3.5.50 \(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [450]

Optimal. Leaf size=148 \[ -\frac {(2 A-3 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {(3 A-3 B+4 C) \tan (c+d x)}{a d}-\frac {(2 A-3 B+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(3 A-3 B+4 C) \tan ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4169, 3872, 3853, 3855, 3852} \begin {gather*} \frac {(3 A-3 B+4 C) \tan ^3(c+d x)}{3 a d}+\frac {(3 A-3 B+4 C) \tan (c+d x)}{a d}-\frac {(2 A-3 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}-\frac {(2 A-3 B+3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \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]),x]

[Out]

-1/2*((2*A - 3*B + 3*C)*ArcTanh[Sin[c + d*x]])/(a*d) + ((3*A - 3*B + 4*C)*Tan[c + d*x])/(a*d) - ((2*A - 3*B +
3*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - ((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) +
((3*A - 3*B + 4*C)*Tan[c + d*x]^3)/(3*a*d)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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)} \, dx &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {\int \sec ^3(c+d x) (-a (2 A-3 B+3 C)+a (3 A-3 B+4 C) \sec (c+d x)) \, dx}{a^2}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(2 A-3 B+3 C) \int \sec ^3(c+d x) \, dx}{a}+\frac {(3 A-3 B+4 C) \int \sec ^4(c+d x) \, dx}{a}\\ &=-\frac {(2 A-3 B+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(2 A-3 B+3 C) \int \sec (c+d x) \, dx}{2 a}-\frac {(3 A-3 B+4 C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a d}\\ &=-\frac {(2 A-3 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {(3 A-3 B+4 C) \tan (c+d x)}{a d}-\frac {(2 A-3 B+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(3 A-3 B+4 C) \tan ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(898\) vs. \(2(148)=296\).
time = 6.34, size = 898, normalized size = 6.07 \begin {gather*} \frac {2 (2 A-3 B+3 C) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))}-\frac {2 (2 A-3 B+3 C) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))}+\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-6 A \sin \left (\frac {d x}{2}\right )+6 B \sin \left (\frac {d x}{2}\right )+6 C \sin \left (\frac {d x}{2}\right )+30 A \sin \left (\frac {3 d x}{2}\right )-27 B \sin \left (\frac {3 d x}{2}\right )+39 C \sin \left (\frac {3 d x}{2}\right )-12 A \sin \left (c-\frac {d x}{2}\right )+12 B \sin \left (c-\frac {d x}{2}\right )-24 C \sin \left (c-\frac {d x}{2}\right )-6 A \sin \left (c+\frac {d x}{2}\right )+6 B \sin \left (c+\frac {d x}{2}\right )-6 C \sin \left (c+\frac {d x}{2}\right )-24 A \sin \left (2 c+\frac {d x}{2}\right )+24 B \sin \left (2 c+\frac {d x}{2}\right )-24 C \sin \left (2 c+\frac {d x}{2}\right )+12 A \sin \left (c+\frac {3 d x}{2}\right )-9 B \sin \left (c+\frac {3 d x}{2}\right )+21 C \sin \left (c+\frac {3 d x}{2}\right )+12 A \sin \left (2 c+\frac {3 d x}{2}\right )-9 B \sin \left (2 c+\frac {3 d x}{2}\right )+9 C \sin \left (2 c+\frac {3 d x}{2}\right )-6 A \sin \left (3 c+\frac {3 d x}{2}\right )+9 B \sin \left (3 c+\frac {3 d x}{2}\right )-9 C \sin \left (3 c+\frac {3 d x}{2}\right )+6 A \sin \left (c+\frac {5 d x}{2}\right )-3 B \sin \left (c+\frac {5 d x}{2}\right )+7 C \sin \left (c+\frac {5 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )+C \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )-3 C \sin \left (3 c+\frac {5 d x}{2}\right )-6 A \sin \left (4 c+\frac {5 d x}{2}\right )+9 B \sin \left (4 c+\frac {5 d x}{2}\right )-9 C \sin \left (4 c+\frac {5 d x}{2}\right )+12 A \sin \left (2 c+\frac {7 d x}{2}\right )-12 B \sin \left (2 c+\frac {7 d x}{2}\right )+16 C \sin \left (2 c+\frac {7 d x}{2}\right )+6 A \sin \left (3 c+\frac {7 d x}{2}\right )-6 B \sin \left (3 c+\frac {7 d x}{2}\right )+10 C \sin \left (3 c+\frac {7 d x}{2}\right )+6 A \sin \left (4 c+\frac {7 d x}{2}\right )-6 B \sin \left (4 c+\frac {7 d x}{2}\right )+6 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{24 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (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]),x]

[Out]

(2*(2*A - 3*B + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(A + B*Sec
[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (2
*(2*A - 3*B + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(A + B*Sec[c
+ d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (Cos
[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-6*A*Sin[(d*x)/2] + 6*
B*Sin[(d*x)/2] + 6*C*Sin[(d*x)/2] + 30*A*Sin[(3*d*x)/2] - 27*B*Sin[(3*d*x)/2] + 39*C*Sin[(3*d*x)/2] - 12*A*Sin
[c - (d*x)/2] + 12*B*Sin[c - (d*x)/2] - 24*C*Sin[c - (d*x)/2] - 6*A*Sin[c + (d*x)/2] + 6*B*Sin[c + (d*x)/2] -
6*C*Sin[c + (d*x)/2] - 24*A*Sin[2*c + (d*x)/2] + 24*B*Sin[2*c + (d*x)/2] - 24*C*Sin[2*c + (d*x)/2] + 12*A*Sin[
c + (3*d*x)/2] - 9*B*Sin[c + (3*d*x)/2] + 21*C*Sin[c + (3*d*x)/2] + 12*A*Sin[2*c + (3*d*x)/2] - 9*B*Sin[2*c +
(3*d*x)/2] + 9*C*Sin[2*c + (3*d*x)/2] - 6*A*Sin[3*c + (3*d*x)/2] + 9*B*Sin[3*c + (3*d*x)/2] - 9*C*Sin[3*c + (3
*d*x)/2] + 6*A*Sin[c + (5*d*x)/2] - 3*B*Sin[c + (5*d*x)/2] + 7*C*Sin[c + (5*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2]
+ C*Sin[2*c + (5*d*x)/2] + 3*B*Sin[3*c + (5*d*x)/2] - 3*C*Sin[3*c + (5*d*x)/2] - 6*A*Sin[4*c + (5*d*x)/2] + 9
*B*Sin[4*c + (5*d*x)/2] - 9*C*Sin[4*c + (5*d*x)/2] + 12*A*Sin[2*c + (7*d*x)/2] - 12*B*Sin[2*c + (7*d*x)/2] + 1
6*C*Sin[2*c + (7*d*x)/2] + 6*A*Sin[3*c + (7*d*x)/2] - 6*B*Sin[3*c + (7*d*x)/2] + 10*C*Sin[3*c + (7*d*x)/2] + 6
*A*Sin[4*c + (7*d*x)/2] - 6*B*Sin[4*c + (7*d*x)/2] + 6*C*Sin[4*c + (7*d*x)/2]))/(24*d*(A + 2*C + 2*B*Cos[c + d
*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x]))

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Maple [A]
time = 0.64, size = 207, normalized size = 1.40

method	result	size
derivativedivides	\(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-B +2 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (\frac {3 C}{2}-\frac {3 B}{2}+A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 C}{2}-\frac {3 B}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {B -2 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-\frac {3 C}{2}+\frac {3 B}{2}-A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\frac {5 C}{2}-\frac {3 B}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d a}\)	\(207\)
default	\(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-B +2 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (\frac {3 C}{2}-\frac {3 B}{2}+A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 C}{2}-\frac {3 B}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {B -2 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-\frac {3 C}{2}+\frac {3 B}{2}-A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\frac {5 C}{2}-\frac {3 B}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d a}\)	\(207\)
norman	\(\frac {\frac {\left (A -B +C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (4 C -2 B +3 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\left (9 C -7 B +6 A \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (37 C -27 B +30 A \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (49 C -39 B +36 A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {\left (2 A -3 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {\left (2 A -3 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(216\)
risch	\(\frac {i \left (6 A \,{\mathrm e}^{6 i \left (d x +c \right )}-9 B \,{\mathrm e}^{6 i \left (d x +c \right )}+9 C \,{\mathrm e}^{6 i \left (d x +c \right )}+6 A \,{\mathrm e}^{5 i \left (d x +c \right )}-9 B \,{\mathrm e}^{5 i \left (d x +c \right )}+9 C \,{\mathrm e}^{5 i \left (d x +c \right )}+24 A \,{\mathrm e}^{4 i \left (d x +c \right )}-24 B \,{\mathrm e}^{4 i \left (d x +c \right )}+24 C \,{\mathrm e}^{4 i \left (d x +c \right )}+12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B \,{\mathrm e}^{3 i \left (d x +c \right )}+24 C \,{\mathrm e}^{3 i \left (d x +c \right )}+30 A \,{\mathrm e}^{2 i \left (d x +c \right )}-27 B \,{\mathrm e}^{2 i \left (d x +c \right )}+39 C \,{\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )} A -3 B \,{\mathrm e}^{i \left (d x +c \right )}+7 C \,{\mathrm e}^{i \left (d x +c \right )}+12 A -12 B +16 C \right )}{3 d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a d}\)	\(394\)

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)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(A*tan(1/2*d*x+1/2*c)-B*tan(1/2*d*x+1/2*c)+C*tan(1/2*d*x+1/2*c)-1/3*C/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(-B+2
*C)/(tan(1/2*d*x+1/2*c)-1)^2+(3/2*C-3/2*B+A)*ln(tan(1/2*d*x+1/2*c)-1)-(5/2*C-3/2*B+A)/(tan(1/2*d*x+1/2*c)-1)-1
/3*C/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(B-2*C)/(tan(1/2*d*x+1/2*c)+1)^2+(-3/2*C+3/2*B-A)*ln(tan(1/2*d*x+1/2*c)+1)-(
5/2*C-3/2*B+A)/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 485 vs. \(2 (142) = 284\).
time = 0.28, size = 485, normalized size = 3.28 \begin {gather*} \frac {C {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 6 \, A {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, 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)),x, algorithm="maxima")

[Out]

1/6*(C*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*B*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c
) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*si
n(d*x + c)/(a*(cos(d*x + c) + 1))) - 6*A*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*
x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]
time = 2.91, size = 195, normalized size = 1.32 \begin {gather*} -\frac {3 \, {\left ({\left (2 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (3 \, A - 3 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A - 3 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B - C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\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)),x, algorithm="fricas")

[Out]

-1/12*(3*((2*A - 3*B + 3*C)*cos(d*x + c)^4 + (2*A - 3*B + 3*C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((2*A
- 3*B + 3*C)*cos(d*x + c)^4 + (2*A - 3*B + 3*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(3*A - 3*B + 4*
C)*cos(d*x + c)^3 + (6*A - 3*B + 7*C)*cos(d*x + c)^2 + (3*B - C)*cos(d*x + c) + 2*C)*sin(d*x + c))/(a*d*cos(d*
x + c)^4 + a*d*cos(d*x + c)^3)

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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 {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \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)),x)

[Out]

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

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Giac [A]
time = 0.51, size = 243, normalized size = 1.64 \begin {gather*} -\frac {\frac {3 \, {\left (2 \, A - 3 \, B + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {3 \, {\left (2 \, A - 3 \, B + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C \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} a}}{6 \, 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)),x, algorithm="giac")

[Out]

-1/6*(3*(2*A - 3*B + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*(2*A - 3*B + 3*C)*log(abs(tan(1/2*d*x + 1/2
*c) - 1))/a - 6*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a + 2*(6*A*tan(1/2*
d*x + 1/2*c)^5 - 9*B*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1/2*c)^5 - 12*A*tan(1/2*d*x + 1/2*c)^3 + 12*B
*tan(1/2*d*x + 1/2*c)^3 - 16*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) - 3*B*tan(1/2*d*x + 1/2*c) +
9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d

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Mupad [B]
time = 3.81, size = 165, normalized size = 1.11 \begin {gather*} \frac {\left (2\,A-3\,B+5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,B-4\,A-\frac {16\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A-B+3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-\frac {3\,B}{2}+\frac {3\,C}{2}\right )}{a\,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))),x)

[Out]

(tan(c/2 + (d*x)/2)*(2*A - B + 3*C) + tan(c/2 + (d*x)/2)^5*(2*A - 3*B + 5*C) - tan(c/2 + (d*x)/2)^3*(4*A - 4*B
+ (16*C)/3))/(d*(a - 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^4 - a*tan(c/2 + (d*x)/2)^6)) + (tan(c/
2 + (d*x)/2)*(A - B + C))/(a*d) - (2*atanh(tan(c/2 + (d*x)/2))*(A - (3*B)/2 + (3*C)/2))/(a*d)

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