\(\int \cos (c+d x) (a+b \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [778]

Optimal result

Integrand size = 38, antiderivative size = 86 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^2 B x+\frac {\left (4 a b B+2 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b (2 b B+3 a C) \tan (c+d x)}{2 d}+\frac {b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {4157, 4003, 3855, 3852, 8} \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (2 a^2 C+4 a b B+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+a^2 B x+\frac {b (3 a C+2 b B) \tan (c+d x)}{2 d}+\frac {b C \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]

[In]

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

[Out]

a^2*B*x + ((4*a*b*B + 2*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (b*(2*b*B + 3*a*C)*Tan[c + d*x])/(2*d) +
(b*C*(a + b*Sec[c + d*x])*Tan[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4003

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

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 (a+b \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx \\ & = \frac {b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^2 B+\left (4 a b B+2 a^2 C+b^2 C\right ) \sec (c+d x)+b (2 b B+3 a C) \sec ^2(c+d x)\right ) \, dx \\ & = a^2 B x+\frac {b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac {1}{2} (b (2 b B+3 a C)) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (4 a b B+2 a^2 C+b^2 C\right ) \int \sec (c+d x) \, dx \\ & = a^2 B x+\frac {\left (4 a b B+2 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}-\frac {(b (2 b B+3 a C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d} \\ & = a^2 B x+\frac {\left (4 a b B+2 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b (2 b B+3 a C) \tan (c+d x)}{2 d}+\frac {b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^2 B d x+\left (4 a b B+2 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))+b (2 b B+4 a C+b C \sec (c+d x)) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

(2*a^2*B*d*x + (4*a*b*B + 2*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]] + b*(2*b*B + 4*a*C + b*C*Sec[c + d*x])*Tan[c
+ d*x])/(2*d)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {B \,a^{2} \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 C \tan \left (d x +c \right ) a b +B \tan \left (d x +c \right ) b^{2}+C \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(112\)
default	\(\frac {B \,a^{2} \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 C \tan \left (d x +c \right ) a b +B \tan \left (d x +c \right ) b^{2}+C \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(112\)
parallelrisch	\(\frac {-2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (B a b +\frac {1}{2} C \,a^{2}+\frac {1}{4} C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (B a b +\frac {1}{2} C \,a^{2}+\frac {1}{4} C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+a^{2} B x d \cos \left (2 d x +2 c \right )+\left (B \,b^{2}+2 C a b \right ) \sin \left (2 d x +2 c \right )+a^{2} B x d +C \sin \left (d x +c \right ) b^{2}}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(158\)
risch	\(a^{2} B x -\frac {i b \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 B b -4 C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{2}}{2 d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{2}}{2 d}\)	\(217\)
norman	\(\frac {a^{2} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {b \left (2 B b +4 C a -C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {b \left (2 B b +4 C a +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-a^{2} B x +2 a^{2} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 a^{2} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {b \left (2 B b +4 C a -C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {b \left (2 B b +4 C a +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (4 B a b +2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (4 B a b +2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(276\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.58 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, B a^{2} d x \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b^{2} + 2 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/4*(4*B*a^2*d*x*cos(d*x + c)^2 + (2*C*a^2 + 4*B*a*b + C*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*C*a^2
+ 4*B*a*b + C*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(C*b^2 + 2*(2*C*a*b + B*b^2)*cos(d*x + c))*sin(d*
x + c))/(d*cos(d*x + c)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.63 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (d x + c\right )} B a^{2} - C b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \tan \left (d x + c\right ) + 4 \, B b^{2} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(4*(d*x + c)*B*a^2 - C*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) + 2*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*B*a*b*(log(sin(d*x + c) + 1) - log(sin(d*
x + c) - 1)) + 8*C*a*b*tan(d*x + c) + 4*B*b^2*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (80) = 160\).

Time = 0.32 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.23 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (d x + c\right )} B a^{2} + {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(d*x + c)*B*a^2 + (2*C*a^2 + 4*B*a*b + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*C*a^2 + 4*B*a*b +
C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(4*C*a*b*tan(1/2*d*x + 1/2*c)^3 + 2*B*b^2*tan(1/2*d*x + 1/2*c)^
3 - C*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*b*tan(1/2*d*x + 1/2*c) - 2*B*b^2*tan(1/2*d*x + 1/2*c) - C*b^2*tan(1/2
*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 17.84 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.05 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+2\,B\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {B\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,b^2\,\sin \left (c+d\,x\right )}{2}+C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

[In]

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

[Out]

(2*(B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + C*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (
C*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + 2*B*a*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))
/d + ((B*b^2*sin(2*c + 2*d*x))/2 + (C*b^2*sin(c + d*x))/2 + C*a*b*sin(2*c + 2*d*x))/(d*(cos(2*c + 2*d*x)/2 + 1
/2))