3.5.51 \(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [451]

Optimal. Leaf size=119 \[ \frac {(2 A-2 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(A-2 B+2 C) \tan (c+d x)}{a d}+\frac {(2 A-2 B+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 119, 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, 3872, 3852, 8, 3853, 3855} \begin {gather*} -\frac {(A-2 B+2 C) \tan (c+d x)}{a d}+\frac {(2 A-2 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {(2 A-2 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]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]

[Out]

((2*A - 2*B + 3*C)*ArcTanh[Sin[c + d*x]])/(2*a*d) - ((A - 2*B + 2*C)*Tan[c + d*x])/(a*d) + ((2*A - 2*B + 3*C)*
Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - ((A - B + 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 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 ^2(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 ^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 B+2 C)+a (2 A-2 B+3 C) \sec (c+d x)) \, dx}{a^2}\\ &=-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(A-2 B+2 C) \int \sec ^2(c+d x) \, dx}{a}+\frac {(2 A-2 B+3 C) \int \sec ^3(c+d x) \, dx}{a}\\ &=\frac {(2 A-2 B+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(2 A-2 B+3 C) \int \sec (c+d x) \, dx}{2 a}+\frac {(A-2 B+2 C) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d}\\ &=\frac {(2 A-2 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(A-2 B+2 C) \tan (c+d x)}{a d}+\frac {(2 A-2 B+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(392\) vs. \(2(119)=238\).
time = 4.70, size = 392, normalized size = 3.29 \begin {gather*} \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-2 (2 A-2 B+3 C) \cos \left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (2 A-2 B+3 C) \cos \left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 (A-B+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\frac {C \cos \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (B-C) \cos \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\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 )}-\frac {C \cos \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (B-C) \cos \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\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 )}\right )}{a d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.50, size = 158, normalized size = 1.33

method	result	size
derivativedivides	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-\frac {3 C}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (\frac {3 C}{2}-B +A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {3 C}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-\frac {3 C}{2}+B -A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(158\)
default	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-\frac {3 C}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (\frac {3 C}{2}-B +A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {3 C}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-\frac {3 C}{2}+B -A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(158\)
norman	\(\frac {\frac {\left (A -3 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\left (3 A -5 B +6 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (A -B +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (3 A -7 B +7 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {\left (2 A -2 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\left (2 A -2 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(185\)
risch	\(-\frac {i \left (2 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2 B \,{\mathrm e}^{4 i \left (d x +c \right )}+3 C \,{\mathrm e}^{4 i \left (d x +c \right )}-2 B \,{\mathrm e}^{3 i \left (d x +c \right )}+3 C \,{\mathrm e}^{3 i \left (d x +c \right )}+4 A \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B \,{\mathrm e}^{2 i \left (d x +c \right )}+5 C \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{i \left (d x +c \right )}+C \,{\mathrm e}^{i \left (d x +c \right )}+2 A -4 B +4 C \right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a d}\)	\(296\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(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)-1/2*C/(tan(1/2*d*x+1/2*c)+1)^2-(-3/2*C+
B)/(tan(1/2*d*x+1/2*c)+1)+(3/2*C-B+A)*ln(tan(1/2*d*x+1/2*c)+1)+1/2*C/(tan(1/2*d*x+1/2*c)-1)^2-(-3/2*C+B)/(tan(
1/2*d*x+1/2*c)-1)+(-3/2*C+B-A)*ln(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. 356 vs. \(2 (115) = 230\).
time = 0.29, size = 356, normalized size = 2.99 \begin {gather*} -\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 \, B {\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 )} - 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+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*B*(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))) - 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

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Fricas [A]
time = 3.72, size = 174, normalized size = 1.46 \begin {gather*} \frac {{\left ({\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 2 \, B + 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 - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 2 \, B + 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 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (2 \, B - C\right )} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2/(sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**3/(sec(c + d*x) + 1), x) + Integ
ral(C*sec(c + d*x)**4/(sec(c + d*x) + 1), x))/a

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Giac [A]
time = 0.48, size = 173, normalized size = 1.45 \begin {gather*} \frac {\frac {{\left (2 \, A - 2 \, B + 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 - 2 \, B + 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 ) - 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 (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, 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 )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.43, size = 124, normalized size = 1.04 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-B+\frac {3\,C}{2}\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-3\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-C\right )}{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 )} \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)^2*(a + a/cos(c + d*x))),x)

[Out]

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

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