Integrand size = 21, antiderivative size = 83 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {1}{16} (6 a+b) x+\frac {(6 a+b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {(6 a+b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {b \cos ^5(e+f x) \sin (e+f x)}{6 f} \]
1/16*(6*a+b)*x+1/16*(6*a+b)*cos(f*x+e)*sin(f*x+e)/f+1/24*(6*a+b)*cos(f*x+e )^3*sin(f*x+e)/f-1/6*b*cos(f*x+e)^5*sin(f*x+e)/f
Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {72 a e+72 a f x+12 b f x+3 (16 a+b) \sin (2 (e+f x))+(6 a-3 b) \sin (4 (e+f x))-b \sin (6 (e+f x))}{192 f} \]
(72*a*e + 72*a*f*x + 12*b*f*x + 3*(16*a + b)*Sin[2*(e + f*x)] + (6*a - 3*b )*Sin[4*(e + f*x)] - b*Sin[6*(e + f*x)])/(192*f)
Time = 0.24 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3670, 298, 215, 215, 216}
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.
\(\displaystyle \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^4 \left (a+b \sin (e+f x)^2\right )dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \frac {\int \frac {(a+b) \tan ^2(e+f x)+a}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {1}{6} (6 a+b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)-\frac {b \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {1}{6} (6 a+b) \left (\frac {3}{4} \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {b \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {1}{6} (6 a+b) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)+\frac {\tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {b \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {1}{6} (6 a+b) \left (\frac {3}{4} \left (\frac {1}{2} \arctan (\tan (e+f x))+\frac {\tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {b \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
(-1/6*(b*Tan[e + f*x])/(1 + Tan[e + f*x]^2)^3 + ((6*a + b)*(Tan[e + f*x]/( 4*(1 + Tan[e + f*x]^2)^2) + (3*(ArcTan[Tan[e + f*x]]/2 + Tan[e + f*x]/(2*( 1 + Tan[e + f*x]^2))))/4))/6)/f
3.3.86.3.1 Defintions of rubi rules used
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] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Time = 1.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75
method | result | size |
parallelrisch | \(\frac {\left (48 a +3 b \right ) \sin \left (2 f x +2 e \right )+\left (6 a -3 b \right ) \sin \left (4 f x +4 e \right )-\sin \left (6 f x +6 e \right ) b +72 f \left (a +\frac {b}{6}\right ) x}{192 f}\) | \(62\) |
risch | \(\frac {3 a x}{8}+\frac {b x}{16}-\frac {\sin \left (6 f x +6 e \right ) b}{192 f}+\frac {\sin \left (4 f x +4 e \right ) a}{32 f}-\frac {\sin \left (4 f x +4 e \right ) b}{64 f}+\frac {\sin \left (2 f x +2 e \right ) a}{4 f}+\frac {\sin \left (2 f x +2 e \right ) b}{64 f}\) | \(85\) |
derivativedivides | \(\frac {b \left (-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{6}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{24}+\frac {f x}{16}+\frac {e}{16}\right )+a \left (\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\) | \(92\) |
default | \(\frac {b \left (-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{6}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{24}+\frac {f x}{16}+\frac {e}{16}\right )+a \left (\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\) | \(92\) |
norman | \(\frac {\left (\frac {3 a}{8}+\frac {b}{16}\right ) x +\left (\frac {3 a}{8}+\frac {b}{16}\right ) x \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {9 a}{4}+\frac {3 b}{8}\right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {9 a}{4}+\frac {3 b}{8}\right ) x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {15 a}{2}+\frac {5 b}{4}\right ) x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {45 a}{8}+\frac {15 b}{16}\right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {45 a}{8}+\frac {15 b}{16}\right ) x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {\left (2 a -13 b \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}-\frac {\left (2 a -13 b \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}+\frac {\left (10 a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{8 f}-\frac {\left (10 a -b \right ) \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}+\frac {\left (42 a +47 b \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{24 f}-\frac {\left (42 a +47 b \right ) \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{24 f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{6}}\) | \(283\) |
1/192*((48*a+3*b)*sin(2*f*x+2*e)+(6*a-3*b)*sin(4*f*x+4*e)-sin(6*f*x+6*e)*b +72*f*(a+1/6*b)*x)/f
Time = 0.31 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.76 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {3 \, {\left (6 \, a + b\right )} f x - {\left (8 \, b \cos \left (f x + e\right )^{5} - 2 \, {\left (6 \, a + b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (6 \, a + b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f} \]
1/48*(3*(6*a + b)*f*x - (8*b*cos(f*x + e)^5 - 2*(6*a + b)*cos(f*x + e)^3 - 3*(6*a + b)*cos(f*x + e))*sin(f*x + e))/f
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (76) = 152\).
Time = 0.40 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.01 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\begin {cases} \frac {3 a x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 a x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 a x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {3 a \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {5 a \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac {b x \sin ^{6}{\left (e + f x \right )}}{16} + \frac {3 b x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{16} + \frac {3 b x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{16} + \frac {b x \cos ^{6}{\left (e + f x \right )}}{16} + \frac {b \sin ^{5}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{16 f} + \frac {b \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{6 f} - \frac {b \sin {\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{16 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin ^{2}{\left (e \right )}\right ) \cos ^{4}{\left (e \right )} & \text {otherwise} \end {cases} \]
Piecewise((3*a*x*sin(e + f*x)**4/8 + 3*a*x*sin(e + f*x)**2*cos(e + f*x)**2 /4 + 3*a*x*cos(e + f*x)**4/8 + 3*a*sin(e + f*x)**3*cos(e + f*x)/(8*f) + 5* a*sin(e + f*x)*cos(e + f*x)**3/(8*f) + b*x*sin(e + f*x)**6/16 + 3*b*x*sin( e + f*x)**4*cos(e + f*x)**2/16 + 3*b*x*sin(e + f*x)**2*cos(e + f*x)**4/16 + b*x*cos(e + f*x)**6/16 + b*sin(e + f*x)**5*cos(e + f*x)/(16*f) + b*sin(e + f*x)**3*cos(e + f*x)**3/(6*f) - b*sin(e + f*x)*cos(e + f*x)**5/(16*f), Ne(f, 0)), (x*(a + b*sin(e)**2)*cos(e)**4, True))
Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {3 \, {\left (f x + e\right )} {\left (6 \, a + b\right )} + \frac {3 \, {\left (6 \, a + b\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (6 \, a + b\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (10 \, a - b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \]
1/48*(3*(f*x + e)*(6*a + b) + (3*(6*a + b)*tan(f*x + e)^5 + 8*(6*a + b)*ta n(f*x + e)^3 + 3*(10*a - b)*tan(f*x + e))/(tan(f*x + e)^6 + 3*tan(f*x + e) ^4 + 3*tan(f*x + e)^2 + 1))/f
Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {1}{16} \, {\left (6 \, a + b\right )} x - \frac {b \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac {{\left (2 \, a - b\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac {{\left (16 \, a + b\right )} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \]
1/16*(6*a + b)*x - 1/192*b*sin(6*f*x + 6*e)/f + 1/64*(2*a - b)*sin(4*f*x + 4*e)/f + 1/64*(16*a + b)*sin(2*f*x + 2*e)/f
Time = 14.56 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=x\,\left (\frac {3\,a}{8}+\frac {b}{16}\right )+\frac {\left (\frac {3\,a}{8}+\frac {b}{16}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (a+\frac {b}{6}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {5\,a}{8}-\frac {b}{16}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+3\,{\mathrm {tan}\left (e+f\,x\right )}^4+3\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \]