3.90 \(\int \sec ^{11}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=408 \[ \frac {5 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac {5 a^4 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {5 a^4 \tan (c+d x) \sec (c+d x)}{16 d}+\frac {4 a^3 b \sec ^7(c+d x)}{7 d}-\frac {15 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{64 d}+\frac {3 a^2 b^2 \tan (c+d x) \sec ^7(c+d x)}{4 d}-\frac {a^2 b^2 \tan (c+d x) \sec ^5(c+d x)}{8 d}-\frac {5 a^2 b^2 \tan (c+d x) \sec ^3(c+d x)}{32 d}-\frac {15 a^2 b^2 \tan (c+d x) \sec (c+d x)}{64 d}+\frac {4 a b^3 \sec ^9(c+d x)}{9 d}-\frac {4 a b^3 \sec ^7(c+d x)}{7 d}+\frac {3 b^4 \tanh ^{-1}(\sin (c+d x))}{256 d}+\frac {b^4 \tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3 b^4 \tan (c+d x) \sec ^7(c+d x)}{80 d}+\frac {b^4 \tan (c+d x) \sec ^5(c+d x)}{160 d}+\frac {b^4 \tan (c+d x) \sec ^3(c+d x)}{128 d}+\frac {3 b^4 \tan (c+d x) \sec (c+d x)}{256 d} \]

[Out]

5/16*a^4*arctanh(sin(d*x+c))/d-15/64*a^2*b^2*arctanh(sin(d*x+c))/d+3/256*b^4*arctanh(sin(d*x+c))/d+4/7*a^3*b*s
ec(d*x+c)^7/d-4/7*a*b^3*sec(d*x+c)^7/d+4/9*a*b^3*sec(d*x+c)^9/d+5/16*a^4*sec(d*x+c)*tan(d*x+c)/d-15/64*a^2*b^2
*sec(d*x+c)*tan(d*x+c)/d+3/256*b^4*sec(d*x+c)*tan(d*x+c)/d+5/24*a^4*sec(d*x+c)^3*tan(d*x+c)/d-5/32*a^2*b^2*sec
(d*x+c)^3*tan(d*x+c)/d+1/128*b^4*sec(d*x+c)^3*tan(d*x+c)/d+1/6*a^4*sec(d*x+c)^5*tan(d*x+c)/d-1/8*a^2*b^2*sec(d
*x+c)^5*tan(d*x+c)/d+1/160*b^4*sec(d*x+c)^5*tan(d*x+c)/d+3/4*a^2*b^2*sec(d*x+c)^7*tan(d*x+c)/d-3/80*b^4*sec(d*
x+c)^7*tan(d*x+c)/d+1/10*b^4*sec(d*x+c)^7*tan(d*x+c)^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.40, antiderivative size = 408, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3090, 3768, 3770, 2606, 30, 2611, 14} \[ -\frac {15 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{64 d}+\frac {3 a^2 b^2 \tan (c+d x) \sec ^7(c+d x)}{4 d}-\frac {a^2 b^2 \tan (c+d x) \sec ^5(c+d x)}{8 d}-\frac {5 a^2 b^2 \tan (c+d x) \sec ^3(c+d x)}{32 d}-\frac {15 a^2 b^2 \tan (c+d x) \sec (c+d x)}{64 d}+\frac {4 a^3 b \sec ^7(c+d x)}{7 d}+\frac {5 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac {5 a^4 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {5 a^4 \tan (c+d x) \sec (c+d x)}{16 d}+\frac {4 a b^3 \sec ^9(c+d x)}{9 d}-\frac {4 a b^3 \sec ^7(c+d x)}{7 d}+\frac {3 b^4 \tanh ^{-1}(\sin (c+d x))}{256 d}+\frac {b^4 \tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3 b^4 \tan (c+d x) \sec ^7(c+d x)}{80 d}+\frac {b^4 \tan (c+d x) \sec ^5(c+d x)}{160 d}+\frac {b^4 \tan (c+d x) \sec ^3(c+d x)}{128 d}+\frac {3 b^4 \tan (c+d x) \sec (c+d x)}{256 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^11*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(5*a^4*ArcTanh[Sin[c + d*x]])/(16*d) - (15*a^2*b^2*ArcTanh[Sin[c + d*x]])/(64*d) + (3*b^4*ArcTanh[Sin[c + d*x]
])/(256*d) + (4*a^3*b*Sec[c + d*x]^7)/(7*d) - (4*a*b^3*Sec[c + d*x]^7)/(7*d) + (4*a*b^3*Sec[c + d*x]^9)/(9*d)
+ (5*a^4*Sec[c + d*x]*Tan[c + d*x])/(16*d) - (15*a^2*b^2*Sec[c + d*x]*Tan[c + d*x])/(64*d) + (3*b^4*Sec[c + d*
x]*Tan[c + d*x])/(256*d) + (5*a^4*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) - (5*a^2*b^2*Sec[c + d*x]^3*Tan[c + d*x]
)/(32*d) + (b^4*Sec[c + d*x]^3*Tan[c + d*x])/(128*d) + (a^4*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) - (a^2*b^2*Sec[
c + d*x]^5*Tan[c + d*x])/(8*d) + (b^4*Sec[c + d*x]^5*Tan[c + d*x])/(160*d) + (3*a^2*b^2*Sec[c + d*x]^7*Tan[c +
d*x])/(4*d) - (3*b^4*Sec[c + d*x]^7*Tan[c + d*x])/(80*d) + (b^4*Sec[c + d*x]^7*Tan[c + d*x]^3)/(10*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3768

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 3770

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

Rubi steps

\begin {align*} \int \sec ^{11}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=\int \left (a^4 \sec ^7(c+d x)+4 a^3 b \sec ^7(c+d x) \tan (c+d x)+6 a^2 b^2 \sec ^7(c+d x) \tan ^2(c+d x)+4 a b^3 \sec ^7(c+d x) \tan ^3(c+d x)+b^4 \sec ^7(c+d x) \tan ^4(c+d x)\right ) \, dx\\ &=a^4 \int \sec ^7(c+d x) \, dx+\left (4 a^3 b\right ) \int \sec ^7(c+d x) \tan (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \sec ^7(c+d x) \tan ^2(c+d x) \, dx+\left (4 a b^3\right ) \int \sec ^7(c+d x) \tan ^3(c+d x) \, dx+b^4 \int \sec ^7(c+d x) \tan ^4(c+d x) \, dx\\ &=\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {3 a^2 b^2 \sec ^7(c+d x) \tan (c+d x)}{4 d}+\frac {b^4 \sec ^7(c+d x) \tan ^3(c+d x)}{10 d}+\frac {1}{6} \left (5 a^4\right ) \int \sec ^5(c+d x) \, dx-\frac {1}{4} \left (3 a^2 b^2\right ) \int \sec ^7(c+d x) \, dx-\frac {1}{10} \left (3 b^4\right ) \int \sec ^7(c+d x) \tan ^2(c+d x) \, dx+\frac {\left (4 a^3 b\right ) \operatorname {Subst}\left (\int x^6 \, dx,x,\sec (c+d x)\right )}{d}+\frac {\left (4 a b^3\right ) \operatorname {Subst}\left (\int x^6 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {4 a^3 b \sec ^7(c+d x)}{7 d}+\frac {5 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{8 d}+\frac {3 a^2 b^2 \sec ^7(c+d x) \tan (c+d x)}{4 d}-\frac {3 b^4 \sec ^7(c+d x) \tan (c+d x)}{80 d}+\frac {b^4 \sec ^7(c+d x) \tan ^3(c+d x)}{10 d}+\frac {1}{8} \left (5 a^4\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{8} \left (5 a^2 b^2\right ) \int \sec ^5(c+d x) \, dx+\frac {1}{80} \left (3 b^4\right ) \int \sec ^7(c+d x) \, dx+\frac {\left (4 a b^3\right ) \operatorname {Subst}\left (\int \left (-x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {4 a^3 b \sec ^7(c+d x)}{7 d}-\frac {4 a b^3 \sec ^7(c+d x)}{7 d}+\frac {4 a b^3 \sec ^9(c+d x)}{9 d}+\frac {5 a^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac {5 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}-\frac {5 a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{32 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{8 d}+\frac {b^4 \sec ^5(c+d x) \tan (c+d x)}{160 d}+\frac {3 a^2 b^2 \sec ^7(c+d x) \tan (c+d x)}{4 d}-\frac {3 b^4 \sec ^7(c+d x) \tan (c+d x)}{80 d}+\frac {b^4 \sec ^7(c+d x) \tan ^3(c+d x)}{10 d}+\frac {1}{16} \left (5 a^4\right ) \int \sec (c+d x) \, dx-\frac {1}{32} \left (15 a^2 b^2\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{32} b^4 \int \sec ^5(c+d x) \, dx\\ &=\frac {5 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {4 a^3 b \sec ^7(c+d x)}{7 d}-\frac {4 a b^3 \sec ^7(c+d x)}{7 d}+\frac {4 a b^3 \sec ^9(c+d x)}{9 d}+\frac {5 a^4 \sec (c+d x) \tan (c+d x)}{16 d}-\frac {15 a^2 b^2 \sec (c+d x) \tan (c+d x)}{64 d}+\frac {5 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}-\frac {5 a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{32 d}+\frac {b^4 \sec ^3(c+d x) \tan (c+d x)}{128 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{8 d}+\frac {b^4 \sec ^5(c+d x) \tan (c+d x)}{160 d}+\frac {3 a^2 b^2 \sec ^7(c+d x) \tan (c+d x)}{4 d}-\frac {3 b^4 \sec ^7(c+d x) \tan (c+d x)}{80 d}+\frac {b^4 \sec ^7(c+d x) \tan ^3(c+d x)}{10 d}-\frac {1}{64} \left (15 a^2 b^2\right ) \int \sec (c+d x) \, dx+\frac {1}{128} \left (3 b^4\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {5 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {15 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{64 d}+\frac {4 a^3 b \sec ^7(c+d x)}{7 d}-\frac {4 a b^3 \sec ^7(c+d x)}{7 d}+\frac {4 a b^3 \sec ^9(c+d x)}{9 d}+\frac {5 a^4 \sec (c+d x) \tan (c+d x)}{16 d}-\frac {15 a^2 b^2 \sec (c+d x) \tan (c+d x)}{64 d}+\frac {3 b^4 \sec (c+d x) \tan (c+d x)}{256 d}+\frac {5 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}-\frac {5 a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{32 d}+\frac {b^4 \sec ^3(c+d x) \tan (c+d x)}{128 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{8 d}+\frac {b^4 \sec ^5(c+d x) \tan (c+d x)}{160 d}+\frac {3 a^2 b^2 \sec ^7(c+d x) \tan (c+d x)}{4 d}-\frac {3 b^4 \sec ^7(c+d x) \tan (c+d x)}{80 d}+\frac {b^4 \sec ^7(c+d x) \tan ^3(c+d x)}{10 d}+\frac {1}{256} \left (3 b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac {5 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {15 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{64 d}+\frac {3 b^4 \tanh ^{-1}(\sin (c+d x))}{256 d}+\frac {4 a^3 b \sec ^7(c+d x)}{7 d}-\frac {4 a b^3 \sec ^7(c+d x)}{7 d}+\frac {4 a b^3 \sec ^9(c+d x)}{9 d}+\frac {5 a^4 \sec (c+d x) \tan (c+d x)}{16 d}-\frac {15 a^2 b^2 \sec (c+d x) \tan (c+d x)}{64 d}+\frac {3 b^4 \sec (c+d x) \tan (c+d x)}{256 d}+\frac {5 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}-\frac {5 a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{32 d}+\frac {b^4 \sec ^3(c+d x) \tan (c+d x)}{128 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{8 d}+\frac {b^4 \sec ^5(c+d x) \tan (c+d x)}{160 d}+\frac {3 a^2 b^2 \sec ^7(c+d x) \tan (c+d x)}{4 d}-\frac {3 b^4 \sec ^7(c+d x) \tan (c+d x)}{80 d}+\frac {b^4 \sec ^7(c+d x) \tan ^3(c+d x)}{10 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.31, size = 242, normalized size = 0.59 \[ \frac {10 \sec ^9(c+d x) \left (32768 a b \left (27 a^2+b^2\right )+189 \left (592 a^4+1604 a^2 b^2+739 b^4\right ) \tan (c+d x)\right )-80640 \left (80 a^4-60 a^2 b^2+3 b^4\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 \sec ^{10}(c+d x) \left (983040 a b \left (a^2-b^2\right ) \cos (3 (c+d x))+420 \left (1552 a^4+1908 a^2 b^2-505 b^4\right ) \sin (3 (c+d x))+7 \left (80 a^4-60 a^2 b^2+3 b^4\right ) (628 \sin (5 (c+d x))+145 \sin (7 (c+d x))+15 \sin (9 (c+d x)))\right )}{20643840 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^11*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(-80640*(80*a^4 - 60*a^2*b^2 + 3*b^4)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]]) + 3*Sec[c + d*x]^10*(983040*a*b*(a^2 - b^2)*Cos[3*(c + d*x)] + 420*(1552*a^4 + 1908*a^2*b^2 - 50
5*b^4)*Sin[3*(c + d*x)] + 7*(80*a^4 - 60*a^2*b^2 + 3*b^4)*(628*Sin[5*(c + d*x)] + 145*Sin[7*(c + d*x)] + 15*Si
n[9*(c + d*x)])) + 10*Sec[c + d*x]^9*(32768*a*b*(27*a^2 + b^2) + 189*(592*a^4 + 1604*a^2*b^2 + 739*b^4)*Tan[c
+ d*x]))/(20643840*d)

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 251, normalized size = 0.62 \[ \frac {315 \, {\left (80 \, a^{4} - 60 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{10} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (80 \, a^{4} - 60 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{10} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 71680 \, a b^{3} \cos \left (d x + c\right ) + 92160 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 42 \, {\left (15 \, {\left (80 \, a^{4} - 60 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{8} + 10 \, {\left (80 \, a^{4} - 60 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{6} + 8 \, {\left (80 \, a^{4} - 60 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 384 \, b^{4} + 48 \, {\left (60 \, a^{2} b^{2} - 11 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{161280 \, d \cos \left (d x + c\right )^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/161280*(315*(80*a^4 - 60*a^2*b^2 + 3*b^4)*cos(d*x + c)^10*log(sin(d*x + c) + 1) - 315*(80*a^4 - 60*a^2*b^2 +
3*b^4)*cos(d*x + c)^10*log(-sin(d*x + c) + 1) + 71680*a*b^3*cos(d*x + c) + 92160*(a^3*b - a*b^3)*cos(d*x + c)
^3 + 42*(15*(80*a^4 - 60*a^2*b^2 + 3*b^4)*cos(d*x + c)^8 + 10*(80*a^4 - 60*a^2*b^2 + 3*b^4)*cos(d*x + c)^6 + 8
*(80*a^4 - 60*a^2*b^2 + 3*b^4)*cos(d*x + c)^4 + 384*b^4 + 48*(60*a^2*b^2 - 11*b^4)*cos(d*x + c)^2)*sin(d*x + c
))/(d*cos(d*x + c)^10)

________________________________________________________________________________________

giac [B]  time = 0.52, size = 880, normalized size = 2.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/80640*(315*(80*a^4 - 60*a^2*b^2 + 3*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 315*(80*a^4 - 60*a^2*b^2 + 3*b
^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(55440*a^4*tan(1/2*d*x + 1/2*c)^19 + 18900*a^2*b^2*tan(1/2*d*x + 1/
2*c)^19 - 945*b^4*tan(1/2*d*x + 1/2*c)^19 - 322560*a^3*b*tan(1/2*d*x + 1/2*c)^18 - 213360*a^4*tan(1/2*d*x + 1/
2*c)^17 + 462420*a^2*b^2*tan(1/2*d*x + 1/2*c)^17 + 9135*b^4*tan(1/2*d*x + 1/2*c)^17 + 967680*a^3*b*tan(1/2*d*x
+ 1/2*c)^16 - 645120*a*b^3*tan(1/2*d*x + 1/2*c)^16 + 450240*a^4*tan(1/2*d*x + 1/2*c)^15 + 146160*a^2*b^2*tan(
1/2*d*x + 1/2*c)^15 + 218484*b^4*tan(1/2*d*x + 1/2*c)^15 - 2580480*a^3*b*tan(1/2*d*x + 1/2*c)^14 - 430080*a*b^
3*tan(1/2*d*x + 1/2*c)^14 - 624960*a^4*tan(1/2*d*x + 1/2*c)^13 + 468720*a^2*b^2*tan(1/2*d*x + 1/2*c)^13 + 6539
40*b^4*tan(1/2*d*x + 1/2*c)^13 + 5160960*a^3*b*tan(1/2*d*x + 1/2*c)^12 - 2150400*a*b^3*tan(1/2*d*x + 1/2*c)^12
+ 332640*a^4*tan(1/2*d*x + 1/2*c)^11 - 1096200*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 + 1183770*b^4*tan(1/2*d*x + 1/
2*c)^11 - 5806080*a^3*b*tan(1/2*d*x + 1/2*c)^10 + 1290240*a*b^3*tan(1/2*d*x + 1/2*c)^10 + 332640*a^4*tan(1/2*d
*x + 1/2*c)^9 - 1096200*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 1183770*b^4*tan(1/2*d*x + 1/2*c)^9 + 4515840*a^3*b*ta
n(1/2*d*x + 1/2*c)^8 - 624960*a^4*tan(1/2*d*x + 1/2*c)^7 + 468720*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 653940*b^4*
tan(1/2*d*x + 1/2*c)^7 - 2949120*a^3*b*tan(1/2*d*x + 1/2*c)^6 + 1658880*a*b^3*tan(1/2*d*x + 1/2*c)^6 + 450240*
a^4*tan(1/2*d*x + 1/2*c)^5 + 146160*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 218484*b^4*tan(1/2*d*x + 1/2*c)^5 + 11059
20*a^3*b*tan(1/2*d*x + 1/2*c)^4 + 184320*a*b^3*tan(1/2*d*x + 1/2*c)^4 - 213360*a^4*tan(1/2*d*x + 1/2*c)^3 + 46
2420*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 9135*b^4*tan(1/2*d*x + 1/2*c)^3 - 138240*a^3*b*tan(1/2*d*x + 1/2*c)^2 +
102400*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 55440*a^4*tan(1/2*d*x + 1/2*c) + 18900*a^2*b^2*tan(1/2*d*x + 1/2*c) - 94
5*b^4*tan(1/2*d*x + 1/2*c) + 46080*a^3*b - 10240*a*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^10)/d

________________________________________________________________________________________

maple [A]  time = 74.03, size = 590, normalized size = 1.45 \[ \frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{10 d \cos \left (d x +c \right )^{10}}+\frac {a^{4} \left (\sec ^{5}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{6 d}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{32 d \cos \left (d x +c \right )^{6}}+\frac {4 a^{3} b}{7 d \cos \left (d x +c \right )^{7}}-\frac {15 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{64 d}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{128 d \cos \left (d x +c \right )^{4}}+\frac {3 b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{256 d}+\frac {5 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {5 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}-\frac {8 a \,b^{3} \cos \left (d x +c \right )}{63 d}+\frac {15 a^{2} b^{2} \sin \left (d x +c \right )}{64 d}-\frac {b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{256 d}-\frac {3 b^{4} \sin \left (d x +c \right )}{256 d}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right ) a \,b^{3}}{63 d}-\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )}+\frac {5 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{6}}+\frac {15 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{64 d \cos \left (d x +c \right )^{2}}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )^{3}}+\frac {15 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{32 d \cos \left (d x +c \right )^{4}}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{21 d \cos \left (d x +c \right )^{5}}+\frac {3 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{8}}+\frac {20 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{63 d \cos \left (d x +c \right )^{7}}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{9 d \cos \left (d x +c \right )^{9}}+\frac {5 a^{4} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{24 d}-\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{256 d \cos \left (d x +c \right )^{2}}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^11*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

4/7/d*a^3*b/cos(d*x+c)^7+1/10/d*b^4*sin(d*x+c)^5/cos(d*x+c)^10-15/64/d*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))-1/256
/d*b^4*sin(d*x+c)^5/cos(d*x+c)^2+3/256/d*b^4*ln(sec(d*x+c)+tan(d*x+c))+5/16/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+5/
16*a^4*sec(d*x+c)*tan(d*x+c)/d+5/24*a^4*sec(d*x+c)^3*tan(d*x+c)/d+1/6*a^4*sec(d*x+c)^5*tan(d*x+c)/d-8/63*a*b^3
*cos(d*x+c)/d+15/64*a^2*b^2*sin(d*x+c)/d+20/63/d*a*b^3*sin(d*x+c)^4/cos(d*x+c)^7+15/64/d*a^2*b^2*sin(d*x+c)^3/
cos(d*x+c)^2-4/63/d*sin(d*x+c)^2*cos(d*x+c)*a*b^3-1/256*b^4*sin(d*x+c)^3/d-3/256*b^4*sin(d*x+c)/d+1/32/d*b^4*s
in(d*x+c)^5/cos(d*x+c)^6+4/21/d*a*b^3*sin(d*x+c)^4/cos(d*x+c)^5+4/63/d*a*b^3*sin(d*x+c)^4/cos(d*x+c)^3-4/63/d*
a*b^3*sin(d*x+c)^4/cos(d*x+c)+15/32/d*a^2*b^2*sin(d*x+c)^3/cos(d*x+c)^4+5/8/d*a^2*b^2*sin(d*x+c)^3/cos(d*x+c)^
6+1/16/d*b^4*sin(d*x+c)^5/cos(d*x+c)^8+3/4/d*a^2*b^2*sin(d*x+c)^3/cos(d*x+c)^8+4/9/d*a*b^3*sin(d*x+c)^4/cos(d*
x+c)^9+1/128/d*b^4*sin(d*x+c)^5/cos(d*x+c)^4

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 382, normalized size = 0.94 \[ -\frac {63 \, b^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{9} - 70 \, \sin \left (d x + c\right )^{7} + 128 \, \sin \left (d x + c\right )^{5} + 70 \, \sin \left (d x + c\right )^{3} - 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{10} - 5 \, \sin \left (d x + c\right )^{8} + 10 \, \sin \left (d x + c\right )^{6} - 10 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 1260 \, a^{2} b^{2} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{5} + 73 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 1680 \, a^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {92160 \, a^{3} b}{\cos \left (d x + c\right )^{7}} + \frac {10240 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 7\right )} a b^{3}}{\cos \left (d x + c\right )^{9}}}{161280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/161280*(63*b^4*(2*(15*sin(d*x + c)^9 - 70*sin(d*x + c)^7 + 128*sin(d*x + c)^5 + 70*sin(d*x + c)^3 - 15*sin(
d*x + c))/(sin(d*x + c)^10 - 5*sin(d*x + c)^8 + 10*sin(d*x + c)^6 - 10*sin(d*x + c)^4 + 5*sin(d*x + c)^2 - 1)
- 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 1260*a^2*b^2*(2*(15*sin(d*x + c)^7 - 55*sin(d*x + c)^
5 + 73*sin(d*x + c)^3 + 15*sin(d*x + c))/(sin(d*x + c)^8 - 4*sin(d*x + c)^6 + 6*sin(d*x + c)^4 - 4*sin(d*x + c
)^2 + 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) + 1680*a^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x
+ c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1
) + 15*log(sin(d*x + c) - 1)) - 92160*a^3*b/cos(d*x + c)^7 + 10240*(9*cos(d*x + c)^2 - 7)*a*b^3/cos(d*x + c)^9
)/d

________________________________________________________________________________________

mupad [B]  time = 5.15, size = 703, normalized size = 1.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^4/cos(c + d*x)^11,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((5*a^4)/8 + (3*b^4)/128 - (15*a^2*b^2)/32))/d + (tan(c/2 + (d*x)/2)^19*((11*a^4)/8
- (3*b^4)/128 + (15*a^2*b^2)/32) + tan(c/2 + (d*x)/2)^7*((519*b^4)/32 - (31*a^4)/2 + (93*a^2*b^2)/8) + tan(c/
2 + (d*x)/2)^13*((519*b^4)/32 - (31*a^4)/2 + (93*a^2*b^2)/8) + tan(c/2 + (d*x)/2)^3*((29*b^4)/128 - (127*a^4)/
24 + (367*a^2*b^2)/32) + tan(c/2 + (d*x)/2)^17*((29*b^4)/128 - (127*a^4)/24 + (367*a^2*b^2)/32) + tan(c/2 + (d
*x)/2)^5*((67*a^4)/6 + (867*b^4)/160 + (29*a^2*b^2)/8) + tan(c/2 + (d*x)/2)^15*((67*a^4)/6 + (867*b^4)/160 + (
29*a^2*b^2)/8) + tan(c/2 + (d*x)/2)^9*((33*a^4)/4 + (1879*b^4)/64 - (435*a^2*b^2)/16) + tan(c/2 + (d*x)/2)^11*
((33*a^4)/4 + (1879*b^4)/64 - (435*a^2*b^2)/16) - (16*a*b^3)/63 + (8*a^3*b)/7 + tan(c/2 + (d*x)/2)*((11*a^4)/8
- (3*b^4)/128 + (15*a^2*b^2)/32) - tan(c/2 + (d*x)/2)^16*(16*a*b^3 - 24*a^3*b) - tan(c/2 + (d*x)/2)^14*((32*a
*b^3)/3 + 64*a^3*b) + tan(c/2 + (d*x)/2)^10*(32*a*b^3 - 144*a^3*b) + tan(c/2 + (d*x)/2)^4*((32*a*b^3)/7 + (192
*a^3*b)/7) + tan(c/2 + (d*x)/2)^2*((160*a*b^3)/63 - (24*a^3*b)/7) - tan(c/2 + (d*x)/2)^12*((160*a*b^3)/3 - 128
*a^3*b) + tan(c/2 + (d*x)/2)^6*((288*a*b^3)/7 - (512*a^3*b)/7) + 112*a^3*b*tan(c/2 + (d*x)/2)^8 - 8*a^3*b*tan(
c/2 + (d*x)/2)^18)/(d*(45*tan(c/2 + (d*x)/2)^4 - 10*tan(c/2 + (d*x)/2)^2 - 120*tan(c/2 + (d*x)/2)^6 + 210*tan(
c/2 + (d*x)/2)^8 - 252*tan(c/2 + (d*x)/2)^10 + 210*tan(c/2 + (d*x)/2)^12 - 120*tan(c/2 + (d*x)/2)^14 + 45*tan(
c/2 + (d*x)/2)^16 - 10*tan(c/2 + (d*x)/2)^18 + tan(c/2 + (d*x)/2)^20 + 1))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**11*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

________________________________________________________________________________________