3.91 \(\int \sec ^{12}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=254 \[ \frac {a^4 \tan (c+d x)}{d}+\frac {2 a^3 b \tan ^2(c+d x)}{d}+\frac {b^2 \left (2 a^2+b^2\right ) \tan ^9(c+d x)}{3 d}+\frac {a b \left (a^2+3 b^2\right ) \tan ^8(c+d x)}{2 d}+\frac {2 a b \left (a^2+b^2\right ) \tan ^6(c+d x)}{d}+\frac {a b \left (3 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac {a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{d}+\frac {\left (a^4+18 a^2 b^2+3 b^4\right ) \tan ^7(c+d x)}{7 d}+\frac {\left (3 a^4+18 a^2 b^2+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac {2 a b^3 \tan ^{10}(c+d x)}{5 d}+\frac {b^4 \tan ^{11}(c+d x)}{11 d} \]

[Out]

a^4*tan(d*x+c)/d+2*a^3*b*tan(d*x+c)^2/d+a^2*(a^2+2*b^2)*tan(d*x+c)^3/d+a*b*(3*a^2+b^2)*tan(d*x+c)^4/d+1/5*(3*a
^4+18*a^2*b^2+b^4)*tan(d*x+c)^5/d+2*a*b*(a^2+b^2)*tan(d*x+c)^6/d+1/7*(a^4+18*a^2*b^2+3*b^4)*tan(d*x+c)^7/d+1/2
*a*b*(a^2+3*b^2)*tan(d*x+c)^8/d+1/3*b^2*(2*a^2+b^2)*tan(d*x+c)^9/d+2/5*a*b^3*tan(d*x+c)^10/d+1/11*b^4*tan(d*x+
c)^11/d

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Rubi [A]  time = 0.22, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3088, 948} \[ \frac {b^2 \left (2 a^2+b^2\right ) \tan ^9(c+d x)}{3 d}+\frac {a b \left (a^2+3 b^2\right ) \tan ^8(c+d x)}{2 d}+\frac {\left (18 a^2 b^2+a^4+3 b^4\right ) \tan ^7(c+d x)}{7 d}+\frac {2 a b \left (a^2+b^2\right ) \tan ^6(c+d x)}{d}+\frac {\left (18 a^2 b^2+3 a^4+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac {a b \left (3 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac {a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{d}+\frac {2 a^3 b \tan ^2(c+d x)}{d}+\frac {a^4 \tan (c+d x)}{d}+\frac {2 a b^3 \tan ^{10}(c+d x)}{5 d}+\frac {b^4 \tan ^{11}(c+d x)}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^12*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(a^4*Tan[c + d*x])/d + (2*a^3*b*Tan[c + d*x]^2)/d + (a^2*(a^2 + 2*b^2)*Tan[c + d*x]^3)/d + (a*b*(3*a^2 + b^2)*
Tan[c + d*x]^4)/d + ((3*a^4 + 18*a^2*b^2 + b^4)*Tan[c + d*x]^5)/(5*d) + (2*a*b*(a^2 + b^2)*Tan[c + d*x]^6)/d +
((a^4 + 18*a^2*b^2 + 3*b^4)*Tan[c + d*x]^7)/(7*d) + (a*b*(a^2 + 3*b^2)*Tan[c + d*x]^8)/(2*d) + (b^2*(2*a^2 +
b^2)*Tan[c + d*x]^9)/(3*d) + (2*a*b^3*Tan[c + d*x]^10)/(5*d) + (b^4*Tan[c + d*x]^11)/(11*d)

Rule 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

\begin {align*} \int \sec ^{12}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(b+a x)^4 \left (1+x^2\right )^3}{x^{12}} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {b^4}{x^{12}}+\frac {4 a b^3}{x^{11}}+\frac {3 \left (2 a^2 b^2+b^4\right )}{x^{10}}+\frac {4 a b \left (a^2+3 b^2\right )}{x^9}+\frac {a^4+18 a^2 b^2+3 b^4}{x^8}+\frac {12 a b \left (a^2+b^2\right )}{x^7}+\frac {3 a^4+18 a^2 b^2+b^4}{x^6}+\frac {4 a b \left (3 a^2+b^2\right )}{x^5}+\frac {3 \left (a^4+2 a^2 b^2\right )}{x^4}+\frac {4 a^3 b}{x^3}+\frac {a^4}{x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {a^4 \tan (c+d x)}{d}+\frac {2 a^3 b \tan ^2(c+d x)}{d}+\frac {a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{d}+\frac {a b \left (3 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac {\left (3 a^4+18 a^2 b^2+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac {2 a b \left (a^2+b^2\right ) \tan ^6(c+d x)}{d}+\frac {\left (a^4+18 a^2 b^2+3 b^4\right ) \tan ^7(c+d x)}{7 d}+\frac {a b \left (a^2+3 b^2\right ) \tan ^8(c+d x)}{2 d}+\frac {b^2 \left (2 a^2+b^2\right ) \tan ^9(c+d x)}{3 d}+\frac {2 a b^3 \tan ^{10}(c+d x)}{5 d}+\frac {b^4 \tan ^{11}(c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]  time = 1.80, size = 175, normalized size = 0.69 \[ \frac {\frac {1}{3} \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^9-\frac {1}{2} a \left (5 a^2+3 b^2\right ) (a+b \tan (c+d x))^8+\frac {3}{7} \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^7-a \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^6+\frac {1}{5} \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^5+\frac {1}{11} (a+b \tan (c+d x))^{11}-\frac {3}{5} a (a+b \tan (c+d x))^{10}}{b^7 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^12*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(((a^2 + b^2)^3*(a + b*Tan[c + d*x])^5)/5 - a*(a^2 + b^2)^2*(a + b*Tan[c + d*x])^6 + (3*(a^2 + b^2)*(5*a^2 + b
^2)*(a + b*Tan[c + d*x])^7)/7 - (a*(5*a^2 + 3*b^2)*(a + b*Tan[c + d*x])^8)/2 + ((5*a^2 + b^2)*(a + b*Tan[c + d
*x])^9)/3 - (3*a*(a + b*Tan[c + d*x])^10)/5 + (a + b*Tan[c + d*x])^11/11)/(b^7*d)

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fricas [A]  time = 0.48, size = 194, normalized size = 0.76 \[ \frac {924 \, a b^{3} \cos \left (d x + c\right ) + 1155 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (16 \, {\left (33 \, a^{4} - 22 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{10} + 8 \, {\left (33 \, a^{4} - 22 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{8} + 6 \, {\left (33 \, a^{4} - 22 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} + 5 \, {\left (33 \, a^{4} - 22 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 105 \, b^{4} + 70 \, {\left (11 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2310 \, d \cos \left (d x + c\right )^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/2310*(924*a*b^3*cos(d*x + c) + 1155*(a^3*b - a*b^3)*cos(d*x + c)^3 + 2*(16*(33*a^4 - 22*a^2*b^2 + b^4)*cos(d
*x + c)^10 + 8*(33*a^4 - 22*a^2*b^2 + b^4)*cos(d*x + c)^8 + 6*(33*a^4 - 22*a^2*b^2 + b^4)*cos(d*x + c)^6 + 5*(
33*a^4 - 22*a^2*b^2 + b^4)*cos(d*x + c)^4 + 105*b^4 + 70*(11*a^2*b^2 - 2*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d
*cos(d*x + c)^11)

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giac [A]  time = 0.54, size = 284, normalized size = 1.12 \[ \frac {210 \, b^{4} \tan \left (d x + c\right )^{11} + 924 \, a b^{3} \tan \left (d x + c\right )^{10} + 1540 \, a^{2} b^{2} \tan \left (d x + c\right )^{9} + 770 \, b^{4} \tan \left (d x + c\right )^{9} + 1155 \, a^{3} b \tan \left (d x + c\right )^{8} + 3465 \, a b^{3} \tan \left (d x + c\right )^{8} + 330 \, a^{4} \tan \left (d x + c\right )^{7} + 5940 \, a^{2} b^{2} \tan \left (d x + c\right )^{7} + 990 \, b^{4} \tan \left (d x + c\right )^{7} + 4620 \, a^{3} b \tan \left (d x + c\right )^{6} + 4620 \, a b^{3} \tan \left (d x + c\right )^{6} + 1386 \, a^{4} \tan \left (d x + c\right )^{5} + 8316 \, a^{2} b^{2} \tan \left (d x + c\right )^{5} + 462 \, b^{4} \tan \left (d x + c\right )^{5} + 6930 \, a^{3} b \tan \left (d x + c\right )^{4} + 2310 \, a b^{3} \tan \left (d x + c\right )^{4} + 2310 \, a^{4} \tan \left (d x + c\right )^{3} + 4620 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 4620 \, a^{3} b \tan \left (d x + c\right )^{2} + 2310 \, a^{4} \tan \left (d x + c\right )}{2310 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/2310*(210*b^4*tan(d*x + c)^11 + 924*a*b^3*tan(d*x + c)^10 + 1540*a^2*b^2*tan(d*x + c)^9 + 770*b^4*tan(d*x +
c)^9 + 1155*a^3*b*tan(d*x + c)^8 + 3465*a*b^3*tan(d*x + c)^8 + 330*a^4*tan(d*x + c)^7 + 5940*a^2*b^2*tan(d*x +
c)^7 + 990*b^4*tan(d*x + c)^7 + 4620*a^3*b*tan(d*x + c)^6 + 4620*a*b^3*tan(d*x + c)^6 + 1386*a^4*tan(d*x + c)
^5 + 8316*a^2*b^2*tan(d*x + c)^5 + 462*b^4*tan(d*x + c)^5 + 6930*a^3*b*tan(d*x + c)^4 + 2310*a*b^3*tan(d*x + c
)^4 + 2310*a^4*tan(d*x + c)^3 + 4620*a^2*b^2*tan(d*x + c)^3 + 4620*a^3*b*tan(d*x + c)^2 + 2310*a^4*tan(d*x + c
))/d

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maple [A]  time = 78.72, size = 300, normalized size = 1.18 \[ \frac {-a^{4} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )+\frac {a^{3} b}{2 \cos \left (d x +c \right )^{8}}+6 a^{2} b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+4 a \,b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{40 \cos \left (d x +c \right )^{4}}\right )+b^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{11 \cos \left (d x +c \right )^{11}}+\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{33 \cos \left (d x +c \right )^{9}}+\frac {8 \left (\sin ^{5}\left (d x +c \right )\right )}{231 \cos \left (d x +c \right )^{7}}+\frac {16 \left (\sin ^{5}\left (d x +c \right )\right )}{1155 \cos \left (d x +c \right )^{5}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^12*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

1/d*(-a^4*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c)+1/2*a^3*b/cos(d*x+c)^8+6*a^
2*b^2*(1/9*sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x+c)^3/cos(d*x+c)^5+16/315*sin
(d*x+c)^3/cos(d*x+c)^3)+4*a*b^3*(1/10*sin(d*x+c)^4/cos(d*x+c)^10+3/40*sin(d*x+c)^4/cos(d*x+c)^8+1/20*sin(d*x+c
)^4/cos(d*x+c)^6+1/40*sin(d*x+c)^4/cos(d*x+c)^4)+b^4*(1/11*sin(d*x+c)^5/cos(d*x+c)^11+2/33*sin(d*x+c)^5/cos(d*
x+c)^9+8/231*sin(d*x+c)^5/cos(d*x+c)^7+16/1155*sin(d*x+c)^5/cos(d*x+c)^5))

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maxima [A]  time = 0.33, size = 233, normalized size = 0.92 \[ \frac {66 \, {\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} a^{4} + 44 \, {\left (35 \, \tan \left (d x + c\right )^{9} + 135 \, \tan \left (d x + c\right )^{7} + 189 \, \tan \left (d x + c\right )^{5} + 105 \, \tan \left (d x + c\right )^{3}\right )} a^{2} b^{2} + 2 \, {\left (105 \, \tan \left (d x + c\right )^{11} + 385 \, \tan \left (d x + c\right )^{9} + 495 \, \tan \left (d x + c\right )^{7} + 231 \, \tan \left (d x + c\right )^{5}\right )} b^{4} - \frac {231 \, {\left (5 \, \sin \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\sin \left (d x + c\right )^{10} - 5 \, \sin \left (d x + c\right )^{8} + 10 \, \sin \left (d x + c\right )^{6} - 10 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} - 1} + \frac {1155 \, a^{3} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{4}}}{2310 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/2310*(66*(5*tan(d*x + c)^7 + 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*a^4 + 44*(35*tan(d*x +
c)^9 + 135*tan(d*x + c)^7 + 189*tan(d*x + c)^5 + 105*tan(d*x + c)^3)*a^2*b^2 + 2*(105*tan(d*x + c)^11 + 385*t
an(d*x + c)^9 + 495*tan(d*x + c)^7 + 231*tan(d*x + c)^5)*b^4 - 231*(5*sin(d*x + c)^2 - 1)*a*b^3/(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) + 1155*a^3*b/(sin(d*x +
c)^2 - 1)^4)/d

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mupad [B]  time = 4.82, size = 560, normalized size = 2.20 \[ -\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {226\,a^4}{5}-\frac {64\,a^2\,b^2}{5}+\frac {32\,b^4}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}\,\left (\frac {226\,a^4}{5}-\frac {64\,a^2\,b^2}{5}+\frac {32\,b^4}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {1308\,a^4}{7}-\frac {3008\,a^2\,b^2}{21}+\frac {992\,b^4}{21}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {1308\,a^4}{7}-\frac {3008\,a^2\,b^2}{21}+\frac {992\,b^4}{21}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (-\frac {3952\,a^4}{35}+\frac {3008\,a^2\,b^2}{35}+\frac {576\,b^4}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}\,\left (-\frac {3952\,a^4}{35}+\frac {3008\,a^2\,b^2}{35}+\frac {576\,b^4}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (-\frac {1528\,a^4}{7}+\frac {2272\,a^2\,b^2}{21}+\frac {10624\,b^4}{231}\right )+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{21}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^4-16\,a^2\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{19}\,\left (12\,a^4-16\,a^2\,b^2\right )+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (16\,a\,b^3-24\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}\,\left (16\,a\,b^3-24\,a^3\,b\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (80\,a^3\,b+16\,a\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}\,\left (80\,a^3\,b+16\,a\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (80\,a\,b^3-176\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (80\,a\,b^3-176\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {112\,a\,b^3}{5}-224\,a^3\,b\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {112\,a\,b^3}{5}-224\,a^3\,b\right )+8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{20}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^{11}} \]

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)^12,x)

[Out]

-(tan(c/2 + (d*x)/2)^5*((226*a^4)/5 + (32*b^4)/5 - (64*a^2*b^2)/5) + tan(c/2 + (d*x)/2)^17*((226*a^4)/5 + (32*
b^4)/5 - (64*a^2*b^2)/5) + tan(c/2 + (d*x)/2)^9*((1308*a^4)/7 + (992*b^4)/21 - (3008*a^2*b^2)/21) + tan(c/2 +
(d*x)/2)^13*((1308*a^4)/7 + (992*b^4)/21 - (3008*a^2*b^2)/21) + tan(c/2 + (d*x)/2)^7*((576*b^4)/35 - (3952*a^4
)/35 + (3008*a^2*b^2)/35) + tan(c/2 + (d*x)/2)^15*((576*b^4)/35 - (3952*a^4)/35 + (3008*a^2*b^2)/35) + tan(c/2
+ (d*x)/2)^11*((10624*b^4)/231 - (1528*a^4)/7 + (2272*a^2*b^2)/21) + 2*a^4*tan(c/2 + (d*x)/2)^21 - tan(c/2 +
(d*x)/2)^3*(12*a^4 - 16*a^2*b^2) - tan(c/2 + (d*x)/2)^19*(12*a^4 - 16*a^2*b^2) + 2*a^4*tan(c/2 + (d*x)/2) + ta
n(c/2 + (d*x)/2)^4*(16*a*b^3 - 24*a^3*b) - tan(c/2 + (d*x)/2)^18*(16*a*b^3 - 24*a^3*b) + tan(c/2 + (d*x)/2)^6*
(16*a*b^3 + 80*a^3*b) - tan(c/2 + (d*x)/2)^16*(16*a*b^3 + 80*a^3*b) + tan(c/2 + (d*x)/2)^8*(80*a*b^3 - 176*a^3
*b) - tan(c/2 + (d*x)/2)^14*(80*a*b^3 - 176*a^3*b) - tan(c/2 + (d*x)/2)^10*((112*a*b^3)/5 - 224*a^3*b) + tan(c
/2 + (d*x)/2)^12*((112*a*b^3)/5 - 224*a^3*b) + 8*a^3*b*tan(c/2 + (d*x)/2)^2 - 8*a^3*b*tan(c/2 + (d*x)/2)^20)/(
d*(tan(c/2 + (d*x)/2)^2 - 1)^11)

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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)**12*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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