\(\int \frac {\cos (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [389]

Optimal result

Integrand size = 40, antiderivative size = 91 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

2*B*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)-(B-C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a
+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4157, 4005, 3859, 209, 3880} \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d} \]

[In]

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

[Out]

(2*B*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) - (Sqrt[2]*(B - C)*ArcTan[(Sqrt[a]*T
an[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3880

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

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

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*} \text {integral}& = \int \frac {B+C \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx \\ & = \frac {B \int \sqrt {a+a \sec (c+d x)} \, dx}{a}-(B-C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx \\ & = -\frac {(2 B) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {(2 (B-C)) \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = \frac {2 B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 \left (\sqrt {2} B \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+(-B+C) \arctan \left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {\cos (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(2*(Sqrt[2]*B*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + (-B + C)*ArcTan[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]])*Cos[(c
+ d*x)/2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(202\) vs. \(2(76)=152\).

Time = 0.43 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.23

method	result	size
default	\(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-C \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )\right )}{d a}\)	\(203\)

[In]

int(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/d/a*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(B*ln(csc(d*x+c)
-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-C*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2
-1)^(1/2))-B*2^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.72 (sec) , antiderivative size = 307, normalized size of antiderivative = 3.37 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {\sqrt {2} {\left (B - C\right )} a \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 2 \, B \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right )}{2 \, a d}, \frac {\sqrt {2} {\left (B - C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, B \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{a d}\right ] \]

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(2)*(B - C)*a*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x
+ c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 2*B*sqrt(
-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a
*cos(d*x + c) - a)/(cos(d*x + c) + 1)))/(a*d), (sqrt(2)*(B - C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*B*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))))/(a*d)]

Sympy [F]

\[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral((B + C*sec(c + d*x))*cos(c + d*x)*sec(c + d*x)/sqrt(a*(sec(c + d*x) + 1)), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 699, normalized size of antiderivative = 7.68 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {{\left (\sqrt {2} \sqrt {a} \arctan \left (\frac {{\left ({\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{4} + 16 \, \cos \left (d x + c\right )^{4} + 16 \, \sin \left (d x + c\right )^{4} + 8 \, {\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1\right )} {\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2} - 64 \, \cos \left (d x + c\right )^{3} + 32 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )^{2} + 96 \, \cos \left (d x + c\right )^{2} - 64 \, \cos \left (d x + c\right ) + 16\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\frac {8 \, {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2}}, \frac {{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2} + 4 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) + 4}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2}}\right )\right ) + 2 \, \sin \left (d x + c\right )}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}}, \frac {{\left ({\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{4} + 16 \, \cos \left (d x + c\right )^{4} + 16 \, \sin \left (d x + c\right )^{4} + 8 \, {\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1\right )} {\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2} - 64 \, \cos \left (d x + c\right )^{3} + 32 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )^{2} + 96 \, \cos \left (d x + c\right )^{2} - 64 \, \cos \left (d x + c\right ) + 16\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\frac {8 \, {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2}}, \frac {{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2} + 4 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) + 4}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2}}\right )\right ) + 2 \, \cos \left (d x + c\right ) - 2}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}}\right ) - \sqrt {a} \arctan \left ({\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \sin \left (d x + c\right ), {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \cos \left (d x + c\right )\right )\right )} B}{a d} \]

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-(sqrt(2)*sqrt(a)*arctan2(((abs(2*e^(I*d*x + I*c) + 2)^4 + 16*cos(d*x + c)^4 + 16*sin(d*x + c)^4 + 8*(cos(d*x
+ c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(2*e^(I*d*x + I*c) + 2)^2 - 64*cos(d*x + c)^3 + 32*(cos(d*x +
c)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 96*cos(d*x + c)^2 - 64*cos(d*x + c) + 16)^(1/4)*sin(1/2*arctan2(8
*(cos(d*x + c) - 1)*sin(d*x + c)/abs(2*e^(I*d*x + I*c) + 2)^2, (abs(2*e^(I*d*x + I*c) + 2)^2 + 4*cos(d*x + c)^
2 - 4*sin(d*x + c)^2 - 8*cos(d*x + c) + 4)/abs(2*e^(I*d*x + I*c) + 2)^2)) + 2*sin(d*x + c))/abs(2*e^(I*d*x + I
*c) + 2), ((abs(2*e^(I*d*x + I*c) + 2)^4 + 16*cos(d*x + c)^4 + 16*sin(d*x + c)^4 + 8*(cos(d*x + c)^2 - sin(d*x
+ c)^2 - 2*cos(d*x + c) + 1)*abs(2*e^(I*d*x + I*c) + 2)^2 - 64*cos(d*x + c)^3 + 32*(cos(d*x + c)^2 - 2*cos(d*
x + c) + 1)*sin(d*x + c)^2 + 96*cos(d*x + c)^2 - 64*cos(d*x + c) + 16)^(1/4)*cos(1/2*arctan2(8*(cos(d*x + c) -
1)*sin(d*x + c)/abs(2*e^(I*d*x + I*c) + 2)^2, (abs(2*e^(I*d*x + I*c) + 2)^2 + 4*cos(d*x + c)^2 - 4*sin(d*x +
c)^2 - 8*cos(d*x + c) + 4)/abs(2*e^(I*d*x + I*c) + 2)^2)) + 2*cos(d*x + c) - 2)/abs(2*e^(I*d*x + I*c) + 2)) -
sqrt(a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2
*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c)))*B/(a*d)

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

int((cos(c + d*x)*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2),x)

[Out]

int((cos(c + d*x)*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2), x)