3.682 \(\int \frac {\cos ^3(c+d x) (A+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\)
Optimal. Leaf size=175 \[ -\frac {A b \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {2 b^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {b x \left (a^2 (A+2 C)+2 A b^2\right )}{2 a^4}+\frac {\left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]
[Out]
-1/2*b*(2*A*b^2+a^2*(A+2*C))*x/a^4+1/3*(3*A*b^2+a^2*(2*A+3*C))*sin(d*x+c)/a^3/d-1/2*A*b*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+C*a^2)*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 = 0.61, antiderivative size = 173, normalized size of antiderivative = 0.99,
number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used =
{4105, 4104, 3919, 3831, 2659, 208} \[ \frac {\left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac {2 b^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {b x \left (\frac {2 A b^2}{a^2}+A+2 C\right )}{2 a^2}-\frac {A b \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
[Out]
-(b*(A + (2*A*b^2)/a^2 + 2*C)*x)/(2*a^2) + (2*b^2*(A*b^2 + a^2*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 + a^2*(2*A + 3*C))*Sin[c + d*x])/(3*a^3*d) - (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 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]
Rule 4105
Int[((A_.) + 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*(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, 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+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 (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 \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+\frac {1}{2} a^2 (4 A+6 C)\right )+a A b \sec (c+d x)-3 A b^2 \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^2}\\ &=\frac {\left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {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 b \left (2 A b^2+a^2 (A+2 C)\right )+3 a A b^2 \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^3}\\ &=-\frac {b \left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}+\frac {\left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {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^2 C\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4}\\ &=-\frac {b \left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}+\frac {\left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {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^2 C\right )\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^4}\\ &=-\frac {b \left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}+\frac {\left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {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^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 )}{a^4 d}\\ &=-\frac {b \left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}+\frac {2 b^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 )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {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 = 0.51, size = 149, normalized size = 0.85 \[ \frac {a^3 A \sin (3 (c+d x))-6 b (c+d x) \left (a^2 (A+2 C)+2 A b^2\right )+3 a \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x)-\frac {24 b^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-3 a^2 A b \sin (2 (c+d x))}{12 a^4 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
[Out]
(-6*b*(2*A*b^2 + a^2*(A + 2*C))*(c + d*x) - (24*b^2*(A*b^2 + a^2*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 + a^2*(3*A + 4*C))*Sin[c + d*x] - 3*a^2*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 = 0.53, size = 483, normalized size = 2.76 \[ \left [-\frac {3 \, {\left ({\left (A + 2 \, C\right )} a^{4} b + {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, A b^{5}\right )} d x - 3 \, {\left (C a^{2} b^{2} + A b^{4}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (2 \, {\left (2 \, A + 3 \, C\right )} a^{5} + 2 \, {\left (A - 3 \, C\right )} a^{3} b^{2} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (A a^{4} b - A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}, -\frac {3 \, {\left ({\left (A + 2 \, C\right )} a^{4} b + {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, A b^{5}\right )} d x - 6 \, {\left (C a^{2} b^{2} + A b^{4}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (2 \, {\left (2 \, A + 3 \, C\right )} a^{5} + 2 \, {\left (A - 3 \, C\right )} a^{3} b^{2} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (A a^{4} b - A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")
[Out]
[-1/6*(3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*d*x - 3*(C*a^2*b^2 + 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 + 2*(A - 3*C)*a^3*b^2 - 6*A*a*b^4 +
2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 - 3*(A*a^4*b - A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d
), -1/6*(3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*d*x - 6*(C*a^2*b^2 + 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 + 2*(A - 3*C)*a^3*b^2
- 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 - 3*(A*a^4*b - 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 = 0.26, size = 326, normalized size = 1.86 \[ -\frac {\frac {3 \, {\left (A a^{2} b + 2 \, C a^{2} b + 2 \, A b^{3}\right )} {\left (d x + c\right )}}{a^{4}} - \frac {12 \, {\left (C a^{2} b^{2} + A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{4}} - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")
[Out]
-1/6*(3*(A*a^2*b + 2*C*a^2*b + 2*A*b^3)*(d*x + c)/a^4 - 12*(C*a^2*b^2 + 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 + 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*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 + 12*C*a^2*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) + 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*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 = 1.45, size = 551, normalized size = 3.15 \[ \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^{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 ) C}{d \,a^{2} \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 {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{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}{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 ) C}{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}{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 ) C}{d a \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 {A b \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{d \,a^{4}}-\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(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^2/a^2/((a-b)*(a+
b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*C+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+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*C+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/(1+tan(1/2*d*x
+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*C+2/d/a/(1+tan(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/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*C-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/a^2*b*arctan(tan(1/2*d*x+1/2*c))-2/d/a^4*A*arctan(tan(1/2*d*x+
1/2*c))*b^3-2/d/a^2*C*arctan(tan(1/2*d*x+1/2*c))*b
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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+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 = 7.61, size = 3942, normalized size = 22.53 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^3*(A + C/cos(c + d*x)^2))/(a + b/cos(c + d*x)),x)
[Out]
((tan(c/2 + (d*x)/2)^5*(2*A*a^2 + 2*A*b^2 + 2*C*a^2 + A*a*b))/a^3 + (4*tan(c/2 + (d*x)/2)^3*(A*a^2 + 3*A*b^2 +
3*C*a^2))/(3*a^3) + (tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + 2*C*a^2 - A*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 - 16*A^2*a
*b^8 + 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*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*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))/a^6 + (((8*(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*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 + (8*tan(c/2 + (d*x)/2)*(A*b^3*1i + (a^
2*b*(A + 2*C)*1i)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/a^10)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2))/a^4)*(A*b
^3*1i + (a^2*b*(A + 2*C)*1i)/2)*1i)/a^4 + (((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - 16*A^2*a*b^8 + 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*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*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))/a^6 - (((8*(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*C*a^10*b^3
- 8*C*a^11*b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 - (8*tan(c/2 + (d*x)/2)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2)*(8
*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/a^10)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2))/a^4)*(A*b^3*1i + (a^2*b*(A + 2*C
)*1i)/2)*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*C^3*a^6*b^5 - 4*C^3*a^7*b^4 + 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))/a
^9 + (((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - 16*A^2*a*b^8 + 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*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*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))/a^6 + (((
8*(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*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 4*C*a^
12*b))/a^9 + (8*tan(c/2 + (d*x)/2)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/a^
10)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2))/a^4)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2))/a^4 - (((8*tan(c/2 + (d*x)/
2)*(8*A^2*b^9 - 16*A^2*a*b^8 + 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*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*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))/a^6 - (((8*(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*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 - (8*tan(c/2 +
(d*x)/2)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/a^10)*(A*b^3*1i + (a^2*b*(A
+ 2*C)*1i)/2))/a^4)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2))/a^4))*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2)*2i)/(a^4*d)
- (b^2*atan(((b^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - 16*A^2*a*b^8 +
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*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*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))/a^6 + (b^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*(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*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 +
(8*b^2*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a
^6 - a^4*b^2))))/(a^6 - a^4*b^2))*1i)/(a^6 - a^4*b^2) + (b^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*tan(c
/2 + (d*x)/2)*(8*A^2*b^9 - 16*A^2*a*b^8 + 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*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*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))/a^6 - (b^2*((a + b)*(a -
b))^(1/2)*(A*b^2 + C*a^2)*((8*(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*C*a^10*b^3 - 8*C*a
^11*b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 - (8*b^2*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(8
*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2))))/(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*C^3*a^6*b^5 - 4*C^3*a^7*b
^4 + 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))/a^9 + (b^2*((a + b)*(a - b))^(1/
2)*(A*b^2 + C*a^2)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - 16*A^2*a*b^8 + 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*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*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))/a^6 + (b^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*(4*A*a^8*b^5 - 6*A*a^9*b^4 + 2*A*a^10*b^3 - 2*A*a^1
1*b^2 + 4*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 + (8*b^2*tan(c/2 + (d*x)/2)*((a + b)*(a -
b))^(1/2)*(A*b^2 + C*a^2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2))))/(a^6 - a^4*b^2)))/(a^6
- a^4*b^2) - (b^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - 16*A^2*a*b^8 + 1
6*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*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*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))/a^6 - (b^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*(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*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 - (
8*b^2*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^
6 - a^4*b^2))))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2)))*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*2i)/(d*(a^6 - a^4*
b^2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)),x)
[Out]
Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**3/(a + b*sec(c + d*x)), x)
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