3.9.3 \(\int \frac {\sec (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [803]

Optimal. Leaf size=131 \[ \frac {C \tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac {2 \left (b^3 B+a^3 C-2 a b^2 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}+\frac {a (b B-a C) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

C*arctanh(sin(d*x+c))/b^2/d-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*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 = 0.23, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {4157, 4094, 4083, 3855, 3916, 2738, 214} \begin {gather*} -\frac {2 \left (a^3 C-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 {a (b B-a C) \tan (c+d x)}{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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*(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*(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*B - a*C)*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c + d
*x]))

Rule 214

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

Rule 2738

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 3855

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

Rule 3916

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

Rule 4083

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 4094

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

Rule 4157

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\sec ^2(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\\ &=\frac {a (b B-a C) \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 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 {a (b B-a C) \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-2 a b^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 {a (b B-a C) \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-2 a b^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 {a (b B-a C) \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-2 a b^2 C\right )\right ) \text {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 (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 {a (b B-a C) \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 [A]
time = 0.75, size = 191, normalized size = 1.46 \begin {gather*} \frac {\cos (c+d x) (B+C \sec (c+d x)) \left (\frac {2 \left (b^3 B+a^3 C-2 a b^2 C\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a b (-b B+a C) \sin (c+d x)}{(-a+b) (a+b) (b+a \cos (c+d x))}\right )}{b^2 d (C+B \cos (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.31, size = 185, normalized size = 1.41

method	result	size
derivativedivides	\(\frac {-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}+\frac {-\frac {2 a b \left (b B -a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {2 \left (b^{3} B +a^{3} C -2 C \,b^{2} a \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{2}}}{d}\)	\(185\)
default	\(\frac {-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}+\frac {-\frac {2 a b \left (b B -a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {2 \left (b^{3} B +a^{3} C -2 C \,b^{2} a \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{2}}}{d}\)	\(185\)
risch	\(-\frac {2 i \left (-b B +a C \right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{\left (a^{2}-b^{2}\right ) d b \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{b^{2} d}\)	\(619\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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 de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (123) = 246\).
time = 9.02, size = 694, normalized size = 5.30 \begin {gather*} \left [\frac {{\left (C a^{3} b - 2 \, C a b^{3} + B b^{4} + {\left (C a^{4} - 2 \, C 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} - C a^{2} b^{3} + B a b^{4}\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 - 2 \, C a b^{3} + B b^{4} + {\left (C a^{4} - 2 \, C 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} - C a^{2} b^{3} + B a b^{4}\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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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 - 2*C*a*b^3 + B*b^4 + (C*a^4 - 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 - C*a^2*b^3 + B*a*b^4)*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 - 2*C*a*b^3 + B*b^4 +
(C*a^4 - 2*C*a^2*b^2 + B*a*b^3)*cos(d*x + c))*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))) - (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))*lo
g(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 - C*a^2*b^3 + B*a*b^4)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*c
os(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b^7)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (B + C \sec {\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.50, size = 229, normalized size = 1.75 \begin {gather*} \frac {\frac {2 \, {\left (C a^{3} - 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 )\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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 - 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*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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Mupad [B]
time = 11.90, size = 2500, normalized size = 19.08 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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*((C*((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + 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 - (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 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*
C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2 - (C*((C*((32*(B*a^2*b^7 - C*b^9 - B*b^9
- B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + 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 + (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 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 -
a^3*b^2))*1i)/b^2)/((C*((C*((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4
+ 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 - (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 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*
b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/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 - 3*B*C^2*a*b^4 + B*C^2*a^2*b^3 + B*C^2*a^3*b^2
))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (C*((C*((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^
3*b^6 + C*a^5*b^4 + 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
+ (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 + 3*C^2*a^2*b^4 + 4*C^2*a^
3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*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 + 3*C^2*a^2*b^4 + 4
*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (((32*(B*a^2*
b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + 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 - 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 - 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 - 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 + 3*C^2*a^2*b^4 + 4
*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (((32*(B*a^2*
b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + 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 - 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 - 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 - 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 - 3*B*C^
2*a*b^4 + B*C^2*a^2*b^3 + B*C^2*a^3*b^2))/(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 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*
b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2
*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + 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 - 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 - 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 - 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 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5
+ 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^
7 - 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8 + 2*C*a*b...

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