\(\int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx\) [11]

Optimal result

Integrand size = 21, antiderivative size = 96 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {7 a^2 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {7 a^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d} \]

[Out]

7/8*a^2*arctanh(sin(d*x+c))/d+2*a^2*tan(d*x+c)/d+7/8*a^2*sec(d*x+c)*tan(d*x+c)/d+1/4*a^2*sec(d*x+c)^3*tan(d*x+
c)/d+2/3*a^2*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3873, 3852, 4131, 3853, 3855} \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {7 a^2 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {7 a^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^2,x]

[Out]

(7*a^2*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a^2*Tan[c + d*x])/d + (7*a^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^2*
Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (2*a^2*Tan[c + d*x]^3)/(3*d)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rule 3873

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

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \left (2 a^2\right ) \int \sec ^4(c+d x) \, dx+\int \sec ^3(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \left (7 a^2\right ) \int \sec ^3(c+d x) \, dx-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {2 a^2 \tan (c+d x)}{d}+\frac {7 a^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}+\frac {1}{8} \left (7 a^2\right ) \int \sec (c+d x) \, dx \\ & = \frac {7 a^2 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {7 a^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {a^2 \left (21 \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (21 \sec (c+d x)+6 \sec ^3(c+d x)+16 \left (3+\tan ^2(c+d x)\right )\right )\right )}{24 d} \]

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^2,x]

[Out]

(a^2*(21*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(21*Sec[c + d*x] + 6*Sec[c + d*x]^3 + 16*(3 + Tan[c + d*x]^2))))
/(24*d)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(112\)
default	\(\frac {a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(112\)
parts	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {2 a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(117\)
norman	\(\frac {\frac {25 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {83 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {77 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {7 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {7 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {7 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(133\)
risch	\(-\frac {i a^{2} \left (21 \,{\mathrm e}^{7 i \left (d x +c \right )}+45 \,{\mathrm e}^{5 i \left (d x +c \right )}-96 \,{\mathrm e}^{4 i \left (d x +c \right )}-45 \,{\mathrm e}^{3 i \left (d x +c \right )}-128 \,{\mathrm e}^{2 i \left (d x +c \right )}-21 \,{\mathrm e}^{i \left (d x +c \right )}-32\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(134\)
parallelrisch	\(\frac {a^{2} \left (21 \left (-\cos \left (4 d x +4 c \right )-4 \cos \left (2 d x +2 c \right )-3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+21 \left (3+\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 \sin \left (4 d x +4 c \right )+90 \sin \left (d x +c \right )+128 \sin \left (2 d x +2 c \right )+42 \sin \left (3 d x +3 c \right )\right )}{24 d \left (3+\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )\right )}\)	\(149\)

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-2*a^2*(-2/3-1/3*sec(d*
x+c)^2)*tan(d*x+c)+a^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 = 111, normalized size of antiderivative = 1.16 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {21 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, a^{2} \cos \left (d x + c\right )^{3} + 21 \, a^{2} \cos \left (d x + c\right )^{2} + 16 \, a^{2} \cos \left (d x + c\right ) + 6 \, a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(21*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 21*a^2*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(32*a^2*c
os(d*x + c)^3 + 21*a^2*cos(d*x + c)^2 + 16*a^2*cos(d*x + c) + 6*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx=a^{2} \left (\int \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 \sec ^{4}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(sec(c + d*x)**3, x) + Integral(2*sec(c + d*x)**4, x) + Integral(sec(c + d*x)**5, x))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.51 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 3 \, a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 - 3*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4
- 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*a^2*(2*sin(d*x + c)/(sin(d*x
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {21 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 21 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 77 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(21*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 21*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(21*a^2*tan(1/
2*d*x + 1/2*c)^7 - 77*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*a^2*tan(1/2*d*x + 1/2*c)^3 - 75*a^2*tan(1/2*d*x + 1/2*c)
)/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 17.01 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.47 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {7\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {77\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {83\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}-\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int((a + a/cos(c + d*x))^2/cos(c + d*x)^3,x)

[Out]

(7*a^2*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((83*a^2*tan(c/2 + (d*x)/2)^3)/12 - (77*a^2*tan(c/2 + (d*x)/2)^5)/12
+ (7*a^2*tan(c/2 + (d*x)/2)^7)/4 - (25*a^2*tan(c/2 + (d*x)/2))/4)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d
*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))