3.5.94 \(\int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx\) [494]

Optimal. Leaf size=110 \[ \frac {\left (a^2+2 b^2\right ) x}{2 a^3}-\frac {2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}-\frac {b \sin (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3938, 4189, 4004, 3916, 2738, 214} \begin {gather*} -\frac {2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {b \sin (c+d x)}{a^2 d}+\frac {x \left (a^2+2 b^2\right )}{2 a^3}+\frac {\sin (c+d x) \cos (c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 + 2*b^2)*x)/(2*a^3) - (2*b^3*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*Sqrt[a - b]*Sqrt[
a + b]*d) - (b*Sin[c + d*x])/(a^2*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*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 3938

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

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

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Mathematica [A]
time = 0.25, size = 97, normalized size = 0.88 \begin {gather*} \frac {2 \left (a^2+2 b^2\right ) (c+d x)+\frac {8 b^3 \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}}-4 a b \sin (c+d x)+a^2 \sin (2 (c+d x))}{4 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 + 2*b^2)*(c + d*x) + (8*b^3*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 4*
a*b*Sin[c + d*x] + a^2*Sin[2*(c + d*x)])/(4*a^3*d)

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Maple [A]
time = 0.13, size = 138, normalized size = 1.25

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2}-b a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{2}-b a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (a^{2}+2 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 b^{3} \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(138\)
default	\(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2}-b a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{2}-b a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (a^{2}+2 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 b^{3} \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(138\)
risch	\(\frac {x}{2 a}+\frac {x \,b^{2}}{a^{3}}+\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}-\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}+\frac {\sin \left (2 d x +2 c \right )}{4 a d}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(2/a^3*(((-1/2*a^2-b*a)*tan(1/2*d*x+1/2*c)^3+(1/2*a^2-b*a)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+
1/2*(a^2+2*b^2)*arctan(tan(1/2*d*x+1/2*c)))-2*b^3/a^3/((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)^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 = 2.88, size = 334, normalized size = 3.04 \begin {gather*} \left [\frac {\sqrt {a^{2} - b^{2}} b^{3} \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 (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} d x - {\left (2 \, a^{3} b - 2 \, a b^{3} - {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}, -\frac {2 \, \sqrt {-a^{2} + b^{2}} b^{3} \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 (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} d x + {\left (2 \, a^{3} b - 2 \, a b^{3} - {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a^2 - b^2)*b^3*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)) + (a^4 + a^2*b^2 - 2*
b^4)*d*x - (2*a^3*b - 2*a*b^3 - (a^4 - a^2*b^2)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d), -1/2*(2*sqrt(
-a^2 + b^2)*b^3*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (a^4 + a^2*b^2 - 2
*b^4)*d*x + (2*a^3*b - 2*a*b^3 - (a^4 - a^2*b^2)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\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)**2/(a+b*sec(d*x+c)),x)

[Out]

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

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Giac [A]
time = 0.47, size = 178, normalized size = 1.62 \begin {gather*} -\frac {\frac {4 \, {\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 )} b^{3}}{\sqrt {-a^{2} + b^{2}} a^{3}} - \frac {{\left (a^{2} + 2 \, b^{2}\right )} {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 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 )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(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)))*b^3/(sqrt(-a^2 + b^2)*a^3) - (a^2 + 2*b^2)*(d*x + c)/a^3 + 2*(a*tan(1/2*d*x + 1/2*c
)^3 + 2*b*tan(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x + 1/2*c) + 2*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2
+ 1)^2*a^2))/d

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Mupad [B]
time = 1.81, size = 592, normalized size = 5.38 \begin {gather*} \frac {a\,\left (\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {\sin \left (2\,c+2\,d\,x\right )}{4}\right )}{d\,\left (a^2-b^2\right )}-\frac {b\,\sin \left (c+d\,x\right )}{d\,\left (a^2-b^2\right )}+\frac {b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}}{a\,d\,\left (a^2-b^2\right )}+\frac {b^3\,\sin \left (c+d\,x\right )}{a^2\,d\,\left (a^2-b^2\right )}-\frac {2\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^3\,d\,\left (a^2-b^2\right )}-\frac {b^3\,\mathrm {atan}\left (\frac {\left (8\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}-a^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+8\,b^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-8\,a^2\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-3\,a^4\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+3\,a^5\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+2\,a^6\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-2\,a^7\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^8\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (4\,b^5\,\left (a^2-b^2\right )+2\,a\,b^6-a^7+4\,b^7-2\,a^2\,b^5+a^3\,b^4-2\,a^4\,b^3-2\,a^5\,b^2+2\,a^2\,b^3\,\left (a^2-b^2\right )+2\,a\,b^4\,\left (a^2-b^2\right )\right )}\right )\,2{}\mathrm {i}}{a^3\,d\,\sqrt {a^2-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*(atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + sin(2*c + 2*d*x)/4))/(d*(a^2 - b^2)) - (b*sin(c + d*x))/(d*(
a^2 - b^2)) + (b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (b^2*sin(2*c + 2*d*x))/4)/(a*d*(a^2 - b^2)) -
(b^3*atan(((8*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2) - a^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 8*b^9*sin
(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 8*a^2*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 3*a^4*b^5*sin(c/2 + (d*x)
/2)*(a^2 - b^2)^(1/2) + 3*a^5*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 2*a^6*b^3*sin(c/2 + (d*x)/2)*(a^2 - b
^2)^(1/2) - 2*a^7*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + a^8*b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))*1i)/(
cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(4*b^5*(a^2 - b^2) + 2*a*b^6 - a^7 + 4*b^7 - 2*a^2*b^5 + a^3*b^4 - 2*a^4*b^3
- 2*a^5*b^2 + 2*a^2*b^3*(a^2 - b^2) + 2*a*b^4*(a^2 - b^2))))*2i)/(a^3*d*(a^2 - b^2)^(1/2)) + (b^3*sin(c + d*x)
)/(a^2*d*(a^2 - b^2)) - (2*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^3*d*(a^2 - b^2))

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