3.318 \(\int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=178 \[ \frac {2 b^3 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {(A b-a B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {x \left (a^2+2 b^2\right ) (A b-a B)}{2 a^4}+\frac {\left (2 a^2 A-3 a b B+3 A b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]

[Out]

-1/2*(a^2+2*b^2)*(A*b-B*a)*x/a^4+1/3*(2*A*a^2+3*A*b^2-3*B*a*b)*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^3*(A*b-B*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 = 0.64, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4034, 4104, 3919, 3831, 2659, 208} \[ \frac {\left (2 a^2 A-3 a b B+3 A b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac {2 b^3 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {x \left (a^2+2 b^2\right ) (A b-a B)}{2 a^4}-\frac {(A b-a B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2 + 2*b^2)*(A*b - a*B)*x)/(2*a^4) + (2*b^3*(A*b - a*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]
])/(a^4*Sqrt[a - b]*Sqrt[a + b]*d) + ((2*a^2*A + 3*A*b^2 - 3*a*b*B)*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 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 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 3919

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 4034

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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*(n +
1)*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, m}, x] && NeQ[A*b
- a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4104

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) (A+B \sec (c+d x))}{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)-2 a A \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 (2 a^2 A+3 A b^2-3 a b B\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 (2 a^2 A+3 A b^2-3 a b B\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 (a^2+2 b^2\right ) (A b-a B)+3 a b (A b-a B) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^3}\\ &=-\frac {\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac {\left (2 a^2 A+3 A b^2-3 a b B\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^3 (A b-a B)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4}\\ &=-\frac {\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac {\left (2 a^2 A+3 A b^2-3 a b B\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 (A b-a B)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^4}\\ &=-\frac {\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac {\left (2 a^2 A+3 A b^2-3 a b B\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^2 (A b-a B)\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 )}{a^4 d}\\ &=-\frac {\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac {2 b^3 (A b-a B) \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 (2 a^2 A+3 A b^2-3 a b B\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 = 0.54, size = 152, normalized size = 0.85 \[ \frac {a^3 A \sin (3 (c+d x))+6 \left (a^2+2 b^2\right ) (c+d x) (a B-A b)+3 a \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x)-\frac {24 b^3 (A b-a B) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+3 a^2 (a B-A b) \sin (2 (c+d x))}{12 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(a^2 + 2*b^2)*(-(A*b) + a*B)*(c + d*x) - (24*b^3*(A*b - a*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 -
b^2]])/Sqrt[a^2 - b^2] + 3*a*(3*a^2*A + 4*A*b^2 - 4*a*b*B)*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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fricas [A]  time = 0.56, size = 547, normalized size = 3.07 \[ \left [\frac {3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x - 3 \, {\left (B a b^{3} - A b^{4}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (4 \, A a^{5} - 6 \, B a^{4} b + 2 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}, \frac {3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x - 6 \, {\left (B a b^{3} - A b^{4}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (4 \, A a^{5} - 6 \, B a^{4} b + 2 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*d*x - 3*(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)) + (4*A*a^5 - 6*B*a^4*b + 2*A*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*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*
b^5)*d*x - 6*(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))) + (4*A*a^5 - 6*B*a^4*b + 2*A*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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giac [B]  time = 0.31, size = 360, normalized size = 2.02 \[ \frac {\frac {3 \, {\left (B a^{3} - A a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} {\left (d x + c\right )}}{a^{4}} - \frac {12 \, {\left (B a b^{3} - A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{4}} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(B*a^3 - A*a^2*b + 2*B*a*b^2 - 2*A*b^3)*(d*x + c)/a^4 - 12*(B*a*b^3 - 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)))/(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 + 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*tan(1/2*d*x + 1/2*c)^5 + 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 1
2*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) - 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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maple [B]  time = 1.22, size = 641, normalized size = 3.60 \[ \frac {2 b^{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}{d \,a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b^{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 ) B}{d \,a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A b}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A \,b^{2}}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b B}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A \,b^{2}}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b B}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A \,b^{2}}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b B}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A b}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d \,a^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A \,b^{3}}{d \,a^{4}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B \,b^{2}}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*b^4/a^4/((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^3/a^3/((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/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x
+1/2*c)^5*A+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A*b+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan
(1/2*d*x+1/2*c)^5*A*b^2-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*B-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^
2)^3*tan(1/2*d*x+1/2*c)^5*b*B+4/3/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A+4/d/a^3/(1+tan(1/2*d*x
+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A*b^2-4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*b*B+2/d/a/(1+t
an(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b^2-2/d/a^
2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*b*B-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b+
1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*B-1/d/a^2*A*arctan(tan(1/2*d*x+1/2*c))*b-2/d/a^4*arctan(ta
n(1/2*d*x+1/2*c))*A*b^3+1/a/d*arctan(tan(1/2*d*x+1/2*c))*B+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B*b^2

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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(cos(d*x+c)^3*(A+B*sec(d*x+c))/(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 = 6.90, size = 4572, normalized size = 25.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + B*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 + A*a*b - 2*B*a*b))/a^3 + (4*tan(c/2 + (d*x)/2)^3*(A*a^2 + 3*A*b^2 - 3*B*a*b))/(3*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((((a^2 + 2*b^2)*(A*b - B*a)*((
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 - 16*A*B*a*b^8 + 2*A*B*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))/a^6 - ((a^2 + 2*b^2)*(A*b - B*a)*((8*(2*
B*a^13 - 4*A*a^8*b^5 + 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 -
2*A*a^12*b - 2*B*a^12*b))/a^9 - (tan(c/2 + (d*x)/2)*(a^2 + 2*b^2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*
b^2)*4i)/a^10)*1i)/(2*a^4)))/(2*a^4) + ((a^2 + 2*b^2)*(A*b - B*a)*((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 - 16*A*B*a*b^8 + 2*A*B*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))/a^6 + ((a^2 + 2*b^2)*(A*b - B*a)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 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 - 2*A*a^12*b - 2*B*a^12*b))/a^9 + (tan(c/2
+ (d*x)/2)*(a^2 + 2*b^2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*4i)/a^10)*1i)/(2*a^4)))/(2*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 - 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))/a^9 + ((a^2 + 2*b^2)*(A*b - B*a
)*((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 - 16*A*B*a*b^8 + 2*A*B*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))/a^6 - ((a^2 + 2*b^2)*(A*b - B*a)*((8
*(2*B*a^13 - 4*A*a^8*b^5 + 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 - 2*A*a^12*b - 2*B*a^12*b))/a^9 - (tan(c/2 + (d*x)/2)*(a^2 + 2*b^2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 16*
a^9*b^2)*4i)/a^10)*1i)/(2*a^4))*1i)/(2*a^4) - ((a^2 + 2*b^2)*(A*b - B*a)*((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 - 16*A*B*a*b^8 + 2*A*B*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))/a^6 + ((a^2 + 2*b^2)*(A*b - B*a)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 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 - 2*A*a^12*b - 2*B*a^12*b))/a^9 + (t
an(c/2 + (d*x)/2)*(a^2 + 2*b^2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*4i)/a^10)*1i)/(2*a^4))*1i)/(2*
a^4)))*(a^2 + 2*b^2)*(A*b - B*a))/(a^4*d) - (b^3*atan(((b^3*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((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 - 16*A*B*a*b^8 + 2*A*B*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))/a^6 + (b^3*((a + b)*(a - b))^(1/2)*((8*(2*B*a^13 - 4
*A*a^8*b^5 + 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 - 2*A*a^12*
b - 2*B*a^12*b))/a^9 + (8*b^3*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 1
6*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(A*b - B*a))/(a^6 - a^4*b^2))*1i)/(a^6 - a^4*b^2) + (b^3*((a + b)*(a - b))^
(1/2)*(A*b - B*a)*((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 - 16*A*B*a*b^8 + 2*A*B*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))/a^6 - (b^3*((a + b)*
(a - b))^(1/2)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 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^1
0*b^3 + 2*B*a^11*b^2 - 2*A*a^12*b - 2*B*a^12*b))/a^9 - (8*b^3*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b
- B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(A*b - B*a))/(a^6 - a^4*b^2))*1i)/(a^6 - a^
4*b^2))/((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 - 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))/a^9 - (b^3*((a + b)
*(a - b))^(1/2)*(A*b - B*a)*((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 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 3
2*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))/a^6 + (b^3
*((a + b)*(a - b))^(1/2)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 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 - 2*A*a^12*b - 2*B*a^12*b))/a^9 + (8*b^3*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(
1/2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(A*b - B*a))/(a^6 - a^4*b^2)))/(a
^6 - a^4*b^2) + (b^3*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((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 - 1
6*A*B*a*b^8 + 2*A*B*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))/a^6 - (b^3*((a + b)*(a - b))^(1/2)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 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 - 2*A*a^12*b - 2*B*a^12*b))/a^9 - (8*b^3*tan(c/2 +
(d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(A*b
- B*a))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2)))*((a + b)*(a - b))^(1/2)*(A*b - B*a)*2i)/(d*(a^6 - a^4*b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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