3.10.6 \(\int \frac {\cos ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\) [906]

Optimal. Leaf size=205 \[ -\frac {\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]
time = 0.53, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4189, 4004, 3916, 2738, 214} \begin {gather*} \frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {(A b-a B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {\sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{3 a^3 d}-\frac {x \left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right )}{2 a^4}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]

[Out]

-1/2*((2*A*b^3 - a^3*B - 2*a*b^2*B + a^2*b*(A + 2*C))*x)/a^4 + (2*b^2*(A*b^2 - a*(b*B - a*C))*ArcTanh[(Sqrt[a
- b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*Sqrt[a - b]*Sqrt[a + b]*d) + ((3*A*b^2 - 3*a*b*B + a^2*(2*A + 3*C))*
Sin[c + d*x])/(3*a^3*d) - ((A*b - a*B)*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) + (A*Cos[c + d*x]^2*Sin[c + d*x])/
(3*a*d)

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 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 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4189

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (3 (A b-a B)-a (2 A+3 C) \sec (c+d x)-2 A b \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a}\\ &=-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac {\int \frac {\cos (c+d x) \left (2 \left (3 A b^2-3 a b B+\frac {1}{2} a^2 (4 A+6 C)\right )+a (A b+3 a B) \sec (c+d x)-3 b (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^2}\\ &=\frac {\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac {\int \frac {3 \left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right )+3 a b (A b-a B) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^3}\\ &=-\frac {\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac {\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac {\left (b^2 \left (A b^2-a (b B-a C)\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4}\\ &=-\frac {\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac {\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac {\left (b \left (A b^2-a (b B-a C)\right )\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^4}\\ &=-\frac {\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac {\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac {\left (2 b \left (A b^2-a (b B-a 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 )}{a^4 d}\\ &=-\frac {\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d}\\ \end {align*}

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Mathematica [A]
time = 0.64, size = 178, normalized size = 0.87 \begin {gather*} \frac {6 \left (-2 A b^3+a^3 B+2 a b^2 B-a^2 b (A+2 C)\right ) (c+d x)-\frac {24 b^2 \left (A b^2+a (-b B+a C)\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+3 a \left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \sin (c+d x)+3 a^2 (-A b+a B) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))}{12 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]

[Out]

(6*(-2*A*b^3 + a^3*B + 2*a*b^2*B - a^2*b*(A + 2*C))*(c + d*x) - (24*b^2*(A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((-
a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 3*a*(4*A*b^2 - 4*a*b*B + a^2*(3*A + 4*C))*Sin[c +
d*x] + 3*a^2*(-(A*b) + a*B)*Sin[2*(c + d*x)] + a^3*A*Sin[3*(c + d*x)])/(12*a^4*d)

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Maple [A]
time = 0.39, size = 275, normalized size = 1.34

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\left (-A \,a^{3}-\frac {1}{2} A \,a^{2} b -a A \,b^{2}+\frac {1}{2} a^{3} B +a^{2} b B -a^{3} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {2}{3} A \,a^{3}-2 a A \,b^{2}+2 a^{2} b B -2 a^{3} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,a^{3}-a A \,b^{2}+a^{2} b B -a^{3} C +\frac {1}{2} A \,a^{2} b -\frac {1}{2} a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\left (A \,a^{2} b +2 A \,b^{3}-a^{3} B -2 a \,b^{2} B +2 a^{2} b C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}+\frac {2 b^{2} \left (A \,b^{2}-a b B +a^{2} C \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(275\)
default	\(\frac {-\frac {2 \left (\frac {\left (-A \,a^{3}-\frac {1}{2} A \,a^{2} b -a A \,b^{2}+\frac {1}{2} a^{3} B +a^{2} b B -a^{3} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {2}{3} A \,a^{3}-2 a A \,b^{2}+2 a^{2} b B -2 a^{3} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,a^{3}-a A \,b^{2}+a^{2} b B -a^{3} C +\frac {1}{2} A \,a^{2} b -\frac {1}{2} a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\left (A \,a^{2} b +2 A \,b^{3}-a^{3} B -2 a \,b^{2} B +2 a^{2} b C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}+\frac {2 b^{2} \left (A \,b^{2}-a b B +a^{2} C \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(275\)
risch	\(-\frac {A b x}{2 a^{2}}-\frac {b^{3} x A}{a^{4}}+\frac {B x}{2 a}+\frac {x \,b^{2} B}{a^{3}}-\frac {b x C}{a^{2}}-\frac {3 i A \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}+\frac {3 i A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} b B}{2 a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{2}}{2 a^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{2}}{2 a^{3} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} b B}{2 a^{2} d}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}+\frac {A \sin \left (3 d x +3 c \right )}{12 a d}-\frac {A b \sin \left (2 d x +2 c \right )}{4 a^{2} d}+\frac {\sin \left (2 d x +2 c \right ) B}{4 a d}\)	\(712\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/a^4*(((-A*a^3-1/2*A*a^2*b-a*A*b^2+1/2*a^3*B+a^2*b*B-a^3*C)*tan(1/2*d*x+1/2*c)^5+(-2/3*A*a^3-2*a*A*b^2+
2*a^2*b*B-2*a^3*C)*tan(1/2*d*x+1/2*c)^3+(-A*a^3-a*A*b^2+a^2*b*B-a^3*C+1/2*A*a^2*b-1/2*a^3*B)*tan(1/2*d*x+1/2*c
))/(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*(A*a^2*b+2*A*b^3-B*a^3-2*B*a*b^2+2*C*a^2*b)*arctan(tan(1/2*d*x+1/2*c)))+2*b^
2*(A*b^2-B*a*b+C*a^2)/a^4/((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(cos(d*x+c)^3*(A+B*sec(d*x+c)+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 de

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Fricas [A]
time = 4.81, size = 599, normalized size = 2.92 \begin {gather*} \left [\frac {3 \, {\left (B a^{5} - {\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} - {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x + 3 \, {\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\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 (2 \, {\left (2 \, A + 3 \, C\right )} a^{5} - 6 \, B a^{4} b + 2 \, {\left (A - 3 \, C\right )} a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}, \frac {3 \, {\left (B a^{5} - {\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} - {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x + 6 \, {\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\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 (2 \, {\left (2 \, A + 3 \, C\right )} a^{5} - 6 \, B a^{4} b + 2 \, {\left (A - 3 \, C\right )} a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/6*(3*(B*a^5 - (A + 2*C)*a^4*b + B*a^3*b^2 - (A - 2*C)*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*d*x + 3*(C*a^2*b^2 - B
*a*b^3 + A*b^4)*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)) + (2*(2*A + 3*C
)*a^5 - 6*B*a^4*b + 2*(A - 3*C)*a^3*b^2 + 6*B*a^2*b^3 - 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 + 3*(
B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d), 1/6*(3*(B*a^5 - (A +
2*C)*a^4*b + B*a^3*b^2 - (A - 2*C)*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*d*x + 6*(C*a^2*b^2 - B*a*b^3 + A*b^4)*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))) + (2*(2*A + 3*C)*a^5 - 6
*B*a^4*b + 2*(A - 3*C)*a^3*b^2 + 6*B*a^2*b^3 - 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 + 3*(B*a^5 - A
*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**3/(a + b*sec(c + d*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (187) = 374\).
time = 0.48, size = 424, normalized size = 2.07 \begin {gather*} \frac {\frac {3 \, {\left (B a^{3} - A a^{2} b - 2 \, C a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} {\left (d x + c\right )}}{a^{4}} + \frac {12 \, {\left (C a^{2} b^{2} - B a b^{3} + 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 )}}{\sqrt {-a^{2} + b^{2}} a^{4}} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \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 )}^{3} a^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(3*(B*a^3 - A*a^2*b - 2*C*a^2*b + 2*B*a*b^2 - 2*A*b^3)*(d*x + c)/a^4 + 12*(C*a^2*b^2 - B*a*b^3 + A*b^4)*(p
i*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))/sq
rt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)*a^4) + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^5 +
6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*ta
n(1/2*d*x + 1/2*c)^5 + 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*B*a*b*tan(1/2*d*x
+ 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) +
6*C*a^2*tan(1/2*d*x + 1/2*c) - 3*A*a*b*tan(1/2*d*x + 1/2*c) - 6*B*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d
*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d

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Mupad [B]
time = 13.19, size = 2500, normalized size = 12.20 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x)),x)

[Out]

((4*tan(c/2 + (d*x)/2)^3*(A*a^2 + 3*A*b^2 + 3*C*a^2 - 3*B*a*b))/(3*a^3) + (tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b
^2 + B*a^2 + 2*C*a^2 - A*a*b - 2*B*a*b))/a^3 + (tan(c/2 + (d*x)/2)^5*(2*A*a^2 + 2*A*b^2 - B*a^2 + 2*C*a^2 + A*
a*b - 2*B*a*b))/a^3)/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (atan(
-((((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6
+ 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b
^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 + 8*C^2*a^4*b^5 - 16*C^2*a^5*b^4 + 12*C^2*a^6*b^3 - 4*C^2
*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 4*B*C*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*
B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2 + 16*A*C*a^2*b^7 - 32*A*C*a^3*b^6 + 28*A*C*a^4*b^5 - 20*A*C*a^5*b^4
+ 12*A*C*a^6*b^3 - 4*A*C*a^7*b^2 - 16*B*C*a^3*b^6 + 32*B*C*a^4*b^5 - 28*B*C*a^5*b^4 + 20*B*C*a^6*b^3 - 12*B*C
*a^7*b^2))/a^6 + (((8*(4*A*a^8*b^5 - 2*B*a^13 - 6*A*a^9*b^4 + 2*A*a^10*b^3 - 2*A*a^11*b^2 - 4*B*a^9*b^4 + 6*B*
a^10*b^3 - 2*B*a^11*b^2 + 4*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 2*B*a^12*b + 4*C*a^12*b))/a^9 + (8*tan(c/
2 + (d*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*(A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^
2*1i))/a^10)*(A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i))/a^4)*(A*b^3*1i - (B*a^3*1i)/2
+ a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i)*1i)/a^4 + (((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b
^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*
b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 + 8*C^
2*a^4*b^5 - 16*C^2*a^5*b^4 + 12*C^2*a^6*b^3 - 4*C^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 4*B*C*a^8*b + 32*A*
B*a^2*b^7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2 + 16*A*C*a^2*b^7
- 32*A*C*a^3*b^6 + 28*A*C*a^4*b^5 - 20*A*C*a^5*b^4 + 12*A*C*a^6*b^3 - 4*A*C*a^7*b^2 - 16*B*C*a^3*b^6 + 32*B*C
*a^4*b^5 - 28*B*C*a^5*b^4 + 20*B*C*a^6*b^3 - 12*B*C*a^7*b^2))/a^6 - (((8*(4*A*a^8*b^5 - 2*B*a^13 - 6*A*a^9*b^4
+ 2*A*a^10*b^3 - 2*A*a^11*b^2 - 4*B*a^9*b^4 + 6*B*a^10*b^3 - 2*B*a^11*b^2 + 4*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A
*a^12*b + 2*B*a^12*b + 4*C*a^12*b))/a^9 - (8*tan(c/2 + (d*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*(A*b^3*1i
- (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i))/a^10)*(A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C
*b*1i) - B*a*b^2*1i))/a^4)*(A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i)*1i)/a^4)/((16*(4
*A^3*b^11 - 6*A^3*a*b^10 + 6*A^3*a^2*b^9 - 5*A^3*a^3*b^8 + 2*A^3*a^4*b^7 - A^3*a^5*b^6 - 4*B^3*a^3*b^8 + 6*B^3
*a^4*b^7 - 6*B^3*a^5*b^6 + 5*B^3*a^6*b^5 - 2*B^3*a^7*b^4 + B^3*a^8*b^3 + 4*C^3*a^6*b^5 - 4*C^3*a^7*b^4 - 12*A^
2*B*a*b^10 + 12*A*B^2*a^2*b^9 - 18*A*B^2*a^3*b^8 + 18*A*B^2*a^4*b^7 - 15*A*B^2*a^5*b^6 + 6*A*B^2*a^6*b^5 - 3*A
*B^2*a^7*b^4 + 18*A^2*B*a^2*b^9 - 18*A^2*B*a^3*b^8 + 15*A^2*B*a^4*b^7 - 6*A^2*B*a^5*b^6 + 3*A^2*B*a^6*b^5 + 12
*A*C^2*a^4*b^7 - 14*A*C^2*a^5*b^6 + 6*A*C^2*a^6*b^5 - 4*A*C^2*a^7*b^4 + 12*A^2*C*a^2*b^9 - 16*A^2*C*a^3*b^8 +
12*A^2*C*a^4*b^7 - 9*A^2*C*a^5*b^6 + 2*A^2*C*a^6*b^5 - A^2*C*a^7*b^4 - 12*B*C^2*a^5*b^6 + 14*B*C^2*a^6*b^5 - 6
*B*C^2*a^7*b^4 + 4*B*C^2*a^8*b^3 + 12*B^2*C*a^4*b^7 - 16*B^2*C*a^5*b^6 + 12*B^2*C*a^6*b^5 - 9*B^2*C*a^7*b^4 +
2*B^2*C*a^8*b^3 - B^2*C*a^9*b^2 - 24*A*B*C*a^3*b^8 + 32*A*B*C*a^4*b^7 - 24*A*B*C*a^5*b^6 + 18*A*B*C*a^6*b^5 -
4*A*B*C*a^7*b^4 + 2*A*B*C*a^8*b^3))/a^9 + (((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*
a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2
*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 + 8*C^2*a^4*b^5 -
16*C^2*a^5*b^4 + 12*C^2*a^6*b^3 - 4*C^2*a^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 4*B*C*a^8*b + 32*A*B*a^2*b^7 -
32*A*B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2 + 16*A*C*a^2*b^7 - 32*A*C*a
^3*b^6 + 28*A*C*a^4*b^5 - 20*A*C*a^5*b^4 + 12*A*C*a^6*b^3 - 4*A*C*a^7*b^2 - 16*B*C*a^3*b^6 + 32*B*C*a^4*b^5 -
28*B*C*a^5*b^4 + 20*B*C*a^6*b^3 - 12*B*C*a^7*b^2))/a^6 + (((8*(4*A*a^8*b^5 - 2*B*a^13 - 6*A*a^9*b^4 + 2*A*a^10
*b^3 - 2*A*a^11*b^2 - 4*B*a^9*b^4 + 6*B*a^10*b^3 - 2*B*a^11*b^2 + 4*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 2
*B*a^12*b + 4*C*a^12*b))/a^9 + (8*tan(c/2 + (d*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*(A*b^3*1i - (B*a^3*1i
)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i))/a^10)*(A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*
a*b^2*1i))/a^4)*(A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i))/a^4 - (((8*tan(c/2 + (d*x)
/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A
^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 +
13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 + 8*C^2*a^4*b^5 ...

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