx=-\frac {2 \, {\left (9 \, d^{2} \cos \left (b x + a\right )^{2} - 5 \, d^{2}\right )}}{45 \, \left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} b d^{3}} \] Input:
integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x, algorithm="maxima")
Output:
-2/45*(9*d^2*cos(b*x + a)^2 - 5*d^2)/((d*cos(b*x + a))^(9/2)*b*d^3)
\[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{3}}{\left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}}} \,d x } \] Input:
integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x, algorithm="giac")
Output:
integrate(sin(b*x + a)^3/(d*cos(b*x + a))^(11/2), x)
Time = 31.73 (sec) , antiderivative size = 279, normalized size of antiderivative = 6.20 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx=\frac {16\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}}{5\,b\,d^6\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}\,464{}\mathrm {i}}{45\,b\,d^6\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^3}-\frac {128\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}}{9\,b\,d^6\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^4}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}\,64{}\mathrm {i}}{9\,b\,d^6\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^5} \] Input:
int(sin(a + b*x)^3/(d*cos(a + b*x))^(11/2),x)
Output:
(16*exp(a*1i + b*x*1i)*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2)) ^(1/2))/(5*b*d^6*(exp(a*2i + b*x*2i)*1i + 1i)^2) - (exp(a*1i + b*x*1i)*(d* (exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2)*464i)/(45*b*d^6*(ex p(a*2i + b*x*2i)*1i + 1i)^3) - (128*exp(a*1i + b*x*1i)*(d*(exp(- a*1i - b* x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2))/(9*b*d^6*(exp(a*2i + b*x*2i)*1i + 1i)^4) + (exp(a*1i + b*x*1i)*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1 i)/2))^(1/2)*64i)/(9*b*d^6*(exp(a*2i + b*x*2i)*1i + 1i)^5)
\[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{3}}{\cos \left (b x +a \right )^{6}}d x \right )}{d^{6}} \] Input:
int(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x)
Output:
(sqrt(d)*int((sqrt(cos(a + b*x))*sin(a + b*x)**3)/cos(a + b*x)**6,x))/d**6