Integrand size = 23, antiderivative size = 38 \[ \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{\sqrt {b} f} \] Output:
arctanh(b^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2))/b^(1/2)/f
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{\sqrt {b} f} \] Input:
Integrate[Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]^2],x]
Output:
ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]/(Sqrt[b]*f)
Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3669, 224, 219}
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.
\(\displaystyle \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)}{\sqrt {a+b \sin (e+f x)^2}}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\int \frac {1}{1-\frac {b \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{\sqrt {b} f}\) |
Input:
Int[Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]^2],x]
Output:
ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]/(Sqrt[b]*f)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 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 = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{f \sqrt {b}}\) | \(34\) |
default | \(\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{f \sqrt {b}}\) | \(34\) |
Input:
int(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/f*ln(b^(1/2)*sin(f*x+e)+(a+b*sin(f*x+e)^2)^(1/2))/b^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (32) = 64\).
Time = 0.14 (sec) , antiderivative size = 394, normalized size of antiderivative = 10.37 \[ \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {\log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right )}{8 \, \sqrt {b} f}, -\frac {\sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right )}{4 \, b f}\right ] \] Input:
integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
Output:
[1/8*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32* (5*a^2*b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160*a^2* b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b^2 + 24*a*b^3 + 16*b^4)*co s(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*x + e)^ 4 - a^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 24*b^3)*c os(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)*sin(f*x + e))/(sqrt (b)*f), -1/4*sqrt(-b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2)*c os(f*x + e)^2 + a^2 + 8*a*b + 8*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt( -b)/((2*b^3*cos(f*x + e)^4 + a^2*b + 3*a*b^2 + 2*b^3 - (3*a*b^2 + 4*b^3)*c os(f*x + e)^2)*sin(f*x + e)))/(b*f)]
\[ \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {\cos {\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \] Input:
integrate(cos(f*x+e)/(a+b*sin(f*x+e)**2)**(1/2),x)
Output:
Integral(cos(e + f*x)/sqrt(a + b*sin(e + f*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.55 \[ \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {b} f} \] Input:
integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
Output:
arcsinh(b*sin(f*x + e)/sqrt(a*b))/(sqrt(b)*f)
\[ \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \] Input:
integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
Output:
integrate(cos(f*x + e)/sqrt(b*sin(f*x + e)^2 + a), x)
Time = 35.88 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\ln \left (\sqrt {b}\,\sin \left (e+f\,x\right )+\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}\right )}{\sqrt {b}\,f} \] Input:
int(cos(e + f*x)/(a + b*sin(e + f*x)^2)^(1/2),x)
Output:
log(b^(1/2)*sin(e + f*x) + (a + b*sin(e + f*x)^2)^(1/2))/(b^(1/2)*f)
\[ \int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )}{\sin \left (f x +e \right )^{2} b +a}d x \] Input:
int(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x)
Output:
int((sqrt(sin(e + f*x)**2*b + a)*cos(e + f*x))/(sin(e + f*x)**2*b + a),x)