\(\int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [288]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=a^2 A x+\frac {\left (4 a A b+2 a^2 B+b^2 B\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b (2 A b+3 a B) \tan (c+d x)}{2 d}+\frac {b B (a+b \sec (c+d x)) \tan (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 86, 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 = {4003, 3855, 3852, 8} \[ \int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\left (2 a^2 B+4 a A b+b^2 B\right ) \text {arctanh}(\sin (c+d x))}{2 d}+a^2 A x+\frac {b (3 a B+2 A b) \tan (c+d x)}{2 d}+\frac {b B \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {b B (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^2 A+\left (4 a A b+2 a^2 B+b^2 B\right ) \sec (c+d x)+b (2 A b+3 a B) \sec ^2(c+d x)\right ) \, dx \\ & = a^2 A x+\frac {b B (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac {1}{2} (b (2 A b+3 a B)) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (4 a A b+2 a^2 B+b^2 B\right ) \int \sec (c+d x) \, dx \\ & = a^2 A x+\frac {\left (4 a A b+2 a^2 B+b^2 B\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b B (a+b \sec (c+d x)) \tan (c+d x)}{2 d}-\frac {(b (2 A b+3 a B)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d} \\ & = a^2 A x+\frac {\left (4 a A b+2 a^2 B+b^2 B\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b (2 A b+3 a B) \tan (c+d x)}{2 d}+\frac {b B (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 \[ \int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2 a^2 A d x+\left (4 a A b+2 a^2 B+b^2 B\right ) \text {arctanh}(\sin (c+d x))+b (2 A b+4 a B+b B \sec (c+d x)) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13

method	result	size
parts	\(a^{2} A x +\frac {\left (A \,b^{2}+2 B a b \right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 A a b +B \,a^{2}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(97\)
derivativedivides	\(\frac {A \,a^{2} \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B \tan \left (d x +c \right ) a b +A \tan \left (d x +c \right ) b^{2}+b^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(112\)
default	\(\frac {A \,a^{2} \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B \tan \left (d x +c \right ) a b +A \tan \left (d x +c \right ) b^{2}+b^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(112\)
parallelrisch	\(\frac {-2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A a b +\frac {1}{2} B \,a^{2}+\frac {1}{4} b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A a b +\frac {1}{2} B \,a^{2}+\frac {1}{4} b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+a^{2} A x d \cos \left (2 d x +2 c \right )+\left (A \,b^{2}+2 B a b \right ) \sin \left (2 d x +2 c \right )+a^{2} A x d +B \sin \left (d x +c \right ) b^{2}}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(158\)
norman	\(\frac {a^{2} A x +a^{2} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {b \left (2 A b +4 B a +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-2 a^{2} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b \left (2 A b +4 B a -B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {\left (4 A a b +2 B \,a^{2}+b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (4 A a b +2 B \,a^{2}+b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(184\)
risch	\(a^{2} A x -\frac {i b \left (B b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-B b \,{\mathrm e}^{i \left (d x +c \right )}-2 A b -4 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A a b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} B}{2 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A a b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} B}{2 d}\)	\(217\)

[In]

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

[Out]

a^2*A*x+(A*b^2+2*B*a*b)/d*tan(d*x+c)+(2*A*a*b+B*a^2)/d*ln(sec(d*x+c)+tan(d*x+c))+b^2*B/d*(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 \[ \int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {4 \, A a^{2} d x \cos \left (d x + c\right )^{2} + {\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B b^{2} + 2 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {4 \, {\left (d x + c\right )} A a^{2} - B b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 8 \, A a b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 8 \, B a b \tan \left (d x + c\right ) + 4 \, A b^{2} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(4*(d*x + c)*A*a^2 - B*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)) + 4*B*a^2*log(sec(d*x + c) + tan(d*x + c)) + 8*A*a*b*log(sec(d*x + c) + tan(d*x + c)) + 8*B*a*b*tan(d*x
+ c) + 4*A*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 = 0.33 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.23 \[ \int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (d x + c\right )} A a^{2} + {\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(d*x + c)*A*a^2 + (2*B*a^2 + 4*A*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*B*a^2 + 4*A*a*b +
B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(4*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^2*tan(1/2*d*x + 1/2*c)^
3 - B*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*B*a*b*tan(1/2*d*x + 1/2*c) - 2*A*b^2*tan(1/2*d*x + 1/2*c) - B*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 = 15.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.05 \[ \int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2\,\left (A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {B\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+2\,A\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {A\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,b^2\,\sin \left (c+d\,x\right )}{2}+B\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

[In]

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

[Out]

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