\(\int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) [50]

Optimal result

Integrand size = 28, antiderivative size = 39 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\left (a^2-b^2\right ) x-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \]

[Out]

(a^2-b^2)*x-2*a*b*ln(cos(d*x+c))/d+b^2*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3165, 3558, 3556} \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=x \left (a^2-b^2\right )-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \]

[In]

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

[Out]

(a^2 - b^2)*x - (2*a*b*Log[Cos[c + d*x]])/d + (b^2*Tan[c + d*x])/d

Rule 3165

Int[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(a + b*Tan[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]

Rule 3556

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

Rule 3558

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[b^2*(Tan[c + d*x]/d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \int (a+b \tan (c+d x))^2 \, dx \\ & = \left (a^2-b^2\right ) x+\frac {b^2 \tan (c+d x)}{d}+(2 a b) \int \tan (c+d x) \, dx \\ & = \left (a^2-b^2\right ) x-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.77 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {-i \left ((a+i b)^2 \log (i-\tan (c+d x))-(a-i b)^2 \log (i+\tan (c+d x))\right )+2 b^2 \tan (c+d x)}{2 d} \]

[In]

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

[Out]

((-I)*((a + I*b)^2*Log[I - Tan[c + d*x]] - (a - I*b)^2*Log[I + Tan[c + d*x]]) + 2*b^2*Tan[c + d*x])/(2*d)

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {a^{2} \left (d x +c \right )-2 a b \ln \left (\cos \left (d x +c \right )\right )+b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(44\)
default	\(\frac {a^{2} \left (d x +c \right )-2 a b \ln \left (\cos \left (d x +c \right )\right )+b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(44\)
parts	\(\frac {a^{2} \left (d x +c \right )}{d}+\frac {b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}+\frac {2 a b \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(49\)
risch	\(2 i a b x +a^{2} x -x \,b^{2}+\frac {4 i a b c}{d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(69\)
parallelrisch	\(\frac {2 a b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \cos \left (d x +c \right )-2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+d x \left (a -b \right ) \left (a +b \right ) \cos \left (d x +c \right )+\sin \left (d x +c \right ) b^{2}}{d \cos \left (d x +c \right )}\)	\(107\)
norman	\(\frac {\left (-a^{2}+b^{2}\right ) x +\left (-a^{2}+b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (a^{2}-b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a^{2}-b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {2 a b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(225\)

[In]

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

[Out]

1/d*(a^2*(d*x+c)-2*a*b*ln(cos(d*x+c))+b^2*(tan(d*x+c)-d*x-c))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.54 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} d x \cos \left (d x + c\right ) - 2 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + b^{2} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)**2*(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)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.26 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {{\left (d x + c\right )} a^{2} - {\left (d x + c - \tan \left (d x + c\right )\right )} b^{2} - a b \log \left (-\sin \left (d x + c\right )^{2} + 1\right )}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + b^{2} \tan \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} {\left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 23.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 3.03 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {b^2\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {2\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,a\,b\,\ln \left (\frac {\cos \left (c+d\,x\right )}{\cos \left (c+d\,x\right )+1}\right )}{d}+\frac {2\,a\,b\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d} \]

[In]

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

[Out]

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