\(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [123]

Optimal result

Integrand size = 33, antiderivative size = 107 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {(2 A+3 C) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {(A+2 C) \tan (c+d x)}{a d}+\frac {(2 A+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4170, 3872, 3852, 8, 3853, 3855} \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {(2 A+3 C) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {(A+2 C) \tan (c+d x)}{a d}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {(2 A+3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \]

[In]

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

[Out]

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

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 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 4170

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)*(A + C)*Cot[e + f*x]*(a + b*Csc[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[b*C*n + A*b
*(2*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, C, n}, x
] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {\int \sec ^2(c+d x) (a (A+2 C)-a (2 A+3 C) \sec (c+d x)) \, dx}{a^2} \\ & = -\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(A+2 C) \int \sec ^2(c+d x) \, dx}{a}+\frac {(2 A+3 C) \int \sec ^3(c+d x) \, dx}{a} \\ & = \frac {(2 A+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(2 A+3 C) \int \sec (c+d x) \, dx}{2 a}+\frac {(A+2 C) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d} \\ & = \frac {(2 A+3 C) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {(A+2 C) \tan (c+d x)}{a d}+\frac {(2 A+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(316\) vs. \(2(107)=214\).

Time = 3.64 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.95 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-4 (A+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left (-2 (2 A+3 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 C \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )}{a d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*(-4*(A + C)*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]*(-2
*(2*A + 3*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 6*C*Log
[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + C/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - C/(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2])^2 - (4*C*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x
)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))))/(a*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x]))

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.10

method	result	size
parallelrisch	\(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +2 C \right ) \cos \left (2 d x +2 c \right )+C \cos \left (d x +c \right )+A +C \right )}{a d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(118\)
derivativedivides	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\left (-\frac {3 C}{2}-A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(135\)
default	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\left (-\frac {3 C}{2}-A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3 C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(135\)
norman	\(\frac {\frac {\left (A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}+\frac {3 \left (A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {\left (3 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(166\)
risch	\(-\frac {i \left (2 A \,{\mathrm e}^{4 i \left (d x +c \right )}+3 C \,{\mathrm e}^{4 i \left (d x +c \right )}+3 C \,{\mathrm e}^{3 i \left (d x +c \right )}+4 A \,{\mathrm e}^{2 i \left (d x +c \right )}+5 C \,{\mathrm e}^{2 i \left (d x +c \right )}+C \,{\mathrm e}^{i \left (d x +c \right )}+2 A +4 C \right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a d}\)	\(202\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right ) - C\right )} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (103) = 206\).

Time = 0.21 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.23 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {C {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 2 \, A {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, d} \]

[In]

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

[Out]

-1/2*(C*(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*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 2*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 - sin(d*x + c)/(a*(cos(d*x + c) + 1))))
/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.21 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\frac {{\left (2 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {{\left (2 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \]

[In]

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

[Out]

1/2*((2*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - (2*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 2*(
A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a + 2*(3*C*tan(1/2*d*x + 1/2*c)^3 - C*tan(1/2*d*x + 1/2*c))/(
(tan(1/2*d*x + 1/2*c)^2 - 1)^2*a))/d

Mupad [B] (verification not implemented)

Time = 15.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {3\,C}{2}\right )}{a\,d}-\frac {C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{a\,d} \]

[In]

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

[Out]

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