3.2.14 \(\int \cos (c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\) [114]

Optimal. Leaf size=181 \[ 4 a^4 A x+\frac {a^4 (52 A+35 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}+\frac {5 a^4 (4 A+7 C) \tan (c+d x)}{8 d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {(12 A-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac {(12 A-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d} \]

[Out]

4*a^4*A*x+1/8*a^4*(52*A+35*C)*arctanh(sin(d*x+c))/d+A*(a+a*sec(d*x+c))^4*sin(d*x+c)/d+5/8*a^4*(4*A+7*C)*tan(d*
x+c)/d-1/4*a*(4*A-C)*(a+a*sec(d*x+c))^3*tan(d*x+c)/d-1/12*(12*A-7*C)*(a^2+a^2*sec(d*x+c))^2*tan(d*x+c)/d-1/24*
(12*A-35*C)*(a^4+a^4*sec(d*x+c))*tan(d*x+c)/d

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Rubi [A]
time = 0.24, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4172, 4002, 3999, 3852, 8, 3855} \begin {gather*} \frac {5 a^4 (4 A+7 C) \tan (c+d x)}{8 d}+\frac {a^4 (52 A+35 C) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {(12 A-35 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+4 a^4 A x-\frac {(12 A-7 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}-\frac {a (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

4*a^4*A*x + (a^4*(52*A + 35*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (A*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/d + (5*a
^4*(4*A + 7*C)*Tan[c + d*x])/(8*d) - (a*(4*A - C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) - ((12*A - 7*C)*(
a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) - ((12*A - 35*C)*(a^4 + a^4*Sec[c + d*x])*Tan[c + d*x])/(24*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 3999

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 4002

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] &&
GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4172

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/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}+\frac {\int (a+a \sec (c+d x))^4 (4 a A-a (4 A-C) \sec (c+d x)) \, dx}{a}\\ &=\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\int (a+a \sec (c+d x))^3 \left (16 a^2 A-a^2 (12 A-7 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {(12 A-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}+\frac {\int (a+a \sec (c+d x))^2 \left (48 a^3 A-a^3 (12 A-35 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {(12 A-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac {(12 A-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac {\int (a+a \sec (c+d x)) \left (96 a^4 A+15 a^4 (4 A+7 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=4 a^4 A x+\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {(12 A-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac {(12 A-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac {1}{8} \left (5 a^4 (4 A+7 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (a^4 (52 A+35 C)\right ) \int \sec (c+d x) \, dx\\ &=4 a^4 A x+\frac {a^4 (52 A+35 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {(12 A-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac {(12 A-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}-\frac {\left (5 a^4 (4 A+7 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{8 d}\\ &=4 a^4 A x+\frac {a^4 (52 A+35 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}+\frac {5 a^4 (4 A+7 C) \tan (c+d x)}{8 d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {(12 A-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac {(12 A-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(379\) vs. \(2(181)=362\).
time = 2.39, size = 379, normalized size = 2.09 \begin {gather*} \frac {a^4 \left (C+A \cos ^2(c+d x)\right ) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^4 \left (-24 (52 A+35 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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec (c) (288 A d x \cos (c)+192 A d x \cos (c+2 d x)+192 A d x \cos (3 c+2 d x)+48 A d x \cos (3 c+4 d x)+48 A d x \cos (5 c+4 d x)-288 A \sin (c)-480 C \sin (c)+24 A \sin (d x)+105 C \sin (d x)+24 A \sin (2 c+d x)+105 C \sin (2 c+d x)+288 A \sin (c+2 d x)+544 C \sin (c+2 d x)-96 A \sin (3 c+2 d x)-96 C \sin (3 c+2 d x)+30 A \sin (2 c+3 d x)+81 C \sin (2 c+3 d x)+30 A \sin (4 c+3 d x)+81 C \sin (4 c+3 d x)+96 A \sin (3 c+4 d x)+160 C \sin (3 c+4 d x)+6 A \sin (4 c+5 d x)+6 A \sin (6 c+5 d x))\right )}{1536 d (A+2 C+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(-24*(52*A + 35*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]*(288*A*d*x*Cos[c] + 192
*A*d*x*Cos[c + 2*d*x] + 192*A*d*x*Cos[3*c + 2*d*x] + 48*A*d*x*Cos[3*c + 4*d*x] + 48*A*d*x*Cos[5*c + 4*d*x] - 2
88*A*Sin[c] - 480*C*Sin[c] + 24*A*Sin[d*x] + 105*C*Sin[d*x] + 24*A*Sin[2*c + d*x] + 105*C*Sin[2*c + d*x] + 288
*A*Sin[c + 2*d*x] + 544*C*Sin[c + 2*d*x] - 96*A*Sin[3*c + 2*d*x] - 96*C*Sin[3*c + 2*d*x] + 30*A*Sin[2*c + 3*d*
x] + 81*C*Sin[2*c + 3*d*x] + 30*A*Sin[4*c + 3*d*x] + 81*C*Sin[4*c + 3*d*x] + 96*A*Sin[3*c + 4*d*x] + 160*C*Sin
[3*c + 4*d*x] + 6*A*Sin[4*c + 5*d*x] + 6*A*Sin[6*c + 5*d*x])))/(1536*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]
time = 0.78, size = 237, normalized size = 1.31

method	result	size
derivativedivides	\(\frac {A \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 A \,a^{4} \tan \left (d x +c \right )-4 a^{4} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+6 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 A \,a^{4} \left (d x +c \right )+4 a^{4} C \tan \left (d x +c \right )+A \,a^{4} \sin \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(237\)
default	\(\frac {A \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 A \,a^{4} \tan \left (d x +c \right )-4 a^{4} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+6 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 A \,a^{4} \left (d x +c \right )+4 a^{4} C \tan \left (d x +c \right )+A \,a^{4} \sin \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(237\)
risch	\(4 a^{4} A x -\frac {i A \,a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i A \,a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{4} \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+81 C \,{\mathrm e}^{7 i \left (d x +c \right )}-96 A \,{\mathrm e}^{6 i \left (d x +c \right )}-96 C \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+105 C \,{\mathrm e}^{5 i \left (d x +c \right )}-288 A \,{\mathrm e}^{4 i \left (d x +c \right )}-480 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-105 C \,{\mathrm e}^{3 i \left (d x +c \right )}-288 A \,{\mathrm e}^{2 i \left (d x +c \right )}-544 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 \,{\mathrm e}^{i \left (d x +c \right )} A -81 C \,{\mathrm e}^{i \left (d x +c \right )}-96 A -160 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\)	\(332\)
norman	\(\frac {-4 a^{4} A x +16 a^{4} A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 a^{4} A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 a^{4} A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 a^{4} A x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a^{4} A x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {5 a^{4} \left (4 A +7 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a^{4} \left (44 A +93 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {7 a^{4} \left (60 A +73 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a^{4} \left (-203 C +12 A \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {a^{4} \left (53 C +204 A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a^{4} \left (385 C +156 A \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {a^{4} \left (52 A +35 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{4} \left (52 A +35 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(339\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(A*a^4*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a^4*C*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c)
)*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+4*A*a^4*tan(d*x+c)-4*a^4*C*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+6*A*
a^4*ln(sec(d*x+c)+tan(d*x+c))+6*a^4*C*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+4*A*a^4*(d*x+c
)+4*a^4*C*tan(d*x+c)+A*a^4*sin(d*x+c)+a^4*C*ln(sec(d*x+c)+tan(d*x+c)))

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Maxima [A]
time = 0.30, size = 296, normalized size = 1.64 \begin {gather*} \frac {192 \, {\left (d x + c\right )} A a^{4} + 64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 3 \, C a^{4} {\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^{4} {\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 )} - 72 \, C a^{4} {\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^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 192 \, A a^{4} \tan \left (d x + c\right ) + 192 \, C a^{4} \tan \left (d x + c\right )}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(192*(d*x + c)*A*a^4 + 64*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 - 3*C*a^4*(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^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 72*C*a^4*(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^4*(log(sin(d*x + c) + 1
) - log(sin(d*x + c) - 1)) + 24*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*A*a^4*sin(d*x + c)
+ 192*A*a^4*tan(d*x + c) + 192*C*a^4*tan(d*x + c))/d

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Fricas [A]
time = 3.27, size = 171, normalized size = 0.94 \begin {gather*} \frac {192 \, A a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (52 \, A + 35 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (52 \, A + 35 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 32 \, {\left (3 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 27 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \, C a^{4} \cos \left (d x + c\right ) + 6 \, C a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(192*A*a^4*d*x*cos(d*x + c)^4 + 3*(52*A + 35*C)*a^4*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(52*A + 35*C
)*a^4*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*A*a^4*cos(d*x + c)^4 + 32*(3*A + 5*C)*a^4*cos(d*x + c)^3 +
3*(4*A + 27*C)*a^4*cos(d*x + c)^2 + 32*C*a^4*cos(d*x + c) + 6*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]
time = 0.51, size = 253, normalized size = 1.40 \begin {gather*} \frac {96 \, {\left (d x + c\right )} A a^{4} + \frac {48 \, A a^{4} \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 (52 \, A a^{4} + 35 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (52 \, A a^{4} + 35 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (84 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 276 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 385 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 300 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 511 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 279 \, C a^{4} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(96*(d*x + c)*A*a^4 + 48*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 3*(52*A*a^4 + 35*C*a^4
)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(52*A*a^4 + 35*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(84*A*a^
4*tan(1/2*d*x + 1/2*c)^7 + 105*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 276*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 385*C*a^4*tan
(1/2*d*x + 1/2*c)^5 + 300*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 511*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 108*A*a^4*tan(1/2*
d*x + 1/2*c) - 279*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 2.79, size = 246, normalized size = 1.36 \begin {gather*} \frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {8\,A\,a^4\,\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 {13\,A\,a^4\,\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 {35\,C\,a^4\,\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 )}{4\,d}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {27\,C\,a^4\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {4\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4} \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 + a/cos(c + d*x))^4,x)

[Out]

(A*a^4*sin(c + d*x))/d + (8*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13*A*a^4*atanh(sin(c/2 + (
d*x)/2)/cos(c/2 + (d*x)/2)))/d + (35*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (4*A*a^4*sin(
c + d*x))/(d*cos(c + d*x)) + (A*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (20*C*a^4*sin(c + d*x))/(3*d*cos(c +
d*x)) + (27*C*a^4*sin(c + d*x))/(8*d*cos(c + d*x)^2) + (4*C*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (C*a^4*si
n(c + d*x))/(4*d*cos(c + d*x)^4)

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