Integrand size = 27, antiderivative size = 177 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \]
-3/16*(a+b)*(a^2+7*a*b+8*b^2)*ln(1-sin(d*x+c))/d+3/16*(a-b)*(a^2-7*a*b+8*b ^2)*ln(1+sin(d*x+c))/d-29/8*a*b^2*sin(d*x+c)/d-1/2*b^3*sin(d*x+c)^2/d-1/8* sec(d*x+c)^2*(8*b+5*a*sin(d*x+c))*(a+b*sin(d*x+c))^2/d+1/4*sec(d*x+c)^3*(a +b*sin(d*x+c))^3*tan(d*x+c)/d
Time = 0.41 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.98 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {-3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))+3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}+\frac {(a+b)^2 (5 a+11 b)}{-1+\sin (c+d x)}-48 a b^2 \sin (c+d x)-8 b^3 \sin ^2(c+d x)-\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {(5 a-11 b) (a-b)^2}{1+\sin (c+d x)}}{16 d} \]
(-3*(a + b)*(a^2 + 7*a*b + 8*b^2)*Log[1 - Sin[c + d*x]] + 3*(a - b)*(a^2 - 7*a*b + 8*b^2)*Log[1 + Sin[c + d*x]] + (a + b)^3/(-1 + Sin[c + d*x])^2 + ((a + b)^2*(5*a + 11*b))/(-1 + Sin[c + d*x]) - 48*a*b^2*Sin[c + d*x] - 8*b ^3*Sin[c + d*x]^2 - (a - b)^3/(1 + Sin[c + d*x])^2 + ((5*a - 11*b)*(a - b) ^2)/(1 + Sin[c + d*x]))/(16*d)
Time = 0.59 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3316, 27, 531, 25, 2176, 25, 2160, 2009}
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.
| \(\displaystyle \int \tan ^4(c+d x) \sec (c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4 (a+b \sin (c+d x))^3}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^4(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {b^4 \sin ^4(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle \frac {b \left (\frac {\int -\frac {(a+b \sin (c+d x))^2 \left (4 \sin ^3(c+d x) b^5+4 \sin (c+d x) b^5+4 a \sin ^2(c+d x) b^4+a b^4\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {(a+b \sin (c+d x))^2 \left (4 \sin ^3(c+d x) b^5+4 \sin (c+d x) b^5+4 a \sin ^2(c+d x) b^4+a b^4\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 2176 |
| \(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {(a+b \sin (c+d x)) \left (8 \sin ^2(c+d x) b^6+21 a \sin (c+d x) b^5+\left (3 a^2+16 b^2\right ) b^4\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}+\frac {b^2 \left (5 a b \sin (c+d x)+8 b^2\right ) (a+b \sin (c+d x))^2}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x))^2 \left (5 a b \sin (c+d x)+8 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {(a+b \sin (c+d x)) \left (8 \sin ^2(c+d x) b^6+21 a \sin (c+d x) b^5+\left (3 a^2+16 b^2\right ) b^4\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x))^2 \left (5 a b \sin (c+d x)+8 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \left (-8 \sin (c+d x) b^5-29 a b^4+\frac {3 \left (8 \left (a^2+b^2\right ) \sin (c+d x) b^5+a \left (a^2+15 b^2\right ) b^4\right )}{b^2-b^2 \sin ^2(c+d x)}\right )d(b \sin (c+d x))}{2 b^2}}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x))^2 \left (5 a b \sin (c+d x)+8 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {3 a b^3 \left (a^2+15 b^2\right ) \text {arctanh}(\sin (c+d x))-12 b^4 \left (a^2+b^2\right ) \log \left (b^2-b^2 \sin ^2(c+d x)\right )-29 a b^5 \sin (c+d x)-4 b^6 \sin ^2(c+d x)}{2 b^2}}{4 b^2}\right )}{d}\) |
(b*((b^3*Sin[c + d*x]*(a + b*Sin[c + d*x])^3)/(4*(b^2 - b^2*Sin[c + d*x]^2 )^2) - ((b^2*(a + b*Sin[c + d*x])^2*(8*b^2 + 5*a*b*Sin[c + d*x]))/(2*(b^2 - b^2*Sin[c + d*x]^2)) - (3*a*b^3*(a^2 + 15*b^2)*ArcTanh[Sin[c + d*x]] - 1 2*b^4*(a^2 + b^2)*Log[b^2 - b^2*Sin[c + d*x]^2] - 29*a*b^5*Sin[c + d*x] - 4*b^6*Sin[c + d*x]^2)/(2*b^2))/(4*b^2)))/d
3.16.2.3.1 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[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b *x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 , x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && !(IGtQ[m, 0] && R ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.98 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.60
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(284\) |
| default | \(\frac {a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(284\) |
| parallelrisch | \(\frac {96 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+b^{2}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (a^{2}+7 a b +8 b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-7 a b +8 b^{2}\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-24 a^{2} b -9 b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (18 a^{2} b +12 b^{3}\right ) \cos \left (4 d x +4 c \right )+\left (-10 a^{3}-90 a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+b^{3} \cos \left (6 d x +6 c \right )-12 \sin \left (5 d x +5 c \right ) a \,b^{2}+\left (6 a^{3}-30 a \,b^{2}\right ) \sin \left (d x +c \right )+6 a^{2} b -4 b^{3}}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(305\) |
| norman | \(\frac {\frac {\left (48 a^{2} b +16 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (48 a^{2} b +16 b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (2 a^{2} b +2 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (2 a^{2} b +2 b^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (6 a^{2} b +6 b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (6 a^{2} b +6 b^{3}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a \left (a^{2}+15 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (a^{2}+15 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (a^{2}+15 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 a \left (a^{2}+15 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {3 a \left (5 a^{2}+11 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (35 a^{2}+141 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (35 a^{2}+141 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 \left (a^{3}-8 a^{2} b +15 a \,b^{2}-8 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {3 \left (a^{3}+8 a^{2} b +15 a \,b^{2}+8 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 b \left (a^{2}+b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(489\) |
| risch | \(\frac {3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {6 i b^{3} c}{d}+\frac {b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {6 i a^{2} b c}{d}+3 i a^{2} b x +\frac {b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+3 i b^{3} x -\frac {3 i a \,b^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {{\mathrm e}^{i \left (d x +c \right )} \left (-5 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-27 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+48 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+24 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+48 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+32 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+5 i a^{3}+27 i a \,b^{2}+48 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+24 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{d}+\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{d}-\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{d}\) | \(531\) |
1/d*(a^3*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8* sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+3*a^2*b*(1/4*ta n(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c)))+3*a*b^2*(1/4*sin(d*x+c)^7/cos( d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*sin(d*x+c)^3-1 5/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)))+b^3*(1/4*sin(d*x+c)^8/cos(d *x+c)^4-1/2*sin(d*x+c)^8/cos(d*x+c)^2-1/2*sin(d*x+c)^6-3/4*sin(d*x+c)^4-3/ 2*sin(d*x+c)^2-3*ln(cos(d*x+c))))
Time = 0.30 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.18 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {8 \, b^{3} \cos \left (d x + c\right )^{6} - 4 \, b^{3} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, a^{2} b + 4 \, b^{3} - 24 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (24 \, a b^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{3} - 6 \, a b^{2} + {\left (5 \, a^{3} + 27 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
1/16*(8*b^3*cos(d*x + c)^6 - 4*b^3*cos(d*x + c)^4 + 3*(a^3 - 8*a^2*b + 15* a*b^2 - 8*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(a^3 + 8*a^2*b + 1 5*a*b^2 + 8*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 12*a^2*b + 4*b^3 - 24*(2*a^2*b + b^3)*cos(d*x + c)^2 - 2*(24*a*b^2*cos(d*x + c)^4 - 2*a^3 - 6*a*b^2 + (5*a^3 + 27*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c )^4)
Timed out. \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.07 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {8 \, b^{3} \sin \left (d x + c\right )^{2} + 48 \, a b^{2} \sin \left (d x + c\right ) - 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (5 \, a^{3} + 27 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 18 \, a^{2} b - 10 \, b^{3} + 12 \, {\left (2 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{2} - 3 \, {\left (a^{3} + 7 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
-1/16*(8*b^3*sin(d*x + c)^2 + 48*a*b^2*sin(d*x + c) - 3*(a^3 - 8*a^2*b + 1 5*a*b^2 - 8*b^3)*log(sin(d*x + c) + 1) + 3*(a^3 + 8*a^2*b + 15*a*b^2 + 8*b ^3)*log(sin(d*x + c) - 1) - 2*((5*a^3 + 27*a*b^2)*sin(d*x + c)^3 - 18*a^2* b - 10*b^3 + 12*(2*a^2*b + b^3)*sin(d*x + c)^2 - 3*(a^3 + 7*a*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.39 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.25 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {8 \, b^{3} \sin \left (d x + c\right )^{2} + 48 \, a b^{2} \sin \left (d x + c\right ) - 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, a^{2} b \sin \left (d x + c\right )^{4} + 18 \, b^{3} \sin \left (d x + c\right )^{4} + 5 \, a^{3} \sin \left (d x + c\right )^{3} + 27 \, a b^{2} \sin \left (d x + c\right )^{3} - 12 \, a^{2} b \sin \left (d x + c\right )^{2} - 24 \, b^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \sin \left (d x + c\right ) - 21 \, a b^{2} \sin \left (d x + c\right ) + 8 \, b^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
-1/16*(8*b^3*sin(d*x + c)^2 + 48*a*b^2*sin(d*x + c) - 3*(a^3 - 8*a^2*b + 1 5*a*b^2 - 8*b^3)*log(abs(sin(d*x + c) + 1)) + 3*(a^3 + 8*a^2*b + 15*a*b^2 + 8*b^3)*log(abs(sin(d*x + c) - 1)) - 2*(18*a^2*b*sin(d*x + c)^4 + 18*b^3* sin(d*x + c)^4 + 5*a^3*sin(d*x + c)^3 + 27*a*b^2*sin(d*x + c)^3 - 12*a^2*b *sin(d*x + c)^2 - 24*b^3*sin(d*x + c)^2 - 3*a^3*sin(d*x + c) - 21*a*b^2*si n(d*x + c) + 8*b^3)/(sin(d*x + c)^2 - 1)^2)/d
Time = 12.03 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.54 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (3\,a^2\,b+3\,b^3\right )}{d}-\frac {\left (-\frac {3\,a^3}{4}-\frac {45\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-6\,a^2\,b-6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {5\,a^3}{4}+\frac {75\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (12\,a^2\,b+12\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (\frac {15\,a^3}{2}+\frac {33\,a\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (36\,a^2\,b+4\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {15\,a^3}{2}+\frac {33\,a\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (12\,a^2\,b+12\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {5\,a^3}{4}+\frac {75\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-6\,a^2\,b-6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {3\,a^3}{4}-\frac {45\,a\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+b\right )\,\left (a^2+7\,a\,b+8\,b^2\right )}{8\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-b\right )\,\left (a^2-7\,a\,b+8\,b^2\right )}{8\,d} \]
(log(tan(c/2 + (d*x)/2)^2 + 1)*(3*a^2*b + 3*b^3))/d - (tan(c/2 + (d*x)/2)^ 5*((33*a*b^2)/2 + (15*a^3)/2) - tan(c/2 + (d*x)/2)*((45*a*b^2)/4 + (3*a^3) /4) + tan(c/2 + (d*x)/2)^7*((33*a*b^2)/2 + (15*a^3)/2) - tan(c/2 + (d*x)/2 )^11*((45*a*b^2)/4 + (3*a^3)/4) + tan(c/2 + (d*x)/2)^3*((75*a*b^2)/4 + (5* a^3)/4) + tan(c/2 + (d*x)/2)^9*((75*a*b^2)/4 + (5*a^3)/4) - tan(c/2 + (d*x )/2)^2*(6*a^2*b + 6*b^3) - tan(c/2 + (d*x)/2)^10*(6*a^2*b + 6*b^3) + tan(c /2 + (d*x)/2)^4*(12*a^2*b + 12*b^3) + tan(c/2 + (d*x)/2)^8*(12*a^2*b + 12* b^3) + tan(c/2 + (d*x)/2)^6*(36*a^2*b + 4*b^3))/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 2*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x)/2)^12 - 1)) - (3*log(tan(c/2 + ( d*x)/2) - 1)*(a + b)*(7*a*b + a^2 + 8*b^2))/(8*d) + (3*log(tan(c/2 + (d*x) /2) + 1)*(a - b)*(a^2 - 7*a*b + 8*b^2))/(8*d)