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

Optimal. Leaf size=153 \[ \frac {a \left (a^2 C+A b^2\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {2 \left (-2 a^4 C+3 a^2 b^2 C+A b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {2 a C \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac {C \tan (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.48, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4091, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac {2 \left (3 a^2 b^2 C-2 a^4 C+A b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {a \left (a^2 C+A b^2\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {2 a C \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac {C \tan (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3770

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3998

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 4082

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

Rule 4091

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

Rubi steps

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

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Mathematica [B]  time = 2.75, size = 336, normalized size = 2.20 \[ \frac {2 (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac {a b \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b)}+\frac {2 \left (-2 a^4 C+3 a^2 b^2 C+A b^4\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {b C \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {b C \sin \left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+2 a C (a \cos (c+d x)+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a C (a \cos (c+d x)+b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{b^3 d (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*((2*(A*b^4 - 2*a^4*C + 3*a^2*b^2*C)*ArcTanh[((-a + b)*Tan[(c +
d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x]))/(a^2 - b^2)^(3/2) + 2*a*C*(b + a*Cos[c + d*x])*Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]] - 2*a*C*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b*C*(b + a*
Cos[c + d*x])*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (b*C*(b + a*Cos[c + d*x])*Sin[(c + d*x
)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (a*b*(A*b^2 + a^2*C)*Sin[c + d*x])/((a - b)*(a + b))))/(b^3*d*(A
+ 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)

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fricas [B]  time = 3.34, size = 852, normalized size = 5.57 \[ \left [\frac {{\left ({\left (2 \, C a^{5} - 3 \, C a^{3} b^{2} - A a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{4} b - 3 \, C a^{2} b^{3} - A b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \, {\left ({\left (C a^{6} - 2 \, C a^{4} b^{2} + C a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{5} b - 2 \, C a^{3} b^{3} + C a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (C a^{6} - 2 \, C a^{4} b^{2} + C a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{5} b - 2 \, C a^{3} b^{3} + C a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{4} b^{2} - 2 \, C a^{2} b^{4} + C b^{6} + {\left (2 \, C a^{5} b + {\left (A - 3 \, C\right )} a^{3} b^{3} - {\left (A - C\right )} a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (2 \, C a^{5} - 3 \, C a^{3} b^{2} - A a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{4} b - 3 \, C a^{2} b^{3} - A b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left ({\left (C a^{6} - 2 \, C a^{4} b^{2} + C a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{5} b - 2 \, C a^{3} b^{3} + C a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (C a^{6} - 2 \, C a^{4} b^{2} + C a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{5} b - 2 \, C a^{3} b^{3} + C a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{4} b^{2} - 2 \, C a^{2} b^{4} + C b^{6} + {\left (2 \, C a^{5} b + {\left (A - 3 \, C\right )} a^{3} b^{3} - {\left (A - C\right )} a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((2*C*a^5 - 3*C*a^3*b^2 - A*a*b^4)*cos(d*x + c)^2 + (2*C*a^4*b - 3*C*a^2*b^3 - A*b^5)*cos(d*x + c))*sqrt
(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*si
n(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 2*((C*a^6 - 2*C*a^4*b^2 + C*a^2*b
^4)*cos(d*x + c)^2 + (C*a^5*b - 2*C*a^3*b^3 + C*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*((C*a^6 - 2*C*a
^4*b^2 + C*a^2*b^4)*cos(d*x + c)^2 + (C*a^5*b - 2*C*a^3*b^3 + C*a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) +
2*(C*a^4*b^2 - 2*C*a^2*b^4 + C*b^6 + (2*C*a^5*b + (A - 3*C)*a^3*b^3 - (A - C)*a*b^5)*cos(d*x + c))*sin(d*x + c
))/((a^5*b^3 - 2*a^3*b^5 + a*b^7)*d*cos(d*x + c)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c)), (((2*C*a^5 -
3*C*a^3*b^2 - A*a*b^4)*cos(d*x + c)^2 + (2*C*a^4*b - 3*C*a^2*b^3 - A*b^5)*cos(d*x + c))*sqrt(-a^2 + b^2)*arct
an(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - ((C*a^6 - 2*C*a^4*b^2 + C*a^2*b^4)*cos
(d*x + c)^2 + (C*a^5*b - 2*C*a^3*b^3 + C*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((C*a^6 - 2*C*a^4*b^2 +
C*a^2*b^4)*cos(d*x + c)^2 + (C*a^5*b - 2*C*a^3*b^3 + C*a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) + (C*a^4*b^
2 - 2*C*a^2*b^4 + C*b^6 + (2*C*a^5*b + (A - 3*C)*a^3*b^3 - (A - C)*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^5*b^
3 - 2*a^3*b^5 + a*b^7)*d*cos(d*x + c)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c))]

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giac [B]  time = 0.30, size = 382, normalized size = 2.50 \[ \frac {2 \, {\left (\frac {{\left (2 \, C a^{4} - 3 \, C a^{2} b^{2} - A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} b^{2} - b^{4}\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((2*C*a^4 - 3*C*a^2*b^2 - A*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x
+ 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^2*b^3 - b^5)*sqrt(-a^2 + b^2)) - C*a*log(abs(tan(1/2
*d*x + 1/2*c) + 1))/b^3 + C*a*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3 - (2*C*a^3*tan(1/2*d*x + 1/2*c)^3 - C*a^2
*b*tan(1/2*d*x + 1/2*c)^3 + A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + C*b^3*tan(1/2*d*
x + 1/2*c)^3 - 2*C*a^3*tan(1/2*d*x + 1/2*c) - C*a^2*b*tan(1/2*d*x + 1/2*c) - A*a*b^2*tan(1/2*d*x + 1/2*c) + C*
a*b^2*tan(1/2*d*x + 1/2*c) + C*b^3*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4
- 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)*(a^2*b^2 - b^4)))/d

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maple [B]  time = 0.51, size = 402, normalized size = 2.63 \[ -\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d \,b^{2} \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {2 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a^{4} C}{d \,b^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {6 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C \,a^{2}}{d b \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 a C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{3}}-\frac {C}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {C}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*a/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)*A-2/d/b^2*a^3/(a^2-b^2
)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)*C-2/d*b/(a-b)/(a+b)/((a-b)*(a+b))^(1/
2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A+4/d/b^3/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(tan
(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*a^4*C-6/d/b/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2
*c)*(a-b)/((a-b)*(a+b))^(1/2))*C*a^2+2/d*a*C/b^3*ln(tan(1/2*d*x+1/2*c)-1)-1/d*C/b^2/(tan(1/2*d*x+1/2*c)-1)-1/d
*C/b^2/(tan(1/2*d*x+1/2*c)+1)-2/d*a*C/b^3*ln(tan(1/2*d*x+1/2*c)+1)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 12.17, size = 4105, normalized size = 26.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(C*a*atan(((C*a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5
*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^3*
b^4) - (2*C*a*((32*(A*b^12 - A*a^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 + 2
*C*a^6*b^6 - A*a*b^11 - 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (64*C*a*tan(c/2 + (d*x)/2)*(2*a*b^11
- 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))))/b^3)*
2i)/b^3 + (C*a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*
C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^3*b
^4) + (2*C*a*((32*(A*b^12 - A*a^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 + 2*
C*a^6*b^6 - A*a*b^11 - 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (64*C*a*tan(c/2 + (d*x)/2)*(2*a*b^11 -
2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))))/b^3)*2
i)/b^3)/((64*(8*C^3*a^8 - 4*C^3*a^7*b + 12*C^3*a^4*b^4 + 6*C^3*a^5*b^3 - 20*C^3*a^6*b^2 + 2*A^2*C*a*b^7 + 4*A*
C^2*a^2*b^6 + 8*A*C^2*a^3*b^5 - 4*A*C^2*a^4*b^4 - 4*A*C^2*a^5*b^3))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (2*C*a
*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 +
16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (2*C*a*(
(32*(A*b^12 - A*a^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 + 2*C*a^6*b^6 - A*
a*b^11 - 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (64*C*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 -
4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))))/b^3))/b^3 - (2*C*a*(
(32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16
*C^2*a^5*b^3 - 16*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (2*C*a*((3
2*(A*b^12 - A*a^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 + 2*C*a^6*b^6 - A*a*
b^11 - 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (64*C*a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 - 4*
a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))))/b^3))/b^3))*4i)/(b^3*d
) - ((2*tan(c/2 + (d*x)/2)^3*(2*C*a^3 + C*b^3 + A*a*b^2 - C*a*b^2 - C*a^2*b))/(b^2*(a + b)*(a - b)) - (2*tan(c
/2 + (d*x)/2)*(2*C*a^3 - C*b^3 + A*a*b^2 - C*a*b^2 + C*a^2*b))/(b^2*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d
*x)/2)^4*(a - b) - 2*a*tan(c/2 + (d*x)/2)^2)) + (atan(((((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a
^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C
*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (((32*(A*b^12 - A*a^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3
*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 + 2*C*a^6*b^6 - A*a*b^11 - 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32
*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3
*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a
^6*b^3)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)
)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2)*1i)/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3) + ((
(32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16
*C^2*a^5*b^3 - 16*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (((32*(A*b
^12 - A*a^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 + 2*C*a^6*b^6 - A*a*b^11 -
2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 -
2*C*a^4 + 3*C*a^2*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7
- a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 +
3*C*a^2*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*
b^2)*1i)/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))/((64*(8*C^3*a^8 - 4*C^3*a^7*b + 12*C^3*a^4*b^4 + 6*C^3*a^5*b
^3 - 20*C^3*a^6*b^2 + 2*A^2*C*a*b^7 + 4*A*C^2*a^2*b^6 + 8*A*C^2*a^3*b^5 - 4*A*C^2*a^4*b^4 - 4*A*C^2*a^5*b^3))/
(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a^7*b + 4*C^2*a^2*b^
6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C*a^4*b^4))/(a*b^6 +
b^7 - a^2*b^5 - a^3*b^4) + (((32*(A*b^12 - A*a^2*b^10 + A*a^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3*b^9 - 5*C*a^4*b^8
- C*a^5*b^7 + 2*C*a^6*b^6 - A*a*b^11 - 2*C*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*tan(c/2 + (d*x)/2)
*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 +
2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*((a + b)^
3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*((a + b)^3*(a - b
)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3) + (((32*tan(c/2 + (d*x)/2)
*(A^2*b^8 + 8*C^2*a^8 - 8*C^2*a^7*b + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*
a^6*b^2 + 6*A*C*a^2*b^6 - 4*A*C*a^4*b^4))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (((32*(A*b^12 - A*a^2*b^10 + A*a
^3*b^9 + 3*C*a^2*b^10 + 3*C*a^3*b^9 - 5*C*a^4*b^8 - C*a^5*b^7 + 2*C*a^6*b^6 - A*a*b^11 - 2*C*a*b^11))/(a*b^8 +
b^9 - a^2*b^7 - a^3*b^6) - (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2)
*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(
b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2))/(b^9 - 3
*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2))/(b^9 - 3*a^2*b^7
+ 3*a^4*b^5 - a^6*b^3)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^4 - 2*C*a^4 + 3*C*a^2*b^2)*2i)/(d*(b^9 - 3*a^2*b^7
+ 3*a^4*b^5 - a^6*b^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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