3.5.76 \(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [476]

Optimal. Leaf size=173 \[ \frac {C \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(16 A+12 B-215 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A+3 B-10 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]

[Out]

C*arctanh(sin(d*x+c))/a^4/d-1/105*(8*A+6*B-55*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2+1/105*(16*A+12*B-215*C)*tan
(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A-B+C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/35*(4*A+3*B-10*C)*sec(
d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3

________________________________________________________________________________________

Rubi [A]
time = 0.35, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4169, 4104, 4093, 4083, 3855, 3879} \begin {gather*} \frac {(16 A+12 B-215 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {C \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {(4 A+3 B-10 C) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((8*A + 6*B - 55*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + ((16*
A + 12*B - 215*C)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*
d*(a + a*Sec[c + d*x])^4) + ((4*A + 3*B - 10*C)*Sec[c + d*x]^2*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3)

Rule 3855

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

Rule 3879

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

Rule 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4093

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

Rule 4104

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

Rule 4169

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

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec ^3(c+d x) (a (4 A+3 B-3 C)+7 a C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A+3 B-10 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a^2 (4 A+3 B-10 C)+35 a^2 C \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A+3 B-10 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec (c+d x) \left (-2 a^3 (8 A+6 B-55 C)-105 a^3 C \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A+3 B-10 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(16 A+12 B-215 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}+\frac {C \int \sec (c+d x) \, dx}{a^4}\\ &=\frac {C \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(8 A+6 B-55 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A+3 B-10 C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(16 A+12 B-215 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 2.86, size = 335, normalized size = 1.94 \begin {gather*} -\frac {\left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \left (6720 C \cos ^8\left (\frac {1}{2} (c+d x)\right ) \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 )-\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (70 (2 A+3 B-49 C) \sin \left (\frac {d x}{2}\right )-70 (2 A-31 C) \sin \left (c+\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )+126 B \sin \left (c+\frac {3 d x}{2}\right )-2625 C \sin \left (c+\frac {3 d x}{2}\right )+735 C \sin \left (2 c+\frac {3 d x}{2}\right )+56 A \sin \left (2 c+\frac {5 d x}{2}\right )+42 B \sin \left (2 c+\frac {5 d x}{2}\right )-1015 C \sin \left (2 c+\frac {5 d x}{2}\right )+105 C \sin \left (3 c+\frac {5 d x}{2}\right )+8 A \sin \left (3 c+\frac {7 d x}{2}\right )+6 B \sin \left (3 c+\frac {7 d x}{2}\right )-160 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )\right )}{210 a^4 d (1+\cos (c+d x))^4 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

-1/210*((C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*(6720*C*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*
x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Cos[(c + d*x)/2]*Sec[c/2]*(70*(2*A + 3*B - 49*C)*Sin[(d*x
)/2] - 70*(2*A - 31*C)*Sin[c + (d*x)/2] + 168*A*Sin[c + (3*d*x)/2] + 126*B*Sin[c + (3*d*x)/2] - 2625*C*Sin[c +
(3*d*x)/2] + 735*C*Sin[2*c + (3*d*x)/2] + 56*A*Sin[2*c + (5*d*x)/2] + 42*B*Sin[2*c + (5*d*x)/2] - 1015*C*Sin[
2*c + (5*d*x)/2] + 105*C*Sin[3*c + (5*d*x)/2] + 8*A*Sin[3*c + (7*d*x)/2] + 6*B*Sin[3*c + (7*d*x)/2] - 160*C*Si
n[3*c + (7*d*x)/2])))/(a^4*d*(1 + Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

________________________________________________________________________________________

Maple [A]
time = 0.62, size = 199, normalized size = 1.15

method	result	size
derivativedivides	\(\frac {8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{7}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {11 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-15 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(199\)
default	\(\frac {8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{7}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {11 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-15 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(199\)
risch	\(-\frac {2 i \left (105 C \,{\mathrm e}^{6 i \left (d x +c \right )}+735 C \,{\mathrm e}^{5 i \left (d x +c \right )}-140 A \,{\mathrm e}^{4 i \left (d x +c \right )}+2170 C \,{\mathrm e}^{4 i \left (d x +c \right )}-140 A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 B \,{\mathrm e}^{3 i \left (d x +c \right )}+3430 C \,{\mathrm e}^{3 i \left (d x +c \right )}-168 A \,{\mathrm e}^{2 i \left (d x +c \right )}-126 B \,{\mathrm e}^{2 i \left (d x +c \right )}+2625 C \,{\mathrm e}^{2 i \left (d x +c \right )}-56 \,{\mathrm e}^{i \left (d x +c \right )} A -42 B \,{\mathrm e}^{i \left (d x +c \right )}+1015 C \,{\mathrm e}^{i \left (d x +c \right )}-8 A -6 B +160 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{4} d}\)	\(233\)
norman	\(\frac {-\frac {\left (A -B +C \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}+\frac {\left (13 A +B -15 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 a d}+\frac {\left (29 A -57 B -55 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{840 a d}-\frac {\left (47 A +9 B -1465 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 a d}-\frac {\left (101 A -3 B +605 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{840 a d}+\frac {\left (B -15 C +A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (-1145 C +67 A +39 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 a d}-\frac {\left (-169 C +11 A +9 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} a^{3}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}\)	\(283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.31, size = 313, normalized size = 1.81 \begin {gather*} -\frac {5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} - \frac {3 \, B {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(5*C*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) +
1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) - A*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(
d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1
)^7)/a^4 - 3*B*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^
5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d

________________________________________________________________________________________

Fricas [A]
time = 1.77, size = 248, normalized size = 1.43 \begin {gather*} \frac {105 \, {\left (C \cos \left (d x + c\right )^{4} + 4 \, C \cos \left (d x + c\right )^{3} + 6 \, C \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (C \cos \left (d x + c\right )^{4} + 4 \, C \cos \left (d x + c\right )^{3} + 6 \, C \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (4 \, A + 3 \, B - 80 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (32 \, A + 24 \, B - 535 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (52 \, A + 39 \, B - 620 \, C\right )} \cos \left (d x + c\right ) + 13 \, A + 36 \, B - 260 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/210*(105*(C*cos(d*x + c)^4 + 4*C*cos(d*x + c)^3 + 6*C*cos(d*x + c)^2 + 4*C*cos(d*x + c) + C)*log(sin(d*x + c
) + 1) - 105*(C*cos(d*x + c)^4 + 4*C*cos(d*x + c)^3 + 6*C*cos(d*x + c)^2 + 4*C*cos(d*x + c) + C)*log(-sin(d*x
+ c) + 1) + 2*(2*(4*A + 3*B - 80*C)*cos(d*x + c)^3 + (32*A + 24*B - 535*C)*cos(d*x + c)^2 + (52*A + 39*B - 620
*C)*cos(d*x + c) + 13*A + 36*B - 260*C)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d
*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{4}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4,x)

[Out]

(Integral(A*sec(c + d*x)**3/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)
+ Integral(B*sec(c + d*x)**4/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1),
x) + Integral(C*sec(c + d*x)**5/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1)
, x))/a**4

________________________________________________________________________________________

Giac [A]
time = 0.51, size = 248, normalized size = 1.43 \begin {gather*} \frac {\frac {840 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 63 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(840*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - (15*A*a^2
4*tan(1/2*d*x + 1/2*c)^7 - 15*B*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 21*A*a^24*tan
(1/2*d*x + 1/2*c)^5 - 63*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 105*C*a^24*tan(1/2*d*x + 1/2*c)^5 - 35*A*a^24*tan(1/2
*d*x + 1/2*c)^3 - 105*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan(1/2*d*x + 1/2*c)^3 - 105*A*a^24*tan(1/2*d
*x + 1/2*c) - 105*B*a^24*tan(1/2*d*x + 1/2*c) + 1575*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

________________________________________________________________________________________

Mupad [B]
time = 3.25, size = 198, normalized size = 1.14 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,A-6\,C}{24\,a^4}-\frac {A-B+C}{24\,a^4}+\frac {2\,B-4\,C}{24\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B+C}{8\,a^4}-\frac {2\,A-6\,C}{8\,a^4}-\frac {2\,B-4\,C}{8\,a^4}+\frac {2\,B+4\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A-B+C}{40\,a^4}-\frac {2\,B-4\,C}{40\,a^4}\right )}{d}+\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^4),x)

[Out]

(tan(c/2 + (d*x)/2)^3*((2*A - 6*C)/(24*a^4) - (A - B + C)/(24*a^4) + (2*B - 4*C)/(24*a^4)))/d - (tan(c/2 + (d*
x)/2)*((A - B + C)/(8*a^4) - (2*A - 6*C)/(8*a^4) - (2*B - 4*C)/(8*a^4) + (2*B + 4*C)/(8*a^4)))/d - (tan(c/2 +
(d*x)/2)^5*((A - B + C)/(40*a^4) - (2*B - 4*C)/(40*a^4)))/d + (2*C*atanh(tan(c/2 + (d*x)/2)))/(a^4*d) - (tan(c
/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d)

________________________________________________________________________________________