Integrand size = 26, antiderivative size = 87 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=a b x+\frac {b^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{3 d}-\frac {a b \cos (c+d x) \sin (c+d x)}{3 d}-\frac {(b+a \cos (c+d x))^2 \sin (c+d x)}{3 d} \]
a*b*x+b^2*arctanh(sin(d*x+c))/d+1/3*(a^2-2*b^2)*sin(d*x+c)/d-1/3*a*b*cos(d *x+c)*sin(d*x+c)/d-1/3*(b+a*cos(d*x+c))^2*sin(d*x+c)/d
Time = 0.52 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.34 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {12 a b c+12 a b d x-12 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \left (a^2-4 b^2\right ) \sin (c+d x)-6 a b \sin (2 (c+d x))-a^2 \sin (3 (c+d x))}{12 d} \]
(12*a*b*c + 12*a*b*d*x - 12*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 3*(a^2 - 4*b^2)*Sin[c + d*x] - 6*a*b*Sin[2*(c + d*x)] - a^2*Sin[3*(c + d*x)])/(12*d)
Time = 0.90 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 4897, 3042, 3368, 3042, 3529, 3042, 3512, 27, 3042, 3502, 27, 3042, 3214, 3042, 4257}
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.
| \(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \sin (c+d x) \tan (c+d x) (a \cos (c+d x)+b)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3368 |
| \(\displaystyle \int \left (1-\cos ^2(c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (1-\sin \left (c+d x+\frac {\pi }{2}\right )^2\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {1}{3} \int (b+a \cos (c+d x)) \left (-2 b \cos ^2(c+d x)+a \cos (c+d x)+3 b\right ) \sec (c+d x)dx-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-2 b \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \sin \left (c+d x+\frac {\pi }{2}\right )+3 b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int 2 \left (3 b^2+3 a \cos (c+d x) b+\left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \left (3 b^2+3 a \cos (c+d x) b+\left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {3 b^2+3 a \sin \left (c+d x+\frac {\pi }{2}\right ) b+\left (a^2-2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{3} \left (\int 3 \left (b^2+a \cos (c+d x) b\right ) \sec (c+d x)dx+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (3 \int \left (b^2+a \cos (c+d x) b\right ) \sec (c+d x)dx+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 \int \frac {b^2+a \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{3} \left (3 \left (b^2 \int \sec (c+d x)dx+a b x\right )+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 \left (b^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a b x\right )+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{d}+3 \left (a b x+\frac {b^2 \text {arctanh}(\sin (c+d x))}{d}\right )-\frac {a b \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {\sin (c+d x) (a \cos (c+d x)+b)^2}{3 d}\) |
-1/3*((b + a*Cos[c + d*x])^2*Sin[c + d*x])/d + (3*(a*b*x + (b^2*ArcTanh[Si n[c + d*x]])/d) + ((a^2 - 2*b^2)*Sin[c + d*x])/d - (a*b*Cos[c + d*x]*Sin[c + d*x])/d)/3
3.3.38.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n }, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] : > Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x ])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* (a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f , A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{2} \sin \left (d x +c \right )^{3}}{3}+2 a b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(72\) |
| default | \(\frac {\frac {a^{2} \sin \left (d x +c \right )^{3}}{3}+2 a b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(72\) |
| risch | \(x a b -\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}+\frac {b^{2} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {\sin \left (3 d x +3 c \right ) a^{2}}{12 d}-\frac {a b \sin \left (2 d x +2 c \right )}{2 d}\) | \(152\) |
1/d*(1/3*a^2*sin(d*x+c)^3+2*a*b*(-1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c) +b^2*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {6 \, a b d x + 3 \, b^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 3 \, a b \cos \left (d x + c\right ) - a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]
1/6*(6*a*b*d*x + 3*b^2*log(sin(d*x + c) + 1) - 3*b^2*log(-sin(d*x + c) + 1 ) - 2*(a^2*cos(d*x + c)^2 + 3*a*b*cos(d*x + c) - a^2 + 3*b^2)*sin(d*x + c) )/d
\[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \cos {\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {2 \, a^{2} \sin \left (d x + c\right )^{3} + 3 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a b + 3 \, b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{6 \, d} \]
1/6*(2*a^2*sin(d*x + c)^3 + 3*(2*d*x + 2*c - sin(2*d*x + 2*c))*a*b + 3*b^2 *(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c)))/d
Leaf count of result is larger than twice the leaf count of optimal. 4849 vs. \(2 (81) = 162\).
Time = 2.21 (sec) , antiderivative size = 4849, normalized size of antiderivative = 55.74 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\text {Too large to display} \]
-1/12*a^2*sin(3*d*x + 3*c)/d + 1/4*a^2*sin(d*x + c)/d + 1/2*(2*a*b*d*x*tan (d*x)^2*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(c)^2 - b^2*log(2*(tan(1/2*d*x)^2*t an(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/ 2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(d*x)^2*tan (1/2*d*x)^2*tan(1/2*c)^2*tan(c)^2 + b^2*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x )^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*ta n(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^2 *tan(1/2*c)^2*tan(c)^2 + 2*a*b*d*x*tan(d*x)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*b*d*x*tan(d*x)^2*tan(1/2*d*x)^2*tan(c)^2 + 2*a*b*d*x*tan(d*x)^2*tan( 1/2*c)^2*tan(c)^2 + 2*a*b*d*x*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(c)^2 - b^2*l og(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*tan(1/ 2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*t an(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^ 2 + 1))*tan(d*x)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + b^2*log(2*(tan(1/2*d*x)^2 *tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan( 1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(d*x)^2*t an(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*b*tan(d*x)^2*tan(1/2*d*x)^2*tan(1/2*c)...
Time = 22.84 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.39 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {a^2\,\sin \left (c+d\,x\right )}{4\,d}-\frac {b^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {2\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \]