Integrand size = 23, antiderivative size = 65 \[ \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 \sin (e+f x)}{3 a^2 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:
1/3*sin(f*x+e)/a/f/(a+b*sin(f*x+e)^2)^(3/2)+2/3*sin(f*x+e)/a^2/f/(a+b*sin( f*x+e)^2)^(1/2)
Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72 \[ \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\sin (e+f x) \left (3 a+2 b \sin ^2(e+f x)\right )}{3 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \] Input:
Integrate[Cos[e + f*x]/(a + b*Sin[e + f*x]^2)^(5/2),x]
Output:
(Sin[e + f*x]*(3*a + 2*b*Sin[e + f*x]^2))/(3*a^2*f*(a + b*Sin[e + f*x]^2)^ (3/2))
Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3669, 209, 208}
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.
\(\displaystyle \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)}{\left (a+b \sin (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {1}{\left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin (e+f x)}{3 a}+\frac {\sin (e+f x)}{3 a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {2 \sin (e+f x)}{3 a^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sin (e+f x)}{3 a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{f}\) |
Input:
Int[Cos[e + f*x]/(a + b*Sin[e + f*x]^2)^(5/2),x]
Output:
(Sin[e + f*x]/(3*a*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*Sin[e + f*x])/(3*a^2 *Sqrt[a + b*Sin[e + f*x]^2]))/f
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (f x +e \right )}{3 a \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 a^{2} \sqrt {a +b \sin \left (f x +e \right )^{2}}}}{f}\) | \(56\) |
default | \(\frac {\frac {\sin \left (f x +e \right )}{3 a \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 a^{2} \sqrt {a +b \sin \left (f x +e \right )^{2}}}}{f}\) | \(56\) |
Input:
int(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/f*(1/3*sin(f*x+e)/a/(a+b*sin(f*x+e)^2)^(3/2)+2/3/a^2*sin(f*x+e)/(a+b*sin (f*x+e)^2)^(1/2))
Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.60 \[ \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (2 \, b \cos \left (f x + e\right )^{2} - 3 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{3 \, {\left (a^{2} b^{2} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}} \] Input:
integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")
Output:
-1/3*(2*b*cos(f*x + e)^2 - 3*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin( f*x + e)/(a^2*b^2*f*cos(f*x + e)^4 - 2*(a^3*b + a^2*b^2)*f*cos(f*x + e)^2 + (a^4 + 2*a^3*b + a^2*b^2)*f)
Timed out. \[ \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)/(a+b*sin(f*x+e)**2)**(5/2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {2 \, \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} + \frac {\sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a}}{3 \, f} \] Input:
integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")
Output:
1/3*(2*sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^2) + sin(f*x + e)/((b*si n(f*x + e)^2 + a)^(3/2)*a))/f
Leaf count of result is larger than twice the leaf count of optimal. 35375 vs. \(2 (57) = 114\).
Time = 50.60 (sec) , antiderivative size = 35375, normalized size of antiderivative = 544.23 \[ \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")
Output:
-2/3*(((((((3*a^8*b^4*tan(1/2*e)^57 + 12*a^7*b^5*tan(1/2*e)^57 + 18*a^6*b^ 6*tan(1/2*e)^57 + 12*a^5*b^7*tan(1/2*e)^57 + 3*a^4*b^8*tan(1/2*e)^57 + 84* a^8*b^4*tan(1/2*e)^55 + 356*a^7*b^5*tan(1/2*e)^55 + 584*a^6*b^6*tan(1/2*e) ^55 + 456*a^5*b^7*tan(1/2*e)^55 + 164*a^4*b^8*tan(1/2*e)^55 + 20*a^3*b^9*t an(1/2*e)^55 + 1134*a^8*b^4*tan(1/2*e)^53 + 5056*a^7*b^5*tan(1/2*e)^53 + 8 916*a^6*b^6*tan(1/2*e)^53 + 7784*a^5*b^7*tan(1/2*e)^53 + 3406*a^4*b^8*tan( 1/2*e)^53 + 648*a^3*b^9*tan(1/2*e)^53 + 32*a^2*b^10*tan(1/2*e)^53 + 9828*a ^8*b^4*tan(1/2*e)^51 + 45812*a^7*b^5*tan(1/2*e)^51 + 85736*a^6*b^6*tan(1/2 *e)^51 + 81384*a^5*b^7*tan(1/2*e)^51 + 40436*a^4*b^8*tan(1/2*e)^51 + 9572* a^3*b^9*tan(1/2*e)^51 + 768*a^2*b^10*tan(1/2*e)^51 + 61425*a^8*b^4*tan(1/2 *e)^49 + 297700*a^7*b^5*tan(1/2*e)^49 + 585382*a^6*b^6*tan(1/2*e)^49 + 593 028*a^5*b^7*tan(1/2*e)^49 + 322417*a^4*b^8*tan(1/2*e)^49 + 87328*a^3*b^9*t an(1/2*e)^49 + 8832*a^2*b^10*tan(1/2*e)^49 + 294840*a^8*b^4*tan(1/2*e)^47 + 1478360*a^7*b^5*tan(1/2*e)^47 + 3029808*a^6*b^6*tan(1/2*e)^47 + 3232432* a^5*b^7*tan(1/2*e)^47 + 1879448*a^4*b^8*tan(1/2*e)^47 + 558072*a^3*b^9*tan (1/2*e)^47 + 64768*a^2*b^10*tan(1/2*e)^47 + 1130220*a^8*b^4*tan(1/2*e)^45 + 5836480*a^7*b^5*tan(1/2*e)^45 + 12383752*a^6*b^6*tan(1/2*e)^45 + 1377460 8*a^5*b^7*tan(1/2*e)^45 + 8432812*a^4*b^8*tan(1/2*e)^45 + 2675728*a^3*b^9* tan(1/2*e)^45 + 340032*a^2*b^10*tan(1/2*e)^45 + 3552120*a^8*b^4*tan(1/2*e) ^43 + 18813080*a^7*b^5*tan(1/2*e)^43 + 41091248*a^6*b^6*tan(1/2*e)^43 +...
Time = 45.69 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.52 \[ \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {4\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )\,\sqrt {a+b\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,\left (b\,1{}\mathrm {i}-a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,6{}\mathrm {i}-b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,2{}\mathrm {i}+b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}\right )}{3\,a^2\,f\,{\left (b-4\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-2\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}^2} \] Input:
int(cos(e + f*x)/(a + b*sin(e + f*x)^2)^(5/2),x)
Output:
(4*exp(e*1i + f*x*1i)*(exp(e*2i + f*x*2i) - 1)*(a + b*((exp(- e*1i - f*x*1 i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*(b*1i - a*exp(e*2i + f*x*2i )*6i - b*exp(e*2i + f*x*2i)*2i + b*exp(e*4i + f*x*4i)*1i))/(3*a^2*f*(b - 4 *a*exp(e*2i + f*x*2i) - 2*b*exp(e*2i + f*x*2i) + b*exp(e*4i + f*x*4i))^2)
\[ \int \frac {\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )}{\sin \left (f x +e \right )^{6} b^{3}+3 \sin \left (f x +e \right )^{4} a \,b^{2}+3 \sin \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:
int(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x)
Output:
int((sqrt(sin(e + f*x)**2*b + a)*cos(e + f*x))/(sin(e + f*x)**6*b**3 + 3*s in(e + f*x)**4*a*b**2 + 3*sin(e + f*x)**2*a**2*b + a**3),x)