3.325 \(\int \cos (c+d x) (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=111 \[ \frac {5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac {a^3 (7 B+5 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(3 B+5 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+a^3 B x+\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4072, 3917, 3914, 3767, 8, 3770} \[ \frac {5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac {a^3 (7 B+5 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(3 B+5 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+a^3 B x+\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

a^3*B*x + (a^3*(7*B + 5*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(B + C)*Tan[c + d*x])/(2*d) + (a*C*(a + a*Sec
[c + d*x])^2*Tan[c + d*x])/(3*d) + ((3*B + 5*C)*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 3770

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

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4072

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

Rubi steps

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

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Mathematica [B]  time = 6.45, size = 772, normalized size = 6.95 \[ a^3 \left (\frac {(\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (9 B \sin \left (\frac {d x}{2}\right )+11 C \sin \left (\frac {d x}{2}\right )\right )}{24 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {(\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (9 B \sin \left (\frac {d x}{2}\right )+11 C \sin \left (\frac {d x}{2}\right )\right )}{24 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {(\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-3 B \sin \left (\frac {c}{2}\right )+3 B \cos \left (\frac {c}{2}\right )-8 C \sin \left (\frac {c}{2}\right )+10 C \cos \left (\frac {c}{2}\right )\right )}{96 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {(\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-3 B \sin \left (\frac {c}{2}\right )-3 B \cos \left (\frac {c}{2}\right )-8 C \sin \left (\frac {c}{2}\right )-10 C \cos \left (\frac {c}{2}\right )\right )}{96 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {(-7 B-5 C) (\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{16 d}+\frac {(7 B+5 C) (\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{16 d}+\frac {1}{8} B x (\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )+\frac {C \sin \left (\frac {d x}{2}\right ) (\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{48 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {C \sin \left (\frac {d x}{2}\right ) (\cos (c+d x)+1)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{48 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

a^3*((B*x*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6)/8 + ((-7*B - 5*C)*(1 + Cos[c + d*x])^3*Log[Cos[c/2 + (d*x
)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6)/(16*d) + ((7*B + 5*C)*(1 + Cos[c + d*x])^3*Log[Cos[c/2 + (d*x
)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6)/(16*d) + (C*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*Sin[(d*
x)/2])/(48*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + ((1 + Cos[c + d*x])^3*Sec[c/
2 + (d*x)/2]^6*(3*B*Cos[c/2] + 10*C*Cos[c/2] - 3*B*Sin[c/2] - 8*C*Sin[c/2]))/(96*d*(Cos[c/2] - Sin[c/2])*(Cos[
c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + ((1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(9*B*Sin[(d*x)/2] + 11*C*
Sin[(d*x)/2]))/(24*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (C*(1 + Cos[c + d*x])^
3*Sec[c/2 + (d*x)/2]^6*Sin[(d*x)/2])/(48*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3)
+ ((1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-3*B*Cos[c/2] - 10*C*Cos[c/2] - 3*B*Sin[c/2] - 8*C*Sin[c/2]))/(9
6*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + ((1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)
/2]^6*(9*B*Sin[(d*x)/2] + 11*C*Sin[(d*x)/2]))/(24*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x
)/2])))

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fricas [A]  time = 0.49, size = 141, normalized size = 1.27 \[ \frac {12 \, B a^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (7 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (9 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, C a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(12*B*a^3*d*x*cos(d*x + c)^3 + 3*(7*B + 5*C)*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(7*B + 5*C)*a^3
*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(9*B + 11*C)*a^3*cos(d*x + c)^2 + 3*(B + 3*C)*a^3*cos(d*x + c) +
2*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.34, size = 189, normalized size = 1.70 \[ \frac {6 \, {\left (d x + c\right )} B a^{3} + 3 \, {\left (7 \, B a^{3} + 5 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (7 \, B a^{3} + 5 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*B*a^3 + 3*(7*B*a^3 + 5*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(7*B*a^3 + 5*C*a^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 36*B*a
^3*tan(1/2*d*x + 1/2*c)^3 - 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*B*a^3*tan(1/2*d*x + 1/2*c) + 33*C*a^3*tan(1/2
*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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maple [A]  time = 1.53, size = 158, normalized size = 1.42 \[ a^{3} B x +\frac {a^{3} B c}{d}+\frac {5 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {7 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {11 a^{3} C \tan \left (d x +c \right )}{3 d}+\frac {3 a^{3} B \tan \left (d x +c \right )}{d}+\frac {3 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

a^3*B*x+1/d*a^3*B*c+5/2/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+7/2/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+11/3*a^3*C*tan
(d*x+c)/d+3/d*a^3*B*tan(d*x+c)+3/2/d*C*a^3*sec(d*x+c)*tan(d*x+c)+1/2/d*a^3*B*sec(d*x+c)*tan(d*x+c)+1/3/d*C*a^3
*tan(d*x+c)*sec(d*x+c)^2

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maxima [B]  time = 0.45, size = 212, normalized size = 1.91 \[ \frac {12 \, {\left (d x + c\right )} B a^{3} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 9 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{3} \tan \left (d x + c\right ) + 36 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*B*a^3 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 3*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2
- 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 9*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(si
n(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 18*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*C*a^3*
(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*B*a^3*tan(d*x + c) + 36*C*a^3*tan(d*x + c))/d

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mupad [B]  time = 3.01, size = 209, normalized size = 1.88 \[ \frac {2\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {11\,C\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)

[Out]

(2*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (5*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*B*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (B
*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (11*C*a^3*sin(c + d*x))/(3*d*cos(c + d*x)) + (3*C*a^3*sin(c + d*x))/
(2*d*cos(c + d*x)^2) + (C*a^3*sin(c + d*x))/(3*d*cos(c + d*x)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

a**3*(Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(3*B*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(3*B*
cos(c + d*x)*sec(c + d*x)**3, x) + Integral(B*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(C*cos(c + d*x)*sec(c
+ d*x)**2, x) + Integral(3*C*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(3*C*cos(c + d*x)*sec(c + d*x)**4, x)
+ Integral(C*cos(c + d*x)*sec(c + d*x)**5, x))

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