Integrand size = 21, antiderivative size = 175 \[ \int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\frac {5}{16} \left (a^2-6 b^2\right ) x+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {\left (11 a^2-18 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a b \sin ^3(c+d x)}{3 d}-\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {b^2 \tan (c+d x)}{d} \]
5/16*(a^2-6*b^2)*x+2*a*b*arctanh(sin(d*x+c))/d-2*a*b*sin(d*x+c)/d-1/16*(11 *a^2-18*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/24*(13*a^2-6*b^2)*cos(d*x+c)^3*sin( d*x+c)/d-1/6*a^2*cos(d*x+c)^5*sin(d*x+c)/d-2/3*a*b*sin(d*x+c)^3/d-2/5*a*b* sin(d*x+c)^5/d+b^2*tan(d*x+c)/d
Time = 2.14 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.10 \[ \int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\frac {60 \left (5 \left (a^2-6 b^2\right ) (c+d x)-32 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-2128 a b \sin (c+d x)+\left (-185 a^2+1410 b^2-5 \left (29 a^2-84 b^2\right ) \cos (2 (c+d x))+232 a b \cos (3 (c+d x))+35 a^2 \cos (4 (c+d x))-30 b^2 \cos (4 (c+d x))-24 a b \cos (5 (c+d x))-5 a^2 \cos (6 (c+d x))\right ) \tan (c+d x)}{960 d} \]
(60*(5*(a^2 - 6*b^2)*(c + d*x) - 32*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d* x)/2]] + 32*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 2128*a*b*Sin[c + d*x] + (-185*a^2 + 1410*b^2 - 5*(29*a^2 - 84*b^2)*Cos[2*(c + d*x)] + 23 2*a*b*Cos[3*(c + d*x)] + 35*a^2*Cos[4*(c + d*x)] - 30*b^2*Cos[4*(c + d*x)] - 24*a*b*Cos[5*(c + d*x)] - 5*a^2*Cos[6*(c + d*x)])*Tan[c + d*x])/(960*d)
Time = 0.90 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 4360, 3042, 3390, 3042, 3072, 254, 2009, 4889, 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.
| \(\displaystyle \int \sin ^6(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^6 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \sin ^4(c+d x) \tan ^2(c+d x) (-a \cos (c+d x)-b)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^6 \left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3390 |
| \(\displaystyle \int \left (b^2+a^2 \cos ^2(c+d x)\right ) \sin ^4(c+d x) \tan ^2(c+d x)dx+2 a b \int \sin ^5(c+d x) \tan (c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (b^2+a^2 \cos (c+d x)^2\right ) \sin (c+d x)^4 \tan (c+d x)^2dx+2 a b \int \sin (c+d x)^5 \tan (c+d x)dx\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle \int \left (b^2+a^2 \cos (c+d x)^2\right ) \sin (c+d x)^4 \tan (c+d x)^2dx+\frac {2 a b \int \frac {\sin ^6(c+d x)}{1-\sin ^2(c+d x)}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \int \left (b^2+a^2 \cos (c+d x)^2\right ) \sin (c+d x)^4 \tan (c+d x)^2dx+\frac {2 a b \int \left (-\sin ^4(c+d x)-\sin ^2(c+d x)+\frac {1}{1-\sin ^2(c+d x)}-1\right )d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \left (b^2+a^2 \cos (c+d x)^2\right ) \sin (c+d x)^4 \tan (c+d x)^2dx+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4889 |
| \(\displaystyle \frac {\int \frac {\tan ^6(c+d x) \left (a^2+b^2+b^2 \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right )^4}d\tan (c+d x)}{d}+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle \frac {-\frac {1}{6} \int -\frac {6 b^2 \tan ^6(c+d x)+6 a^2 \tan ^4(c+d x)-6 a^2 \tan ^2(c+d x)+a^2}{\left (\tan ^2(c+d x)+1\right )^3}d\tan (c+d x)-\frac {a^2 \tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}}{d}+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {6 b^2 \tan ^6(c+d x)+6 a^2 \tan ^4(c+d x)-6 a^2 \tan ^2(c+d x)+a^2}{\left (\tan ^2(c+d x)+1\right )^3}d\tan (c+d x)-\frac {a^2 \tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}}{d}+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {\left (13 a^2-6 b^2\right ) \tan (c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}-\frac {1}{4} \int \frac {3 \left (-8 b^2 \tan ^4(c+d x)-8 \left (a^2-b^2\right ) \tan ^2(c+d x)+3 a^2-2 b^2\right )}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)\right )-\frac {a^2 \tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}}{d}+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {\left (13 a^2-6 b^2\right ) \tan (c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}-\frac {3}{4} \int \frac {-8 b^2 \tan ^4(c+d x)-8 \left (a^2-b^2\right ) \tan ^2(c+d x)+3 a^2-2 b^2}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)\right )-\frac {a^2 \tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}}{d}+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 1471 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {\left (13 a^2-6 b^2\right ) \tan (c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}-\frac {3}{4} \left (\frac {\left (11 a^2-18 b^2\right ) \tan (c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}-\frac {1}{2} \int \frac {5 a^2-14 b^2+16 b^2 \tan ^2(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)\right )\right )-\frac {a^2 \tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}}{d}+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {\left (13 a^2-6 b^2\right ) \tan (c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}-\frac {3}{4} \left (\frac {1}{2} \left (-5 \left (a^2-6 b^2\right ) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)-16 b^2 \tan (c+d x)\right )+\frac {\left (11 a^2-18 b^2\right ) \tan (c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )\right )-\frac {a^2 \tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}}{d}+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {\left (13 a^2-6 b^2\right ) \tan (c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}-\frac {3}{4} \left (\frac {1}{2} \left (-5 \left (a^2-6 b^2\right ) \arctan (\tan (c+d x))-16 b^2 \tan (c+d x)\right )+\frac {\left (11 a^2-18 b^2\right ) \tan (c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )\right )-\frac {a^2 \tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}}{d}+\frac {2 a b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
(2*a*b*(ArcTanh[Sin[c + d*x]] - Sin[c + d*x] - Sin[c + d*x]^3/3 - Sin[c + d*x]^5/5))/d + (-1/6*(a^2*Tan[c + d*x])/(1 + Tan[c + d*x]^2)^3 + (((13*a^2 - 6*b^2)*Tan[c + d*x])/(4*(1 + Tan[c + d*x]^2)^2) - (3*((-5*(a^2 - 6*b^2) *ArcTan[Tan[c + d*x]] - 16*b^2*Tan[c + d*x])/2 + ((11*a^2 - 18*b^2)*Tan[c + d*x])/(2*(1 + Tan[c + d*x]^2))))/4)/6)/d
3.2.79.3.1 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[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[2*a*(b/d) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e + f* x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Time = 2.14 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.93
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) | \(163\) |
| default | \(\frac {a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) | \(163\) |
| parts | \(\frac {a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(168\) |
| parallelrisch | \(\frac {-3840 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+3840 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\left (-180 a^{2}+450 b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (40 a^{2}-30 b^{2}\right ) \sin \left (5 d x +5 c \right )-2360 a b \sin \left (2 d x +2 c \right )+256 a b \sin \left (4 d x +4 c \right )-24 a b \sin \left (6 d x +6 c \right )-5 a^{2} \sin \left (7 d x +7 c \right )+600 d x \left (a^{2}-6 b^{2}\right ) \cos \left (d x +c \right )+\left (-225 a^{2}+2400 b^{2}\right ) \sin \left (d x +c \right )}{1920 d \cos \left (d x +c \right )}\) | \(191\) |
| risch | \(\frac {5 a^{2} x}{16}-\frac {15 b^{2} x}{8}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{4 d}-\frac {15 i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}+\frac {11 i a b \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{4 d}+\frac {15 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{128 d}-\frac {11 i a b \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{2} \sin \left (6 d x +6 c \right )}{192 d}-\frac {a b \sin \left (5 d x +5 c \right )}{40 d}+\frac {3 a^{2} \sin \left (4 d x +4 c \right )}{64 d}-\frac {\sin \left (4 d x +4 c \right ) b^{2}}{32 d}+\frac {7 a b \sin \left (3 d x +3 c \right )}{24 d}\) | \(265\) |
| norman | \(\frac {\left (-\frac {5 a^{2}}{16}+\frac {15 b^{2}}{8}\right ) x +\left (-\frac {45 a^{2}}{16}+\frac {135 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {25 a^{2}}{16}+\frac {75 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {25 a^{2}}{16}+\frac {75 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {5 a^{2}}{16}-\frac {15 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {25 a^{2}}{16}-\frac {75 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {25 a^{2}}{16}-\frac {75 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {45 a^{2}}{16}-\frac {135 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-\frac {\left (33 a^{2}+58 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}+\frac {\left (5 a^{2}-32 a b -30 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{8 d}+\frac {\left (5 a^{2}+32 a b -30 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (35 a^{2}-256 a b -210 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 d}+\frac {\left (35 a^{2}+256 a b -210 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {\left (565 a^{2}-5216 a b -3390 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{120 d}+\frac {\left (565 a^{2}+5216 a b -3390 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{120 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(464\) |
1/d*(a^2*(-1/6*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+ 5/16*d*x+5/16*c)+2*a*b*(-1/5*sin(d*x+c)^5-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(s ec(d*x+c)+tan(d*x+c)))+b^2*(sin(d*x+c)^7/cos(d*x+c)+(sin(d*x+c)^5+5/4*sin( d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)-15/8*d*x-15/8*c))
Time = 0.29 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.01 \[ \int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\frac {75 \, {\left (a^{2} - 6 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 240 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 240 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (40 \, a^{2} \cos \left (d x + c\right )^{6} + 96 \, a b \cos \left (d x + c\right )^{5} - 352 \, a b \cos \left (d x + c\right )^{3} - 10 \, {\left (13 \, a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 736 \, a b \cos \left (d x + c\right ) + 15 \, {\left (11 \, a^{2} - 18 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 240 \, b^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )} \]
1/240*(75*(a^2 - 6*b^2)*d*x*cos(d*x + c) + 240*a*b*cos(d*x + c)*log(sin(d* x + c) + 1) - 240*a*b*cos(d*x + c)*log(-sin(d*x + c) + 1) - (40*a^2*cos(d* x + c)^6 + 96*a*b*cos(d*x + c)^5 - 352*a*b*cos(d*x + c)^3 - 10*(13*a^2 - 6 *b^2)*cos(d*x + c)^4 + 736*a*b*cos(d*x + c) + 15*(11*a^2 - 18*b^2)*cos(d*x + c)^2 - 240*b^2)*sin(d*x + c))/(d*cos(d*x + c))
Timed out. \[ \int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.99 \[ \int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 64 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a b - 120 \, {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} b^{2}}{960 \, d} \]
1/960*(5*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*s in(2*d*x + 2*c))*a^2 - 64*(6*sin(d*x + c)^5 + 10*sin(d*x + c)^3 - 15*log(s in(d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 30*sin(d*x + c))*a*b - 120*( 15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*ta n(d*x + c)^2 + 1) - 8*tan(d*x + c))*b^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (163) = 326\).
Time = 0.38 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.17 \[ \int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\frac {480 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 480 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 75 \, {\left (a^{2} - 6 \, b^{2}\right )} {\left (d x + c\right )} - \frac {480 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 480 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 210 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 425 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 3040 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 870 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 990 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8256 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 660 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 990 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8256 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 660 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 425 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3040 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 870 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 480 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 210 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]
1/240*(480*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 480*a*b*log(abs(tan(1/ 2*d*x + 1/2*c) - 1)) + 75*(a^2 - 6*b^2)*(d*x + c) - 480*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(75*a^2*tan(1/2*d*x + 1/2*c)^11 - 480*a*b*tan(1/2*d*x + 1/2*c)^11 - 210*b^2*tan(1/2*d*x + 1/2*c)^11 + 425*a^ 2*tan(1/2*d*x + 1/2*c)^9 - 3040*a*b*tan(1/2*d*x + 1/2*c)^9 - 870*b^2*tan(1 /2*d*x + 1/2*c)^9 + 990*a^2*tan(1/2*d*x + 1/2*c)^7 - 8256*a*b*tan(1/2*d*x + 1/2*c)^7 - 660*b^2*tan(1/2*d*x + 1/2*c)^7 - 990*a^2*tan(1/2*d*x + 1/2*c) ^5 - 8256*a*b*tan(1/2*d*x + 1/2*c)^5 + 660*b^2*tan(1/2*d*x + 1/2*c)^5 - 42 5*a^2*tan(1/2*d*x + 1/2*c)^3 - 3040*a*b*tan(1/2*d*x + 1/2*c)^3 + 870*b^2*t an(1/2*d*x + 1/2*c)^3 - 75*a^2*tan(1/2*d*x + 1/2*c) - 480*a*b*tan(1/2*d*x + 1/2*c) + 210*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d
Time = 16.74 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.32 \[ \int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\frac {\frac {5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,a^2}{8}+4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,a\,b-\frac {15\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,b^2}{4}}{d}-\frac {\frac {15\,a^2\,\sin \left (c+d\,x\right )}{128}-\frac {5\,b^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{32}-\frac {a^2\,\sin \left (5\,c+5\,d\,x\right )}{48}+\frac {a^2\,\sin \left (7\,c+7\,d\,x\right )}{384}-\frac {15\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{64}+\frac {b^2\,\sin \left (5\,c+5\,d\,x\right )}{64}+\frac {59\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{48}-\frac {2\,a\,b\,\sin \left (4\,c+4\,d\,x\right )}{15}+\frac {a\,b\,\sin \left (6\,c+6\,d\,x\right )}{80}}{d\,\cos \left (c+d\,x\right )} \]
((5*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 - (15*b^2*atan(sin( c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + 4*a*b*atanh(sin(c/2 + (d*x)/2)/cos (c/2 + (d*x)/2)))/d - ((15*a^2*sin(c + d*x))/128 - (5*b^2*sin(c + d*x))/4 + (3*a^2*sin(3*c + 3*d*x))/32 - (a^2*sin(5*c + 5*d*x))/48 + (a^2*sin(7*c + 7*d*x))/384 - (15*b^2*sin(3*c + 3*d*x))/64 + (b^2*sin(5*c + 5*d*x))/64 + (59*a*b*sin(2*c + 2*d*x))/48 - (2*a*b*sin(4*c + 4*d*x))/15 + (a*b*sin(6*c + 6*d*x))/80)/(d*cos(c + d*x))