3.243 \(\int (a+a \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=169 \[ \frac {a^2 (9 B+10 C) \tan ^3(c+d x)}{15 d}+\frac {a^2 (9 B+10 C) \tan (c+d x)}{5 d}+\frac {a^2 (6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 (6 B+7 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d} \]

[Out]

1/8*a^2*(6*B+7*C)*arctanh(sin(d*x+c))/d+1/5*a^2*(9*B+10*C)*tan(d*x+c)/d+1/8*a^2*(6*B+7*C)*sec(d*x+c)*tan(d*x+c
)/d+1/20*a^2*(6*B+5*C)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*B*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^4*tan(d*x+c)/d+1/15*a^2
*(9*B+10*C)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.39, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3029, 2975, 2968, 3021, 2748, 3767, 3768, 3770} \[ \frac {a^2 (9 B+10 C) \tan ^3(c+d x)}{15 d}+\frac {a^2 (9 B+10 C) \tan (c+d x)}{5 d}+\frac {a^2 (6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 (6 B+7 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(a^2*(6*B + 7*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(9*B + 10*C)*Tan[c + d*x])/(5*d) + (a^2*(6*B + 7*C)*Sec[c
+ d*x]*Tan[c + d*x])/(8*d) + (a^2*(6*B + 5*C)*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (B*(a^2 + a^2*Cos[c + d*x
])*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (a^2*(9*B + 10*C)*Tan[c + d*x]^3)/(15*d)

Rule 2748

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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2975

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

Rule 3021

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 3029

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

Rule 3767

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 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 (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\int (a+a \cos (c+d x))^2 (B+C \cos (c+d x)) \sec ^6(c+d x) \, dx\\ &=\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+a \cos (c+d x)) (a (6 B+5 C)+a (3 B+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx\\ &=\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \left (a^2 (6 B+5 C)+\left (a^2 (3 B+5 C)+a^2 (6 B+5 C)\right ) \cos (c+d x)+a^2 (3 B+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a^2 (6 B+5 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int \left (4 a^2 (9 B+10 C)+5 a^2 (6 B+7 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a^2 (6 B+5 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (a^2 (6 B+7 C)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (a^2 (9 B+10 C)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {a^2 (6 B+7 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (6 B+5 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^2 (6 B+7 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (9 B+10 C)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {a^2 (6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (9 B+10 C) \tan (c+d x)}{5 d}+\frac {a^2 (6 B+7 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (6 B+5 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a^2 (9 B+10 C) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 1.33, size = 280, normalized size = 1.66 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (6 B+7 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (B+2 C) \sin (2 c+d x)+420 B \sin (c+2 d x)+420 B \sin (3 c+2 d x)+720 B \sin (2 c+3 d x)+90 B \sin (3 c+4 d x)+90 B \sin (5 c+4 d x)+144 B \sin (4 c+5 d x)+80 (15 B+14 C) \sin (d x)+330 C \sin (c+2 d x)+330 C \sin (3 c+2 d x)+800 C \sin (2 c+3 d x)+105 C \sin (3 c+4 d x)+105 C \sin (5 c+4 d x)+160 C \sin (4 c+5 d x))\right )}{7680 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

-1/7680*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*Sec[c + d*x]^5*(240*(6*B + 7*C)*Cos[c + d*x]^5*(Log[Cos[(
c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(80*(15*B + 14*C)*Sin[d*x
] - 240*(B + 2*C)*Sin[2*c + d*x] + 420*B*Sin[c + 2*d*x] + 330*C*Sin[c + 2*d*x] + 420*B*Sin[3*c + 2*d*x] + 330*
C*Sin[3*c + 2*d*x] + 720*B*Sin[2*c + 3*d*x] + 800*C*Sin[2*c + 3*d*x] + 90*B*Sin[3*c + 4*d*x] + 105*C*Sin[3*c +
4*d*x] + 90*B*Sin[5*c + 4*d*x] + 105*C*Sin[5*c + 4*d*x] + 144*B*Sin[4*c + 5*d*x] + 160*C*Sin[4*c + 5*d*x])))/
d

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fricas [A]  time = 0.46, size = 165, normalized size = 0.98 \[ \frac {15 \, {\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (9 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (9 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) + 24 \, B a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/240*(15*(6*B + 7*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(6*B + 7*C)*a^2*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(16*(9*B + 10*C)*a^2*cos(d*x + c)^4 + 15*(6*B + 7*C)*a^2*cos(d*x + c)^3 + 8*(9*B + 10*C)*a^2*c
os(d*x + c)^2 + 30*(2*B + C)*a^2*cos(d*x + c) + 24*B*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [A]  time = 0.80, size = 246, normalized size = 1.46 \[ \frac {15 \, {\left (6 \, B a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (6 \, B a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (90 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 420 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 490 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 864 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 540 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 790 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 390 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="giac")

[Out]

1/120*(15*(6*B*a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(6*B*a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x
+ 1/2*c) - 1)) - 2*(90*B*a^2*tan(1/2*d*x + 1/2*c)^9 + 105*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 420*B*a^2*tan(1/2*d*x
+ 1/2*c)^7 - 490*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 864*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*C*a^2*tan(1/2*d*x + 1/
2*c)^5 - 540*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 790*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 390*B*a^2*tan(1/2*d*x + 1/2*c)
+ 375*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 0.42, size = 235, normalized size = 1.39 \[ \frac {7 a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {7 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {6 a^{2} B \tan \left (d x +c \right )}{5 d}+\frac {3 a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 a^{2} C \tan \left (d x +c \right )}{3 d}+\frac {2 a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{2} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)

[Out]

7/8/d*a^2*C*sec(d*x+c)*tan(d*x+c)+7/8/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+6/5*a^2*B*tan(d*x+c)/d+3/5*a^2*B*sec(d
*x+c)^2*tan(d*x+c)/d+4/3/d*a^2*C*tan(d*x+c)+2/3/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2+1/2*a^2*B*sec(d*x+c)^3*tan(d*x
+c)/d+3/4*a^2*B*sec(d*x+c)*tan(d*x+c)/d+3/4/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*a^2*C*tan(d*x+c)*sec(d*x+c
)^3+1/5/d*a^2*B*tan(d*x+c)*sec(d*x+c)^4

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maxima [A]  time = 0.46, size = 278, normalized size = 1.64 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 30 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, C a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c)
)*B*a^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 30*B*a^2*(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)) - 15*C*a^2*(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*C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 3.72, size = 224, normalized size = 1.33 \[ \frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,B+7\,C\right )}{4\,d}-\frac {\left (\frac {3\,B\,a^2}{2}+\frac {7\,C\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-7\,B\,a^2-\frac {49\,C\,a^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,B\,a^2}{5}+\frac {40\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-9\,B\,a^2-\frac {79\,C\,a^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,B\,a^2}{2}+\frac {25\,C\,a^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^2)/cos(c + d*x)^7,x)

[Out]

(a^2*atanh(tan(c/2 + (d*x)/2))*(6*B + 7*C))/(4*d) - (tan(c/2 + (d*x)/2)*((13*B*a^2)/2 + (25*C*a^2)/4) + tan(c/
2 + (d*x)/2)^9*((3*B*a^2)/2 + (7*C*a^2)/4) - tan(c/2 + (d*x)/2)^7*(7*B*a^2 + (49*C*a^2)/6) - tan(c/2 + (d*x)/2
)^3*(9*B*a^2 + (79*C*a^2)/6) + tan(c/2 + (d*x)/2)^5*((72*B*a^2)/5 + (40*C*a^2)/3))/(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))

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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((a+a*cos(d*x+c))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)

[Out]

Timed out

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