Integrand size = 21, antiderivative size = 102 \[ \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {5}{16} (a-7 b) x-\frac {(11 a-29 b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {(13 a-19 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {(a-b) \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {b \tan (e+f x)}{f} \]
5/16*(a-7*b)*x-1/16*(11*a-29*b)*cos(f*x+e)*sin(f*x+e)/f+1/24*(13*a-19*b)*c os(f*x+e)^3*sin(f*x+e)/f-1/6*(a-b)*cos(f*x+e)^5*sin(f*x+e)/f+b*tan(f*x+e)/ f
Time = 0.69 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {60 a e-420 b e+60 a f x-420 b f x+(-45 a+141 b) \sin (2 (e+f x))+3 (3 a-5 b) \sin (4 (e+f x))-a \sin (6 (e+f x))+b \sin (6 (e+f x))+192 b \tan (e+f x)}{192 f} \]
(60*a*e - 420*b*e + 60*a*f*x - 420*b*f*x + (-45*a + 141*b)*Sin[2*(e + f*x) ] + 3*(3*a - 5*b)*Sin[4*(e + f*x)] - a*Sin[6*(e + f*x)] + b*Sin[6*(e + f*x )] + 192*b*Tan[e + f*x])/(192*f)
Time = 0.39 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4146, 360, 25, 2345, 27, 1471, 299, 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.
| \(\displaystyle \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^6 \left (a+b \tan (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \left (b \tan ^2(e+f x)+a\right )}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle \frac {-\frac {1}{6} \int -\frac {6 b \tan ^6(e+f x)+6 (a-b) \tan ^4(e+f x)-6 (a-b) \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)-\frac {(a-b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {6 b \tan ^6(e+f x)+6 (a-b) \tan ^4(e+f x)-6 (a-b) \tan ^2(e+f x)+a-b}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)-\frac {(a-b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {(13 a-19 b) \tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}-\frac {1}{4} \int \frac {3 \left (-8 b \tan ^4(e+f x)-8 (a-2 b) \tan ^2(e+f x)+3 a-5 b\right )}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)\right )-\frac {(a-b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {(13 a-19 b) \tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}-\frac {3}{4} \int \frac {-8 b \tan ^4(e+f x)-8 (a-2 b) \tan ^2(e+f x)+3 a-5 b}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)\right )-\frac {(a-b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 1471 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {(13 a-19 b) \tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}-\frac {3}{4} \left (\frac {(11 a-29 b) \tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}-\frac {1}{2} \int \frac {16 b \tan ^2(e+f x)+5 a-19 b}{\tan ^2(e+f x)+1}d\tan (e+f x)\right )\right )-\frac {(a-b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {(13 a-19 b) \tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}-\frac {3}{4} \left (\frac {1}{2} \left (-5 (a-7 b) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)-16 b \tan (e+f x)\right )+\frac {(11 a-29 b) \tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )\right )-\frac {(a-b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {(13 a-19 b) \tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}-\frac {3}{4} \left (\frac {1}{2} (-5 (a-7 b) \arctan (\tan (e+f x))-16 b \tan (e+f x))+\frac {(11 a-29 b) \tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )\right )-\frac {(a-b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
(-1/6*((a - b)*Tan[e + f*x])/(1 + Tan[e + f*x]^2)^3 + (((13*a - 19*b)*Tan[ e + f*x])/(4*(1 + Tan[e + f*x]^2)^2) - (3*((-5*(a - 7*b)*ArcTan[Tan[e + f* x]] - 16*b*Tan[e + f*x])/2 + ((11*a - 29*b)*Tan[e + f*x])/(2*(1 + Tan[e + f*x]^2))))/4)/6)/f
3.1.36.3.1 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_.)*(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[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[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Time = 3.97 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.20
| method | result | size |
| derivativedivides | \(\frac {a \left (-\frac {\left (\sin \left (f x +e \right )^{5}+\frac {5 \sin \left (f x +e \right )^{3}}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+b \left (\frac {\sin \left (f x +e \right )^{9}}{\cos \left (f x +e \right )}+\left (\sin \left (f x +e \right )^{7}+\frac {7 \sin \left (f x +e \right )^{5}}{6}+\frac {35 \sin \left (f x +e \right )^{3}}{24}+\frac {35 \sin \left (f x +e \right )}{16}\right ) \cos \left (f x +e \right )-\frac {35 f x}{16}-\frac {35 e}{16}\right )}{f}\) | \(122\) |
| default | \(\frac {a \left (-\frac {\left (\sin \left (f x +e \right )^{5}+\frac {5 \sin \left (f x +e \right )^{3}}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+b \left (\frac {\sin \left (f x +e \right )^{9}}{\cos \left (f x +e \right )}+\left (\sin \left (f x +e \right )^{7}+\frac {7 \sin \left (f x +e \right )^{5}}{6}+\frac {35 \sin \left (f x +e \right )^{3}}{24}+\frac {35 \sin \left (f x +e \right )}{16}\right ) \cos \left (f x +e \right )-\frac {35 f x}{16}-\frac {35 e}{16}\right )}{f}\) | \(122\) |
| risch | \(\frac {5 a x}{16}-\frac {35 b x}{16}+\frac {15 i {\mathrm e}^{2 i \left (f x +e \right )} a}{128 f}-\frac {47 i {\mathrm e}^{2 i \left (f x +e \right )} b}{128 f}-\frac {15 i {\mathrm e}^{-2 i \left (f x +e \right )} a}{128 f}+\frac {47 i {\mathrm e}^{-2 i \left (f x +e \right )} b}{128 f}+\frac {2 i b}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {\sin \left (6 f x +6 e \right ) a}{192 f}+\frac {\sin \left (6 f x +6 e \right ) b}{192 f}+\frac {3 \sin \left (4 f x +4 e \right ) a}{64 f}-\frac {5 \sin \left (4 f x +4 e \right ) b}{64 f}\) | \(154\) |
1/f*(a*(-1/6*(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/ 16*f*x+5/16*e)+b*(sin(f*x+e)^9/cos(f*x+e)+(sin(f*x+e)^7+7/6*sin(f*x+e)^5+3 5/24*sin(f*x+e)^3+35/16*sin(f*x+e))*cos(f*x+e)-35/16*f*x-35/16*e))
Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88 \[ \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {15 \, {\left (a - 7 \, b\right )} f x \cos \left (f x + e\right ) - {\left (8 \, {\left (a - b\right )} \cos \left (f x + e\right )^{6} - 2 \, {\left (13 \, a - 19 \, b\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (11 \, a - 29 \, b\right )} \cos \left (f x + e\right )^{2} - 48 \, b\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )} \]
1/48*(15*(a - 7*b)*f*x*cos(f*x + e) - (8*(a - b)*cos(f*x + e)^6 - 2*(13*a - 19*b)*cos(f*x + e)^4 + 3*(11*a - 29*b)*cos(f*x + e)^2 - 48*b)*sin(f*x + e))/(f*cos(f*x + e))
\[ \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\int \left (a + b \tan ^{2}{\left (e + f x \right )}\right ) \sin ^{6}{\left (e + f x \right )}\, dx \]
Time = 0.35 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09 \[ \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {15 \, {\left (f x + e\right )} {\left (a - 7 \, b\right )} + 48 \, b \tan \left (f x + e\right ) - \frac {3 \, {\left (11 \, a - 29 \, b\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a - 17 \, b\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (5 \, a - 19 \, b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \]
1/48*(15*(f*x + e)*(a - 7*b) + 48*b*tan(f*x + e) - (3*(11*a - 29*b)*tan(f* x + e)^5 + 8*(5*a - 17*b)*tan(f*x + e)^3 + 3*(5*a - 19*b)*tan(f*x + e))/(t an(f*x + e)^6 + 3*tan(f*x + e)^4 + 3*tan(f*x + e)^2 + 1))/f
Leaf count of result is larger than twice the leaf count of optimal. 7296 vs. \(2 (94) = 188\).
Time = 2.56 (sec) , antiderivative size = 7296, normalized size of antiderivative = 71.53 \[ \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]
1/192*(21*pi*b*sgn(2*tan(f*x)^2*tan(e)^2 - 2)*sgn(-2*tan(f*x)^2*tan(e) + 2 *tan(f*x)*tan(e)^2 + 2*tan(f*x) - 2*tan(e))*tan(f*x)^7*tan(e)^7 + 60*a*f*x *tan(f*x)^7*tan(e)^7 - 420*b*f*x*tan(f*x)^7*tan(e)^7 + 21*pi*b*sgn(-2*tan( f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 + 2*tan(f*x) - 2*tan(e))*tan(f*x)^7*ta n(e)^7 + 63*pi*b*sgn(2*tan(f*x)^2*tan(e)^2 - 2)*sgn(-2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 + 2*tan(f*x) - 2*tan(e))*tan(f*x)^7*tan(e)^5 - 21*pi* b*sgn(2*tan(f*x)^2*tan(e)^2 - 2)*sgn(-2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan (e)^2 + 2*tan(f*x) - 2*tan(e))*tan(f*x)^6*tan(e)^6 + 63*pi*b*sgn(2*tan(f*x )^2*tan(e)^2 - 2)*sgn(-2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 + 2*tan(f *x) - 2*tan(e))*tan(f*x)^5*tan(e)^7 + 42*b*arctan((tan(f*x) + tan(e))/(tan (f*x)*tan(e) - 1))*tan(f*x)^7*tan(e)^7 - 42*b*arctan(-(tan(f*x) - tan(e))/ (tan(f*x)*tan(e) + 1))*tan(f*x)^7*tan(e)^7 + 180*a*f*x*tan(f*x)^7*tan(e)^5 - 1260*b*f*x*tan(f*x)^7*tan(e)^5 + 63*pi*b*sgn(-2*tan(f*x)^2*tan(e) + 2*t an(f*x)*tan(e)^2 + 2*tan(f*x) - 2*tan(e))*tan(f*x)^7*tan(e)^5 - 60*a*f*x*t an(f*x)^6*tan(e)^6 + 420*b*f*x*tan(f*x)^6*tan(e)^6 - 21*pi*b*sgn(-2*tan(f* x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 + 2*tan(f*x) - 2*tan(e))*tan(f*x)^6*tan( e)^6 + 180*a*f*x*tan(f*x)^5*tan(e)^7 - 1260*b*f*x*tan(f*x)^5*tan(e)^7 + 63 *pi*b*sgn(-2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 + 2*tan(f*x) - 2*tan( e))*tan(f*x)^5*tan(e)^7 + 63*pi*b*sgn(2*tan(f*x)^2*tan(e)^2 - 2)*sgn(-2*ta n(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 + 2*tan(f*x) - 2*tan(e))*tan(f*x)...
Time = 10.80 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03 \[ \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=x\,\left (\frac {5\,a}{16}-\frac {35\,b}{16}\right )-\frac {\left (\frac {11\,a}{16}-\frac {29\,b}{16}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {5\,a}{6}-\frac {17\,b}{6}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {5\,a}{16}-\frac {19\,b}{16}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+3\,{\mathrm {tan}\left (e+f\,x\right )}^4+3\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}+\frac {b\,\mathrm {tan}\left (e+f\,x\right )}{f} \]