\(\int (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [87]

Optimal result

Integrand size = 23, antiderivative size = 58 \[ \int (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=a A x+\frac {a (2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a C \tan (c+d x)}{d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

a*A*x+1/2*a*(2*A+C)*arctanh(sin(d*x+c))/d+a*C*tan(d*x+c)/d+1/2*a*C*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4134, 3855, 3852, 8} \[ \int (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+a A x+\frac {a C \tan (c+d x)}{d}+\frac {a C \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

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

[Out]

a*A*x + (a*(2*A + C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*C*Tan[c + d*x])/d + (a*C*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 4134

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C
*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Dist[1/2, Int[Simp[2*A*a + b*(2*A + C)*Csc[e + f*x] + 2*a*C*Csc[e + f
*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {a C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a A+a (2 A+C) \sec (c+d x)+2 a C \sec ^2(c+d x)\right ) \, dx \\ & = a A x+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d}+(a C) \int \sec ^2(c+d x) \, dx+\frac {1}{2} (a (2 A+C)) \int \sec (c+d x) \, dx \\ & = a A x+\frac {a (2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d}-\frac {(a C) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = a A x+\frac {a (2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a C \tan (c+d x)}{d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.16 \[ \int (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=a A x+\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {a C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a C \tan (c+d x)}{d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

a*A*x + (a*A*ArcTanh[Sin[c + d*x]])/d + (a*C*ArcTanh[Sin[c + d*x]])/(2*d) + (a*C*Tan[c + d*x])/d + (a*C*Sec[c
+ d*x]*Tan[c + d*x])/(2*d)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28

method	result	size
derivativedivides	\(\frac {a A \left (d x +c \right )+C a \tan \left (d x +c \right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(74\)
default	\(\frac {a A \left (d x +c \right )+C a \tan \left (d x +c \right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(74\)
parts	\(a A x +\frac {a C \tan \left (d x +c \right )}{d}+\frac {C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(75\)
parallelrisch	\(-\frac {a \left (\left (A +\frac {C}{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A +\frac {C}{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-d x A \cos \left (2 d x +2 c \right )-d x A -C \sin \left (d x +c \right )-\sin \left (2 d x +2 c \right ) C \right )}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(120\)
norman	\(\frac {a A x +a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {3 C a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {C a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-2 a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {a \left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(134\)
risch	\(a A x -\frac {i a C \left ({\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-2\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(139\)

[In]

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

[Out]

1/d*(a*A*(d*x+c)+C*a*tan(d*x+c)+a*A*ln(sec(d*x+c)+tan(d*x+c))+C*a*(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 = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.74 \[ \int (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, A a d x \cos \left (d x + c\right )^{2} + {\left (2 \, A + C\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + C\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a \cos \left (d x + c\right ) + C a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A\, dx + \int A \sec {\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int C \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)

[Out]

a*(Integral(A, x) + Integral(A*sec(c + d*x), x) + Integral(C*sec(c + d*x)**2, x) + Integral(C*sec(c + d*x)**3,
x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.52 \[ \int (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (d x + c\right )} A a - C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, C a \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(4*(d*x + c)*A*a - C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1
)) + 4*A*a*log(sec(d*x + c) + tan(d*x + c)) + 4*C*a*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.81 \[ \int (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (d x + c\right )} A a + {\left (2 \, A a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(d*x + c)*A*a + (2*A*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A*a + C*a)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(C*a*tan(1/2*d*x + 1/2*c)^3 - 3*C*a*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 = 16.40 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.21 \[ \int (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]

[In]

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

[Out]

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