3.7.57 \(\int \cos (c+d x) (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [657]

Optimal. Leaf size=167 \[ 3 a^2 A b x+\frac {a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d} \]

[Out]

3*a^2*A*b*x+1/2*a*(6*A*b^2+2*C*a^2+3*C*b^2)*arctanh(sin(d*x+c))/d+A*(a+b*sec(d*x+c))^3*sin(d*x+c)/d-1/3*b*(a^2
*(6*A-8*C)-b^2*(3*A+2*C))*tan(d*x+c)/d-1/6*a*b^2*(6*A-5*C)*sec(d*x+c)*tan(d*x+c)/d-1/3*b*(3*A-C)*(a+b*sec(d*x+
c))^2*tan(d*x+c)/d

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Rubi [A]
time = 0.22, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4180, 4141, 4133, 3855, 3852, 8} \begin {gather*} -\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 d}+\frac {a \left (2 a^2 C+6 A b^2+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+3 a^2 A b x-\frac {a b^2 (6 A-5 C) \tan (c+d x) \sec (c+d x)}{6 d}-\frac {b (3 A-C) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3*a^2*A*b*x + (a*(6*A*b^2 + 2*a^2*C + 3*b^2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (A*(a + b*Sec[c + d*x])^3*Sin[c
+ d*x])/d - (b*(a^2*(6*A - 8*C) - b^2*(3*A + 2*C))*Tan[c + d*x])/(3*d) - (a*b^2*(6*A - 5*C)*Sec[c + d*x]*Tan[c
+ d*x])/(6*d) - (b*(3*A - C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(3*d)

Rule 8

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

Rule 3852

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 3855

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

Rule 4133

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

Rule 4141

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/(m + 1), I
nt[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) +
a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

Rule 4180

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*(C*n + A*(n + 1))*Csc[e +
f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\int (a+b \sec (c+d x))^2 \left (3 A b+a C \sec (c+d x)-b (3 A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+b \sec (c+d x)) \left (9 a A b+\left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sec (c+d x)-a b (6 A-5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (18 a^2 A b+3 a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \sec (c+d x)-2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=3 a^2 A b x+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{2} \left (a \left (6 A b^2+2 a^2 C+3 b^2 C\right )\right ) \int \sec (c+d x) \, dx-\frac {1}{3} \left (b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=3 a^2 A b x+\frac {a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {\left (b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=3 a^2 A b x+\frac {a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 1.66, size = 325, normalized size = 1.95 \begin {gather*} \frac {\sec ^3(c+d x) \left (9 a \cos (c+d x) \left (6 a A b (c+d x)-\left (6 A b^2+2 a^2 C+3 b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\left (6 A b^2+2 a^2 C+3 b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 a \cos (3 (c+d x)) \left (6 a A b (c+d x)-\left (6 A b^2+2 a^2 C+3 b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\left (6 A b^2+2 a^2 C+3 b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+2 \left (6 A b^3+18 a^2 b C+8 b^3 C+9 a \left (a^2 A+2 b^2 C\right ) \cos (c+d x)+2 \left (3 A b^3+9 a^2 b C+2 b^3 C\right ) \cos (2 (c+d x))+3 a^3 A \cos (3 (c+d x))\right ) \sin (c+d x)\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(9*a*Cos[c + d*x]*(6*a*A*b*(c + d*x) - (6*A*b^2 + 2*a^2*C + 3*b^2*C)*Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]] + (6*A*b^2 + 2*a^2*C + 3*b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 3*a*Cos[3*(c + d*x
)]*(6*a*A*b*(c + d*x) - (6*A*b^2 + 2*a^2*C + 3*b^2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + (6*A*b^2 + 2*
a^2*C + 3*b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 2*(6*A*b^3 + 18*a^2*b*C + 8*b^3*C + 9*a*(a^2*A +
2*b^2*C)*Cos[c + d*x] + 2*(3*A*b^3 + 9*a^2*b*C + 2*b^3*C)*Cos[2*(c + d*x)] + 3*a^3*A*Cos[3*(c + d*x)])*Sin[c +
d*x]))/(24*d)

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Maple [A]
time = 0.13, size = 155, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {A \,b^{3} \tan \left (d x +c \right )-C \,b^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,b^{2} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{2} b \left (d x +c \right )+3 a^{2} b C \tan \left (d x +c \right )+A \,a^{3} \sin \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(155\)
default	\(\frac {A \,b^{3} \tan \left (d x +c \right )-C \,b^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,b^{2} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{2} b \left (d x +c \right )+3 a^{2} b C \tan \left (d x +c \right )+A \,a^{3} \sin \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(155\)
risch	\(3 a^{2} A b x -\frac {i A \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i A \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i b \left (9 C a b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-18 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-36 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 C b a \,{\mathrm e}^{i \left (d x +c \right )}-6 A \,b^{2}-18 a^{2} C -4 b^{2} C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} C}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} C}{2 d}\)	\(325\)
norman	\(\frac {\frac {\left (2 A \,a^{3}-2 A \,b^{3}-6 a^{2} b C +3 C \,b^{2} a -2 C \,b^{3}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A \,a^{3}+2 A \,b^{3}+6 a^{2} b C +3 C \,b^{2} a +2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 \left (6 A \,a^{3}-3 A \,b^{3}-9 a^{2} b C -C \,b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 \left (6 A \,a^{3}+3 A \,b^{3}+9 a^{2} b C +C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {6 a \left (2 A \,a^{2}-b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+3 a^{2} A b x -9 a^{2} A b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a^{2} A b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a^{2} A b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 a^{2} A b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{2} A b x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a \left (6 A \,b^{2}+2 a^{2} C +3 b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (6 A \,b^{2}+2 a^{2} C +3 b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(418\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(A*b^3*tan(d*x+c)-C*b^3*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+3*a*A*b^2*ln(sec(d*x+c)+tan(d*x+c))+3*C*b^2*a*(
1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+3*A*a^2*b*(d*x+c)+3*a^2*b*C*tan(d*x+c)+A*a^3*sin(d*x+
c)+a^3*C*ln(sec(d*x+c)+tan(d*x+c)))

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Maxima [A]
time = 0.27, size = 181, normalized size = 1.08 \begin {gather*} \frac {36 \, {\left (d x + c\right )} A a^{2} b + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{3} - 9 \, C a b^{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 )} + 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 )} + 18 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, C a^{2} b \tan \left (d x + c\right ) + 12 \, A b^{3} \tan \left (d x + c\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(36*(d*x + c)*A*a^2*b + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b^3 - 9*C*a*b^2*(2*sin(d*x + c)/(sin(d*x +
c)^2 - 1) - 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)) + 18*A*a*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 36*C*a^2*b*tan(d
*x + c) + 12*A*b^3*tan(d*x + c))/d

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Fricas [A]
time = 2.95, size = 178, normalized size = 1.07 \begin {gather*} \frac {36 \, A a^{2} b d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, C a^{3} + 3 \, {\left (2 \, A + C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, C a^{3} + 3 \, {\left (2 \, A + C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 9 \, C a b^{2} \cos \left (d x + c\right ) + 2 \, C b^{3} + 2 \, {\left (9 \, C a^{2} b + {\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(36*A*a^2*b*d*x*cos(d*x + c)^3 + 3*(2*C*a^3 + 3*(2*A + C)*a*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3
*(2*C*a^3 + 3*(2*A + C)*a*b^2)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(6*A*a^3*cos(d*x + c)^3 + 9*C*a*b^2*c
os(d*x + c) + 2*C*b^3 + 2*(9*C*a^2*b + (3*A + 2*C)*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3} \cos {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**3*cos(c + d*x), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (160) = 320\).
time = 0.52, size = 322, normalized size = 1.93 \begin {gather*} \frac {18 \, {\left (d x + c\right )} A a^{2} b + \frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 3 \, {\left (2 \, C a^{3} + 6 \, A a b^{2} + 3 \, C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, C a^{3} + 6 \, A a b^{2} + 3 \, C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(18*(d*x + c)*A*a^2*b + 12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 3*(2*C*a^3 + 6*A*a*b^
2 + 3*C*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*C*a^3 + 6*A*a*b^2 + 3*C*a*b^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(18*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 9*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^3*tan(1/2*d*x +
1/2*c)^5 + 6*C*b^3*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^3*tan(1/2*d*x + 1/2*c)
^3 - 4*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b*tan(1/2*d*x + 1/2*c) + 9*C*a*b^2*tan(1/2*d*x + 1/2*c) + 6*A*b
^3*tan(1/2*d*x + 1/2*c) + 6*C*b^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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Mupad [B]
time = 5.37, size = 464, normalized size = 2.78 \begin {gather*} \frac {\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {A\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{6}+\frac {A\,b^3\,\sin \left (c+d\,x\right )}{4}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,C\,a^2\,b\,\sin \left (c+d\,x\right )}{4}-\frac {C\,a^3\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{2}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {3\,C\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4}-\frac {C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{2}+\frac {3\,A\,a^2\,b\,\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 (3\,c+3\,d\,x\right )}{2}-\frac {A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{2}-\frac {C\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{4}+\frac {9\,A\,a^2\,b\,\cos \left (c+d\,x\right )\,\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 )}{2}-\frac {A\,a\,b^2\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,9{}\mathrm {i}}{2}-\frac {C\,a\,b^2\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,9{}\mathrm {i}}{4}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*a^3*sin(2*c + 2*d*x))/4 + (A*a^3*sin(4*c + 4*d*x))/8 + (A*b^3*sin(3*c + 3*d*x))/4 + (C*b^3*sin(3*c + 3*d*x
))/6 + (A*b^3*sin(c + d*x))/4 + (C*b^3*sin(c + d*x))/2 + (3*C*a^2*b*sin(c + d*x))/4 - (C*a^3*cos(c + d*x)*atan
((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/2 + (3*C*a*b^2*sin(2*c + 2*d*x))/4 + (3*C*a^2*b*sin(3*c + 3*d
*x))/4 - (C*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/2 + (3*A*a^2*b*atan(sin(
c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 - (A*a*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)
/2))*cos(3*c + 3*d*x)*3i)/2 - (C*a*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*3i)/4
+ (9*A*a^2*b*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (A*a*b^2*cos(c + d*x)*atan((sin(c/
2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*9i)/2 - (C*a*b^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x
)/2))*9i)/4)/(d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4))

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