\(\int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) [51]

Optimal result

Integrand size = 28, antiderivative size = 67 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

a^2*arctanh(sin(d*x+c))/d-1/2*b^2*arctanh(sin(d*x+c))/d+2*a*b*sec(d*x+c)/d+1/2*b^2*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3169, 3855, 2686, 8, 2691} \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a b \sec (c+d x)}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

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

[Out]

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

Rule 8

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

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3169

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 \sec (c+d x)+2 a b \sec (c+d x) \tan (c+d x)+b^2 \sec (c+d x) \tan ^2(c+d x)\right ) \, dx \\ & = a^2 \int \sec (c+d x) \, dx+(2 a b) \int \sec (c+d x) \tan (c+d x) \, dx+b^2 \int \sec (c+d x) \tan ^2(c+d x) \, dx \\ & = \frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} b^2 \int \sec (c+d x) \, dx+\frac {(2 a b) \text {Subst}(\int 1 \, dx,x,\sec (c+d x))}{d} \\ & = \frac {a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \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 = 1.00 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {2 a b}{\cos \left (d x +c \right )}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(83\)
default	\(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {2 a b}{\cos \left (d x +c \right )}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(83\)
parts	\(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {2 a b \sec \left (d x +c \right )}{d}\)	\(86\)
parallelrisch	\(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (\cos \left (d x +c \right ) a -\frac {a \cos \left (2 d x +2 c \right )}{2}+\frac {b \sin \left (d x +c \right )}{4}-\frac {a}{2}\right ) b}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(122\)
risch	\(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{2 i \left (d x +c \right )} a +i b +4 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2}}{d}-\frac {b^{2} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d}\)	\(151\)
norman	\(\frac {\frac {4 a b}{d}+\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {3 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {3 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(223\)

[In]

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

[Out]

1/d*(a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a*b/cos(d*x+c)+b^2*(1/2*sin(d*x+c)^3/cos(d*x+c)^2+1/2*sin(d*x+c)-1/2*ln(s
ec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.43 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a b \cos \left (d x + c\right ) + 2 \, b^{2} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((a*cos(c + d*x) + b*sin(c + d*x))**2*sec(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {8 \, a b}{\cos \left (d x + c\right )}}{4 \, d} \]

[In]

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

[Out]

-1/4*(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*a^2*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 8*a*b/cos(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.82 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 22.22 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.58 \[ \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a\,b}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{d} \]

[In]

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

[Out]

(4*a*b + b^2*tan(c/2 + (d*x)/2)^3 + b^2*tan(c/2 + (d*x)/2) - 4*a*b*tan(c/2 + (d*x)/2)^2)/(d*(tan(c/2 + (d*x)/2
)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (atanh(tan(c/2 + (d*x)/2))*(2*a^2 - b^2))/d