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

Optimal. Leaf size=162 \[ \frac {5 a^3 (4 A+4 B+3 C) \tan (c+d x)}{8 d}+\frac {a^3 (28 A+20 B+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{24 d}+a^3 A x+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 a d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[Out]

a^3*A*x+1/8*a^3*(28*A+20*B+15*C)*arctanh(sin(d*x+c))/d+5/8*a^3*(4*A+4*B+3*C)*tan(d*x+c)/d+1/4*C*(a+a*sec(d*x+c
))^3*tan(d*x+c)/d+1/12*(4*B+3*C)*(a^2+a^2*sec(d*x+c))^2*tan(d*x+c)/a/d+1/24*(12*A+20*B+15*C)*(a^3+a^3*sec(d*x+
c))*tan(d*x+c)/d

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Rubi [A]  time = 0.24, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4054, 3917, 3914, 3767, 8, 3770} \[ \frac {5 a^3 (4 A+4 B+3 C) \tan (c+d x)}{8 d}+\frac {a^3 (28 A+20 B+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{24 d}+a^3 A x+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 a d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*A*x + (a^3*(28*A + 20*B + 15*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*(4*A + 4*B + 3*C)*Tan[c + d*x])/(8*d
) + (C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) + ((4*B + 3*C)*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(12*
a*d) + ((12*A + 20*B + 15*C)*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(24*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 4054

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

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\int (a+a \sec (c+d x))^3 (4 a A+a (4 B+3 C) \sec (c+d x)) \, dx}{4 a}\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 a d}+\frac {\int (a+a \sec (c+d x))^2 \left (12 a^2 A+a^2 (12 A+20 B+15 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac {\int (a+a \sec (c+d x)) \left (24 a^3 A+15 a^3 (4 A+4 B+3 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=a^3 A x+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac {1}{8} \left (5 a^3 (4 A+4 B+3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (a^3 (28 A+20 B+15 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 A x+\frac {a^3 (28 A+20 B+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{24 d}-\frac {\left (5 a^3 (4 A+4 B+3 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{8 d}\\ &=a^3 A x+\frac {a^3 (28 A+20 B+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^3 (4 A+4 B+3 C) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{24 d}\\ \end {align*}

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Mathematica [B]  time = 3.20, size = 464, normalized size = 2.86 \[ \frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (\sec (c) (12 A \sin (2 c+d x)+216 A \sin (c+2 d x)-72 A \sin (3 c+2 d x)+12 A \sin (2 c+3 d x)+12 A \sin (4 c+3 d x)+72 A \sin (3 c+4 d x)+72 A d x \cos (c)+48 A d x \cos (c+2 d x)+48 A d x \cos (3 c+2 d x)+12 A d x \cos (3 c+4 d x)+12 A d x \cos (5 c+4 d x)-216 A \sin (c)+12 A \sin (d x)+36 B \sin (2 c+d x)+280 B \sin (c+2 d x)-72 B \sin (3 c+2 d x)+36 B \sin (2 c+3 d x)+36 B \sin (4 c+3 d x)+88 B \sin (3 c+4 d x)-264 B \sin (c)+36 B \sin (d x)+69 C \sin (2 c+d x)+264 C \sin (c+2 d x)-24 C \sin (3 c+2 d x)+45 C \sin (2 c+3 d x)+45 C \sin (4 c+3 d x)+72 C \sin (3 c+4 d x)-216 C \sin (c)+69 C \sin (d x))-24 (28 A+20 B+15 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{768 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Cos[c + d*x])^3*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^6*Sec[c + d*x]^4*(-24*(28*A
+ 20*B + 15*C)*Cos[c + d*x]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]]) + Sec[c]*(72*A*d*x*Cos[c] + 48*A*d*x*Cos[c + 2*d*x] + 48*A*d*x*Cos[3*c + 2*d*x] + 12*A*d*x*Cos[3*c + 4*
d*x] + 12*A*d*x*Cos[5*c + 4*d*x] - 216*A*Sin[c] - 264*B*Sin[c] - 216*C*Sin[c] + 12*A*Sin[d*x] + 36*B*Sin[d*x]
+ 69*C*Sin[d*x] + 12*A*Sin[2*c + d*x] + 36*B*Sin[2*c + d*x] + 69*C*Sin[2*c + d*x] + 216*A*Sin[c + 2*d*x] + 280
*B*Sin[c + 2*d*x] + 264*C*Sin[c + 2*d*x] - 72*A*Sin[3*c + 2*d*x] - 72*B*Sin[3*c + 2*d*x] - 24*C*Sin[3*c + 2*d*
x] + 12*A*Sin[2*c + 3*d*x] + 36*B*Sin[2*c + 3*d*x] + 45*C*Sin[2*c + 3*d*x] + 12*A*Sin[4*c + 3*d*x] + 36*B*Sin[
4*c + 3*d*x] + 45*C*Sin[4*c + 3*d*x] + 72*A*Sin[3*c + 4*d*x] + 88*B*Sin[3*c + 4*d*x] + 72*C*Sin[3*c + 4*d*x]))
)/(768*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.47, size = 173, normalized size = 1.07 \[ \frac {48 \, A a^{3} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (28 \, A + 20 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (28 \, A + 20 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (9 \, A + 11 \, B + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 12 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*A*a^3*d*x*cos(d*x + c)^4 + 3*(28*A + 20*B + 15*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(28*A
+ 20*B + 15*C)*a^3*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(9*A + 11*B + 9*C)*a^3*cos(d*x + c)^3 + 3*(4*A
+ 12*B + 15*C)*a^3*cos(d*x + c)^2 + 8*(B + 3*C)*a^3*cos(d*x + c) + 6*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.33, size = 301, normalized size = 1.86 \[ \frac {24 \, {\left (d x + c\right )} A a^{3} + 3 \, {\left (28 \, A a^{3} + 20 \, B a^{3} + 15 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (28 \, A a^{3} + 20 \, B a^{3} + 15 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 204 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 228 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 84 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 147 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*(d*x + c)*A*a^3 + 3*(28*A*a^3 + 20*B*a^3 + 15*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(28*A*a^3
+ 20*B*a^3 + 15*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(60*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*B*a^3*tan
(1/2*d*x + 1/2*c)^7 + 45*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 204*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 220*B*a^3*tan(1/2*d
*x + 1/2*c)^5 - 165*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 228*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 292*B*a^3*tan(1/2*d*x +
1/2*c)^3 + 219*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 84*A*a^3*tan(1/2*d*x + 1/2*c) - 132*B*a^3*tan(1/2*d*x + 1/2*c) -
147*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 1.46, size = 262, normalized size = 1.62 \[ a^{3} A x +\frac {A \,a^{3} c}{d}+\frac {5 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} C \tan \left (d x +c \right )}{d}+\frac {7 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {11 a^{3} B \tan \left (d x +c \right )}{3 d}+\frac {15 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {3 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*A*x+1/d*A*a^3*c+5/2/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*C*tan(d*x+c)/d+7/2/d*A*a^3*ln(sec(d*x+c)+tan(d
*x+c))+11/3/d*a^3*B*tan(d*x+c)+15/8/d*C*a^3*sec(d*x+c)*tan(d*x+c)+15/8/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*A
*a^3*tan(d*x+c)+3/2/d*a^3*B*sec(d*x+c)*tan(d*x+c)+1/d*C*a^3*tan(d*x+c)*sec(d*x+c)^2+1/2/d*A*a^3*sec(d*x+c)*tan
(d*x+c)+1/3/d*a^3*B*tan(d*x+c)*sec(d*x+c)^2+1/4/d*C*a^3*tan(d*x+c)*sec(d*x+c)^3

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maxima [B]  time = 0.39, size = 352, normalized size = 2.17 \[ \frac {48 \, {\left (d x + c\right )} A a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 144 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 48 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 144 \, A a^{3} \tan \left (d x + c\right ) + 144 \, B a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*(d*x + c)*A*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 48*(tan(d*x + c)^3 + 3*tan(d*x + c))*C
*a^3 - 3*C*a^3*(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)) - 12*A*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)) - 36*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)) - 36*C*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)) +
144*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 48*B*a^3*log(sec(d*x + c) + tan(d*x + c)) + 144*A*a^3*tan(d*x + c
) + 144*B*a^3*tan(d*x + c) + 48*C*a^3*tan(d*x + c))/d

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mupad [B]  time = 4.91, size = 611, normalized size = 3.77 \[ \frac {9\,A\,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 )+\frac {63\,A\,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 )}{2}+\frac {45\,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 )}{2}+\frac {135\,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 )}{8}+9\,A\,a^3\,\sin \left (2\,c+2\,d\,x\right )+\frac {3\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {9\,A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{2}+13\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )+\frac {9\,B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {11\,B\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{2}+15\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )+\frac {45\,C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {9\,C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{2}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{2}+\frac {9\,B\,a^3\,\sin \left (c+d\,x\right )}{2}+\frac {69\,C\,a^3\,\sin \left (c+d\,x\right )}{8}+12\,A\,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 )\,\cos \left (2\,c+2\,d\,x\right )+3\,A\,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 )\,\cos \left (4\,c+4\,d\,x\right )+42\,A\,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 )\,\cos \left (2\,c+2\,d\,x\right )+\frac {21\,A\,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 )\,\cos \left (4\,c+4\,d\,x\right )}{2}+30\,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 )\,\cos \left (2\,c+2\,d\,x\right )+\frac {15\,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 )\,\cos \left (4\,c+4\,d\,x\right )}{2}+\frac {45\,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 )\,\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {45\,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 )\,\cos \left (4\,c+4\,d\,x\right )}{8}}{12\,d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {\cos \left (4\,c+4\,d\,x\right )}{8}+\frac {3}{8}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (63*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))
/2 + (45*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (135*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2)))/8 + 9*A*a^3*sin(2*c + 2*d*x) + (3*A*a^3*sin(3*c + 3*d*x))/2 + (9*A*a^3*sin(4*c + 4*d*x))/2 + 13*B*
a^3*sin(2*c + 2*d*x) + (9*B*a^3*sin(3*c + 3*d*x))/2 + (11*B*a^3*sin(4*c + 4*d*x))/2 + 15*C*a^3*sin(2*c + 2*d*x
) + (45*C*a^3*sin(3*c + 3*d*x))/8 + (9*C*a^3*sin(4*c + 4*d*x))/2 + (3*A*a^3*sin(c + d*x))/2 + (9*B*a^3*sin(c +
d*x))/2 + (69*C*a^3*sin(c + d*x))/8 + 12*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) +
3*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x) + 42*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(
c/2 + (d*x)/2))*cos(2*c + 2*d*x) + (21*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/2
+ 30*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + (15*B*a^3*atanh(sin(c/2 + (d*x)/2)/
cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/2 + (45*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d
*x))/2 + (45*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/8)/(12*d*(cos(2*c + 2*d*x)/2
+ cos(4*c + 4*d*x)/8 + 3/8))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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