\(\int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx\) [437]

Optimal result

Integrand size = 21, antiderivative size = 169 \[ \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx=\frac {b \left (9 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d} \]

[Out]

1/8*b*(9*a^2+4*b^2)*arctanh(sin(d*x+c))/d+1/5*a*(4*a^2+15*b^2)*tan(d*x+c)/d+1/8*b*(9*a^2+4*b^2)*sec(d*x+c)*tan
(d*x+c)/d+11/20*a^2*b*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a^2*(a+b*cos(d*x+c))*sec(d*x+c)^4*tan(d*x+c)/d+1/15*a*(4*a
^2+15*b^2)*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2871, 3100, 2827, 3852, 3853, 3855} \[ \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx=\frac {b \left (9 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {11 a^2 b \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))}{5 d} \]

[In]

Int[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^6,x]

[Out]

(b*(9*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(4*a^2 + 15*b^2)*Tan[c + d*x])/(5*d) + (b*(9*a^2 + 4*b^2)
*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (11*a^2*b*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (a^2*(a + b*Cos[c + d*x])*
Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (a*(4*a^2 + 15*b^2)*Tan[c + d*x]^3)/(15*d)

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2871

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/
(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e
+ f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 +
c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 +
d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \left (11 a^2 b+a \left (4 a^2+15 b^2\right ) \cos (c+d x)+b \left (3 a^2+5 b^2\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx \\ & = \frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int \left (4 a \left (4 a^2+15 b^2\right )+5 b \left (9 a^2+4 b^2\right ) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (b \left (9 a^2+4 b^2\right )\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (a \left (4 a^2+15 b^2\right )\right ) \int \sec ^4(c+d x) \, dx \\ & = \frac {b \left (9 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (b \left (9 a^2+4 b^2\right )\right ) \int \sec (c+d x) \, dx-\frac {\left (a \left (4 a^2+15 b^2\right )\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {b \left (9 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71 \[ \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx=\frac {15 b \left (9 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 b \left (9 a^2+4 b^2\right ) \sec (c+d x)+90 a^2 b \sec ^3(c+d x)+8 a \left (15 \left (a^2+3 b^2\right )+5 \left (2 a^2+3 b^2\right ) \tan ^2(c+d x)+3 a^2 \tan ^4(c+d x)\right )\right )}{120 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^6,x]

[Out]

(15*b*(9*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*b*(9*a^2 + 4*b^2)*Sec[c + d*x] + 90*a^2*b*Sec[c
+ d*x]^3 + 8*a*(15*(a^2 + 3*b^2) + 5*(2*a^2 + 3*b^2)*Tan[c + d*x]^2 + 3*a^2*Tan[c + d*x]^4)))/(120*d)

Maple [A] (verified)

Time = 4.91 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+3 a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(148\)
default	\(\frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+3 a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(148\)
parts	\(-\frac {a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {3 a \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(156\)
parallelrisch	\(\frac {-1350 b \left (a^{2}+\frac {4 b^{2}}{9}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+1350 b \left (a^{2}+\frac {4 b^{2}}{9}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (1260 a^{2} b +240 b^{3}\right ) \sin \left (2 d x +2 c \right )+\left (320 a^{3}+1200 a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (270 a^{2} b +120 b^{3}\right ) \sin \left (4 d x +4 c \right )+64 \left (\left (a^{2}+\frac {15 b^{2}}{4}\right ) \sin \left (5 d x +5 c \right )+\left (10 a^{2}+15 b^{2}\right ) \sin \left (d x +c \right )\right ) a}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(251\)
risch	\(-\frac {i \left (135 a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}+60 b^{3} {\mathrm e}^{9 i \left (d x +c \right )}+630 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+120 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-720 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-640 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-1680 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-630 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-120 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-320 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-1200 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-135 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-60 b^{3} {\mathrm e}^{i \left (d x +c \right )}-64 a^{3}-240 a \,b^{2}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {9 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {9 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(308\)
norman	\(\frac {-\frac {\left (8 a^{3}-15 a^{2} b +24 a \,b^{2}-4 b^{3}\right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (8 a^{3}+15 a^{2} b +24 a \,b^{2}+4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (40 a^{3}-117 a^{2} b +24 a \,b^{2}-12 b^{3}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {\left (40 a^{3}+117 a^{2} b +24 a \,b^{2}+12 b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {\left (344 a^{3}-405 a^{2} b -600 a \,b^{2}+180 b^{3}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d}-\frac {\left (344 a^{3}+405 a^{2} b -600 a \,b^{2}-180 b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d}-\frac {\left (872 a^{3}-45 a^{2} b +120 a \,b^{2}+180 b^{3}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d}-\frac {\left (872 a^{3}+45 a^{2} b +120 a \,b^{2}-180 b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {b \left (9 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {b \left (9 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(402\)

[In]

int((a+cos(d*x+c)*b)^3*sec(d*x+c)^6,x,method=_RETURNVERBOSE)

[Out]

1/d*(-a^3*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+3*a^2*b*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*t
an(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-3*a*b^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b^3*(1/2*sec(d*x+c)*tan(d*
x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx=\frac {15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 90 \, a^{2} b \cos \left (d x + c\right ) + 15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 24 \, a^{3} + 8 \, {\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(15*(9*a^2*b + 4*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(9*a^2*b + 4*b^3)*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 2*(16*(4*a^3 + 15*a*b^2)*cos(d*x + c)^4 + 90*a^2*b*cos(d*x + c) + 15*(9*a^2*b + 4*b^3)*cos(d
*x + c)^3 + 24*a^3 + 8*(4*a^3 + 15*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**3*sec(d*x+c)**6,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.07 \[ \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{2} - 45 \, a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))
*a*b^2 - 45*a^2*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d
*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1)
+ log(sin(d*x + c) - 1)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 367 vs. \(2 (157) = 314\).

Time = 0.35 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.17 \[ \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx=\frac {15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 960 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 960 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(9*a^2*b + 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(9*a^2*b + 4*b^3)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) - 2*(120*a^3*tan(1/2*d*x + 1/2*c)^9 - 225*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*a*b^2*tan(1/2*d*x + 1/
2*c)^9 - 60*b^3*tan(1/2*d*x + 1/2*c)^9 - 160*a^3*tan(1/2*d*x + 1/2*c)^7 + 90*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 96
0*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*a*b^2*
tan(1/2*d*x + 1/2*c)^5 - 160*a^3*tan(1/2*d*x + 1/2*c)^3 - 90*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*a*b^2*tan(1/2*
d*x + 1/2*c)^3 - 120*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*a^3*tan(1/2*d*x + 1/2*c) + 225*a^2*b*tan(1/2*d*x + 1/2*c
) + 360*a*b^2*tan(1/2*d*x + 1/2*c) + 60*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 17.67 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.54 \[ \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {9\,a^2\,b}{4}+b^3\right )}{d}-\frac {\left (2\,a^3-\frac {15\,a^2\,b}{4}+6\,a\,b^2-b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {8\,a^3}{3}+\frac {3\,a^2\,b}{2}-16\,a\,b^2+2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^3}{15}+20\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,a^3}{3}-\frac {3\,a^2\,b}{2}-16\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^3+\frac {15\,a^2\,b}{4}+6\,a\,b^2+b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

int((a + b*cos(c + d*x))^3/cos(c + d*x)^6,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((9*a^2*b)/4 + b^3))/d - (tan(c/2 + (d*x)/2)^9*(6*a*b^2 - (15*a^2*b)/4 + 2*a^3 - b^
3) - tan(c/2 + (d*x)/2)^3*(16*a*b^2 + (3*a^2*b)/2 + (8*a^3)/3 + 2*b^3) - tan(c/2 + (d*x)/2)^7*(16*a*b^2 - (3*a
^2*b)/2 + (8*a^3)/3 - 2*b^3) + tan(c/2 + (d*x)/2)*(6*a*b^2 + (15*a^2*b)/4 + 2*a^3 + b^3) + tan(c/2 + (d*x)/2)^
5*(20*a*b^2 + (116*a^3)/15))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 -
5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))