3.911 \(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\)
Optimal. Leaf size=148 \[ \frac {2 \left (a^3 (-C)+a A b^2+2 a b^2 C-b^3 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {C \tanh ^{-1}(\sin (c+d x))}{b^2 d} \]
[Out]
C*arctanh(sin(d*x+c))/b^2/d+2*(A*a*b^2-B*b^3-C*a^3+2*C*a*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/
2))/(a-b)^(3/2)/b^2/(a+b)^(3/2)/d-(A*b^2-a*(B*b-C*a))*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))
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Rubi [A] time = 0.30, antiderivative size = 148, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used =
{4080, 3998, 3770, 3831, 2659, 208} \[ \frac {2 \left (a^3 (-C)+a A b^2+2 a b^2 C-b^3 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {C \tanh ^{-1}(\sin (c+d x))}{b^2 d} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]
[Out]
(C*ArcTanh[Sin[c + d*x]])/(b^2*d) + (2*(a*A*b^2 - b^3*B - a^3*C + 2*a*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x
)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^2*(a + b)^(3/2)*d) - ((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(b*(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 4080
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[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f
*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e +
f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Rubi steps
\begin {align*} \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\sec (c+d x) \left (b (b B-a (A+C))-\left (a^2-b^2\right ) C \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {C \int \sec (c+d x) \, dx}{b^2}-\frac {\left (b^3 B+a^3 C-a b^2 (A+2 C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=\frac {C \tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (b^3 B+a^3 C-a b^2 (A+2 C)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=\frac {C \tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (2 \left (b^3 B+a^3 C-a b^2 (A+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^3 \left (a^2-b^2\right ) d}\\ &=\frac {C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac {2 \left (a A b^2-b^3 B-a^3 C+2 a 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^2 (a+b)^{3/2} d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [C] time = 3.14, size = 356, normalized size = 2.41 \[ \frac {2 (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 (\sin (c)+i \cos (c)) \left (a^3 C-a b^2 (A+2 C)+b^3 B\right ) (a \cos (c+d x)+b) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^{3/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {b \left (a (a C-b B)+A b^2\right ) (b \sin (c)-a \sin (d x))}{a (a-b) (a+b) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )}-C (a \cos (c+d x)+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+C (a \cos (c+d x)+b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{b^2 d (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]
[Out]
(2*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-(C*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2]
- Sin[(c + d*x)/2]]) + C*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*(b^3*B + a^3*C -
a*b^2*(A + 2*C))*ArcTan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[
(Cos[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])*(I*Cos[c] + Sin[c]))/((a^2 - b^2)^(3/2)*Sqrt[(Cos[c] - I*Sin[c])
^2]) + (b*(A*b^2 + a*(-(b*B) + a*C))*(b*Sin[c] - a*Sin[d*x]))/(a*(a - b)*(a + b)*(Cos[c/2] - Sin[c/2])*(Cos[c/
2] + Sin[c/2]))))/(b^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)
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fricas [B] time = 9.14, size = 728, normalized size = 4.92 \[ \left [\frac {{\left (C a^{3} b - {\left (A + 2 \, C\right )} a b^{3} + B b^{4} + {\left (C a^{4} - {\left (A + 2 \, C\right )} a^{2} b^{2} + B a b^{3}\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 ) + {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (C a^{4} b - B a^{3} b^{2} + {\left (A - C\right )} a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}, -\frac {2 \, {\left (C a^{3} b - {\left (A + 2 \, C\right )} a b^{3} + B b^{4} + {\left (C a^{4} - {\left (A + 2 \, C\right )} a^{2} b^{2} + B a b^{3}\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 (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{4} b - B a^{3} b^{2} + {\left (A - C\right )} a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
[Out]
[1/2*((C*a^3*b - (A + 2*C)*a*b^3 + B*b^4 + (C*a^4 - (A + 2*C)*a^2*b^2 + B*a*b^3)*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)*sin(d*x + c)
+ 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*
C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1) - (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b
^2 + C*a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(C*a^4*b - B*a^3*b^2 + (A - C)*a^2*b^3 + B*a*b^4 - A*b^
5)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b^7)*d), -1/2*(2*(C*a^
3*b - (A + 2*C)*a*b^3 + B*b^4 + (C*a^4 - (A + 2*C)*a^2*b^2 + B*a*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-s
qrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2
*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1) + (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*
b^2 + C*a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(C*a^4*b - B*a^3*b^2 + (A - C)*a^2*b^3 + B*a*b^4 - A*b
^5)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b^7)*d)]
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giac [A] time = 0.29, size = 250, normalized size = 1.69 \[ \frac {\frac {2 \, {\left (C a^{3} - A a b^{2} - 2 \, C a b^{2} + B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")
[Out]
(2*(C*a^3 - A*a*b^2 - 2*C*a*b^2 + B*b^3)*(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^2 - b^4)*sqrt(-a^2 + b^2)) + C*log(abs(tan(1
/2*d*x + 1/2*c) + 1))/b^2 - C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2 + 2*(C*a^2*tan(1/2*d*x + 1/2*c) - B*a*b*t
an(1/2*d*x + 1/2*c) + A*b^2*tan(1/2*d*x + 1/2*c))/((a^2*b - b^3)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1
/2*c)^2 - a - b)))/d
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maple [B] time = 0.77, size = 470, normalized size = 3.18 \[ \frac {2 b \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 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B 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 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} C}{d b \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 \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 {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 ) B}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 a^{3} \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}{d \,b^{2} \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 ) C a}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,b^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x)
[Out]
2/d*b/(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/(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)*B*a+2/d/b/(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*C+2/d*a/(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-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))*B-2/d*a^3/b^2/(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+4/d/(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-1/d/b^2
*ln(tan(1/2*d*x+1/2*c)-1)*C+1/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)*C
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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)*(A+B*sec(d*x+c)+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 = 13.13, size = 4536, normalized size = 30.65 \[ \text {result too large to display} \]
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)*(a + b/cos(c + d*x))^2),x)
[Out]
- (C*atan(((C*((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4
+ 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 2*A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 +
2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (C*((32*(A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^
9 + B*a^2*b^7 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6
- a^2*b^4 - a^3*b^3) + (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*
a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2)*1i)/b^2 + (C*((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2
*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 2*A
*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (
C*((32*(A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^9 + B*a^2*b^7 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 +
C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*C*tan(c/2 + (d*x)/2)*(2*a*
b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^
2)*1i)/b^2)/((64*(C^3*a^5 - B*C^2*b^5 + B^2*C*b^5 + 2*C^3*a*b^4 - C^3*a^4*b + 2*C^3*a^2*b^3 - 3*C^3*a^3*b^2 +
A*C^2*a*b^4 - A*C^2*a^4*b - 3*B*C^2*a*b^4 + 3*A*C^2*a^2*b^3 - A*C^2*a^3*b^2 + A^2*C*a^2*b^3 + B*C^2*a^2*b^3 +
B*C^2*a^3*b^2 - 2*A*B*C*a*b^4))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (C*((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^
2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 2*
A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) +
(C*((32*(A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^9 + B*a^2*b^7 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 +
C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*C*tan(c/2 + (d*x)/2)*(2*a
*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b
^2))/b^2 + (C*((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4
+ 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 2*A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 +
2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (C*((32*(A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^
9 + B*a^2*b^7 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6
- a^2*b^4 - a^3*b^3) - (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*
a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2))/b^2))*2i)/(b^2*d) - (atan(((((32*tan(c/2 + (d*x)/2)*
(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C
^2*a^4*b^2 - 2*A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^
3 - a^3*b^2) + (((32*(A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^9 + B*a^2*b^7 - B*a^3*b^6 + C*a^2*b^7 -
3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*tan(c/2 + (d
*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*
a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))
*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(
(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2)*1i)/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2) +
(((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2
*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 2*A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 + 2*B*C*a^3*b
^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (((32*(A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^9 + B*a^2*b^7
- B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a
^3*b^3) - (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2)*(2*a*b^9 -
2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^
6 + 3*a^4*b^4 - a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6
+ 3*a^4*b^4 - a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2)*1i)/(b^8 - 3*a^2*b^
6 + 3*a^4*b^4 - a^6*b^2))/((64*(C^3*a^5 - B*C^2*b^5 + B^2*C*b^5 + 2*C^3*a*b^4 - C^3*a^4*b + 2*C^3*a^2*b^3 - 3*
C^3*a^3*b^2 + A*C^2*a*b^4 - A*C^2*a^4*b - 3*B*C^2*a*b^4 + 3*A*C^2*a^2*b^3 - A*C^2*a^3*b^2 + A^2*C*a^2*b^3 + B*
C^2*a^2*b^3 + B*C^2*a^3*b^2 - 2*A*B*C*a*b^4))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (((32*tan(c/2 + (d*x)/2)*(B^
2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*
a^4*b^2 - 2*A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 -
a^3*b^2) + (((32*(A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^9 + B*a^2*b^7 - B*a^3*b^6 + C*a^2*b^7 - 3*C
*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*tan(c/2 + (d*x)
/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4
*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*((
a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*((a
+ b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2) + (((32
*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 +
4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 2*A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 + 2*B*C*a^3*b^3))/(
a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (((32*(A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^9 + B*a^2*b^7 - B*a^
3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3
) - (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2)*(2*a*b^9 - 2*a^2*
b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*
a^4*b^4 - a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a
^4*b^4 - a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4
*b^4 - a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - A*a*b^2 - 2*C*a*b^2)*2i)/(d*(b^8 - 3*a^2*b^6 +
3*a^4*b^4 - a^6*b^2)) - (2*tan(c/2 + (d*x)/2)*(A*b^2 + C*a^2 - B*a*b))/(d*(a + b)*(a*b - b^2)*(a + b - tan(c/2
+ (d*x)/2)^2*(a - b)))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**2,x)
[Out]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)/(a + b*sec(c + d*x))**2, x)
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