Integrand size = 25, antiderivative size = 117 \[ \int \cos ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {a (a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 b^{3/2} f}+\frac {(a+4 b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 b f}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 b f} \]
1/8*a*(a+4*b)*arctanh(sin(f*x+e)*b^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/b^(3/2) /f-1/4*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2)/b/f+1/8*(a+4*b)*sin(f*x+e)*(a+b *sin(f*x+e)^2)^(1/2)/b/f
Time = 0.47 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07 \[ \int \cos ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\sqrt {a+b \sin ^2(e+f x)} \left (\sqrt {a} (a+4 b) \text {arcsinh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a}}\right )-\sqrt {b} \sin (e+f x) \left (a-4 b+2 b \sin ^2(e+f x)\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right )}{8 b^{3/2} f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}} \]
(Sqrt[a + b*Sin[e + f*x]^2]*(Sqrt[a]*(a + 4*b)*ArcSinh[(Sqrt[b]*Sin[e + f* x])/Sqrt[a]] - Sqrt[b]*Sin[e + f*x]*(a - 4*b + 2*b*Sin[e + f*x]^2)*Sqrt[1 + (b*Sin[e + f*x]^2)/a]))/(8*b^(3/2)*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a])
Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3669, 299, 211, 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.
| \(\displaystyle \int \cos ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (e+f x)^3 \sqrt {a+b \sin (e+f x)^2}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {(a+4 b) \int \sqrt {b \sin ^2(e+f x)+a}d\sin (e+f x)}{4 b}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 b}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {(a+4 b) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+\frac {1}{2} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right )}{4 b}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {(a+4 b) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}+\frac {1}{2} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right )}{4 b}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {(a+4 b) \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 \sqrt {b}}+\frac {1}{2} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right )}{4 b}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 b}}{f}\) |
(-1/4*(Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2))/b + ((a + 4*b)*((a*ArcTa nh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(2*Sqrt[b]) + (Sin[ e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/2))/(4*b))/f
3.4.24.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 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 = 0.58 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.25
| method | result | size |
| derivativedivides | \(\frac {\frac {\sin \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2}+\frac {a \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{2 \sqrt {b}}-\frac {\sin \left (f x +e \right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}}}{4 b}+\frac {a \left (\frac {\sin \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2}+\frac {a \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{2 \sqrt {b}}\right )}{4 b}}{f}\) | \(146\) |
| default | \(\frac {\frac {\sin \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2}+\frac {a \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{2 \sqrt {b}}-\frac {\sin \left (f x +e \right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}}}{4 b}+\frac {a \left (\frac {\sin \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2}+\frac {a \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{2 \sqrt {b}}\right )}{4 b}}{f}\) | \(146\) |
1/f*(1/2*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*sin( f*x+e)+(a+b*sin(f*x+e)^2)^(1/2))-1/4*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2)/b +1/4*a/b*(1/2*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2) *sin(f*x+e)+(a+b*sin(f*x+e)^2)^(1/2))))
Time = 0.59 (sec) , antiderivative size = 511, normalized size of antiderivative = 4.37 \[ \int \cos ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [\frac {{\left (a^{2} + 4 \, a b\right )} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) + 8 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{2} - a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{64 \, b^{2} f}, -\frac {{\left (a^{2} + 4 \, a b\right )} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) - 4 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{2} - a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{32 \, b^{2} f}\right ] \]
[1/64*((a^2 + 4*a*b)*sqrt(b)*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)*cos(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)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqr t(b)*sin(f*x + e)) + 8*(2*b^2*cos(f*x + e)^2 - a*b + 2*b^2)*sqrt(-b*cos(f* x + e)^2 + a + b)*sin(f*x + e))/(b^2*f), -1/32*((a^2 + 4*a*b)*sqrt(-b)*arc tan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2)*cos(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)*cos(f*x + e)^2)*sin(f*x + e))) - 4*(2*b^2*cos(f*x + e)^2 - a*b + 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/(b^2*f)]
Timed out. \[ \int \cos ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.02 \[ \int \cos ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\frac {a^{2} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {4 \, a \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {b}} + 4 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right ) - \frac {2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )}{b} + \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )}{b}}{8 \, f} \]
1/8*(a^2*arcsinh(b*sin(f*x + e)/sqrt(a*b))/b^(3/2) + 4*a*arcsinh(b*sin(f*x + e)/sqrt(a*b))/sqrt(b) + 4*sqrt(b*sin(f*x + e)^2 + a)*sin(f*x + e) - 2*( b*sin(f*x + e)^2 + a)^(3/2)*sin(f*x + e)/b + sqrt(b*sin(f*x + e)^2 + a)*a* sin(f*x + e)/b)/f
\[ \int \cos ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \cos ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int {\cos \left (e+f\,x\right )}^3\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]