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

Optimal. Leaf size=95 \[ \frac {2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {a C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac {C \tan (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4083, 3998, 3770, 3831, 2659, 208} \[ \frac {2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {a C \tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac {C \tan (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4083

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + 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) - a*C*Csc[e + f*x], x], x], x] /; FreeQ
[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [C]  time = 2.34, size = 331, normalized size = 3.48 \[ \frac {2 \cos (c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {2 i (\cos (c)-i \sin (c)) \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}+a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-a C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b C \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b C \sin \left (\frac {d x}{2}\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{b^2 d (a+b \sec (c+d x)) (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.07, size = 424, normalized size = 4.46 \[ \left [\frac {{\left (C a^{2} + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (C a^{3} - C a b^{2}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{3} - C a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}, \frac {2 \, {\left (C a^{2} + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (C a^{3} - C a b^{2}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{3} - C a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((C*a^2 + A*b^2)*sqrt(a^2 - b^2)*cos(d*x + c)*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^3 - C*a*b^2)*cos(d*x + c)*log(sin(d*x + c) + 1) + (C*a^3 - C*a*b^2)*cos(d*x + c)*log(-sin(d*x + c)
+ 1) + 2*(C*a^2*b - C*b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c)), 1/2*(2*(C*a^2 + A*b^2)*sqrt(-a^2 +
b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c) - (C*a^3 - C*a*b^2
)*cos(d*x + c)*log(sin(d*x + c) + 1) + (C*a^3 - C*a*b^2)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(C*a^2*b - C*
b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c))]

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giac [A]  time = 0.27, size = 163, normalized size = 1.72 \[ -\frac {\frac {C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} b} + \frac {2 \, {\left (C a^{2} + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.50, size = 183, normalized size = 1.93 \[ \frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a^{2} C}{d \,b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {C}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,b^{2}}-\frac {C}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/((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^2/((a-b)*(a+b))^(1/2)*ar
ctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*a^2*C-1/d/b/(tan(1/2*d*x+1/2*c)-1)*C+1/d*a/b^2*ln(tan(1/2*
d*x+1/2*c)-1)*C-1/d/b/(tan(1/2*d*x+1/2*c)+1)*C-1/d*a/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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),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 = 6.06, size = 934, normalized size = 9.83 \[ -\frac {2\,C\,a\,\mathrm {atanh}\left (\frac {64\,C^3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,C^3\,a^3+128\,A\,C^2\,a^3-\frac {64\,C^3\,a^4}{b}+64\,A^2\,C\,a\,b^2-64\,A^2\,C\,a^2\,b-\frac {128\,A\,C^2\,a^4}{b}}+\frac {64\,C^3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,C\,a^2\,b^2-64\,A^2\,C\,a\,b^3+128\,A\,C^2\,a^4-128\,A\,C^2\,a^3\,b+64\,C^3\,a^4-64\,C^3\,a^3\,b}+\frac {128\,A\,C^2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,C\,a^2\,b^2-64\,A^2\,C\,a\,b^3+128\,A\,C^2\,a^4-128\,A\,C^2\,a^3\,b+64\,C^3\,a^4-64\,C^3\,a^3\,b}+\frac {64\,A^2\,C\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,C\,a^2-\frac {64\,C^3\,a^3}{b}+\frac {64\,C^3\,a^4}{b^2}-64\,A^2\,C\,a\,b-\frac {128\,A\,C^2\,a^3}{b}+\frac {128\,A\,C^2\,a^4}{b^2}}+\frac {128\,A\,C^2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,C^3\,a^3+128\,A\,C^2\,a^3-\frac {64\,C^3\,a^4}{b}+64\,A^2\,C\,a\,b^2-64\,A^2\,C\,a^2\,b-\frac {128\,A\,C^2\,a^4}{b}}-\frac {64\,A^2\,C\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,C\,a^2-\frac {64\,C^3\,a^3}{b}+\frac {64\,C^3\,a^4}{b^2}-64\,A^2\,C\,a\,b-\frac {128\,A\,C^2\,a^3}{b}+\frac {128\,A\,C^2\,a^4}{b^2}}\right )}{b^2\,d}-\frac {\ln \left (b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {a^2-b^2}\right )\,\left (A\,b^2\,\sqrt {a^2-b^2}+C\,a^2\,\sqrt {a^2-b^2}\right )}{b^2\,d\,\left (a^2-b^2\right )}-\frac {\ln \left (\frac {\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (C\,a^2+A\,b^2\right )\,\left (\frac {32\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-b\right )\,\left (A^2\,b^4+2\,A\,C\,a^2\,b^2+2\,C^2\,a^4-2\,C^2\,a^3\,b+C^2\,a^2\,b^2\right )}{b^2}+\frac {32\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (C\,a^2+A\,b^2\right )\,\left (a-b\right )\,\left (A\,a^2\,b^2-A\,b^4+C\,a\,b^3-C\,a^3\,b+2\,C\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+2\,A\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\right )}{\left (b^4-a^2\,b^2\right )\,\left (a+b\right )}\right )}{b^4-a^2\,b^2}+\frac {32\,C\,a\,\left (a-b\right )\,\left (A^2\,b^3+A\,C\,a^2\,b+A\,C\,a\,b^2+C^2\,a^3\right )}{b^3}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (C\,a^2+A\,b^2\right )}{d\,\left (b^4-a^2\,b^2\right )}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*C*a*atanh((64*C^3*a^3*tan(c/2 + (d*x)/2))/(64*C^3*a^3 + 128*A*C^2*a^3 - (64*C^3*a^4)/b + 64*A^2*C*a*b^2 -
64*A^2*C*a^2*b - (128*A*C^2*a^4)/b) + (64*C^3*a^4*tan(c/2 + (d*x)/2))/(64*C^3*a^4 + 128*A*C^2*a^4 - 64*C^3*a^
3*b - 128*A*C^2*a^3*b - 64*A^2*C*a*b^3 + 64*A^2*C*a^2*b^2) + (128*A*C^2*a^4*tan(c/2 + (d*x)/2))/(64*C^3*a^4 +
128*A*C^2*a^4 - 64*C^3*a^3*b - 128*A*C^2*a^3*b - 64*A^2*C*a*b^3 + 64*A^2*C*a^2*b^2) + (64*A^2*C*a^2*tan(c/2 +
(d*x)/2))/(64*A^2*C*a^2 - (64*C^3*a^3)/b + (64*C^3*a^4)/b^2 - 64*A^2*C*a*b - (128*A*C^2*a^3)/b + (128*A*C^2*a^
4)/b^2) + (128*A*C^2*a^3*tan(c/2 + (d*x)/2))/(64*C^3*a^3 + 128*A*C^2*a^3 - (64*C^3*a^4)/b + 64*A^2*C*a*b^2 - 6
4*A^2*C*a^2*b - (128*A*C^2*a^4)/b) - (64*A^2*C*a*b*tan(c/2 + (d*x)/2))/(64*A^2*C*a^2 - (64*C^3*a^3)/b + (64*C^
3*a^4)/b^2 - 64*A^2*C*a*b - (128*A*C^2*a^3)/b + (128*A*C^2*a^4)/b^2)))/(b^2*d) - (log(b*tan(c/2 + (d*x)/2) - a
*tan(c/2 + (d*x)/2) + (a^2 - b^2)^(1/2))*(A*b^2*(a^2 - b^2)^(1/2) + C*a^2*(a^2 - b^2)^(1/2)))/(b^2*d*(a^2 - b^
2)) - (log((((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((32*tan(c/2 + (d*x)/2)*(a - b)*(A^2*b^4 + 2*C^2*a^4 - 2*C
^2*a^3*b + C^2*a^2*b^2 + 2*A*C*a^2*b^2))/b^2 + (32*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(a - b)*(A*a^2*b^2
- A*b^4 + C*a*b^3 - C*a^3*b + 2*C*a^3*tan(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 2*A*a*b^2*tan(c/2 + (d*x)/2)*(a^2
- b^2)^(1/2)))/((b^4 - a^2*b^2)*(a + b))))/(b^4 - a^2*b^2) + (32*C*a*(a - b)*(A^2*b^3 + C^2*a^3 + A*C*a*b^2 +
A*C*a^2*b))/b^3)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2))/(d*(b^4 - a^2*b^2)) - (2*C*tan(c/2 + (d*x)/2))/(b*d
*(tan(c/2 + (d*x)/2)^2 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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