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

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

[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 = 0.19, 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 = {4084, 3787, 3767, 8, 3768, 3770} \[ -\frac {(A-2 B+2 C) \tan (c+d x)}{a d}+\frac {(2 A-2 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {(2 A-2 B+3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \]

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 3767

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 3768

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 3770

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

Rule 3787

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 4084

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*Cs
c[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) \operatorname {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]  time = 4.57, size = 392, normalized size = 3.29 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-4 \sec \left (\frac {c}{2}\right ) (A-B+C) \sin \left (\frac {d x}{2}\right )-2 (2 A-2 B+3 C) \cos \left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (2 A-2 B+3 C) \cos \left (\frac {1}{2} (c+d x)\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 (B-C) \sin \left (\frac {d x}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 (B-C) \sin \left (\frac {d x}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {C \cos \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {C \cos \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{a d (\sec (c+d x)+1) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

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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fricas [A]  time = 0.45, size = 174, normalized size = 1.46 \[ \frac {{\left ({\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (A - 2 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (2 \, B - C\right )} \cos \left (d x + c\right ) - C\right )} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]

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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giac [A]  time = 0.24, size = 173, normalized size = 1.45 \[ \frac {\frac {{\left (2 \, A - 2 \, B + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {{\left (2 \, A - 2 \, B + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \]

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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maple [B]  time = 0.67, size = 311, normalized size = 2.61 \[ -\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {C}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 C}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{2 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{a d}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}-\frac {C}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3 C}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{2 a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{a d}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d} \]

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)

[Out]

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

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maxima [B]  time = 0.36, size = 356, normalized size = 2.99 \[ -\frac {C {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 2 \, B {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 2 \, A {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, d} \]

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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mupad [B]  time = 3.43, size = 124, normalized size = 1.04 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-B+\frac {3\,C}{2}\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-3\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-C\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )} \]

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

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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