\(\int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx\) [640]

Optimal result

Integrand size = 41, antiderivative size = 26 \[ \int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {(b \sec (c+d x)+a \sin (c+d x))^2}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {4470} \[ \int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {(a \sin (c+d x)+b \sec (c+d x))^2}{2 d} \]

[In]

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

[Out]

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

Rule 4470

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[ActivateTrig[y], ActivateTrig[u], x]}, Simp[q*(A
ctivateTrig[y^(m + 1)]/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1] &&  !InertTrigFreeQ[u]

Rubi steps \begin{align*} \text {integral}& = \frac {(b \sec (c+d x)+a \sin (c+d x))^2}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(67\) vs. \(2(26)=52\).

Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.58 \[ \int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=a b x-\frac {a b \arctan (\tan (c+d x))}{d}-\frac {a^2 \cos ^2(c+d x)}{2 d}+\frac {b^2 \sec ^2(c+d x)}{2 d}+\frac {a b \tan (c+d x)}{d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(56\) vs. \(2(24)=48\).

Time = 8.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.19

method	result	size
derivativedivides	\(\frac {-\frac {\cos \left (d x +c \right )^{2} a^{2}}{2}+a b \left (\tan \left (d x +c \right )-d x -c \right )+a b \left (d x +c \right )+\frac {b^{2}}{2 \cos \left (d x +c \right )^{2}}}{d}\)	\(57\)
default	\(\frac {-\frac {\cos \left (d x +c \right )^{2} a^{2}}{2}+a b \left (\tan \left (d x +c \right )-d x -c \right )+a b \left (d x +c \right )+\frac {b^{2}}{2 \cos \left (d x +c \right )^{2}}}{d}\)	\(57\)
parts	\(\frac {-\frac {\cos \left (d x +c \right )^{2} a^{2}}{2}+a b \left (d x +c \right )}{d}+\frac {b^{2} \sec \left (d x +c \right )^{2}}{2 d}+\frac {a b \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(64\)
risch	\(-\frac {a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 b \left (i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}\)	\(84\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.35 \[ \int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=-\frac {2 \, a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, b^{2}}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (20) = 40\).

Time = 0.41 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.81 \[ \int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\begin {cases} \frac {a^{2} \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {a b \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}}{d} + \frac {b^{2} \sec ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + b \sec {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + b \tan {\left (c \right )} \sec {\left (c \right )}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*sin(c + d*x)**2/(2*d) + a*b*sin(c + d*x)*sec(c + d*x)/d + b**2*sec(c + d*x)**2/(2*d), Ne(d, 0)
), (x*(a*sin(c) + b*sec(c))*(a*cos(c) + b*tan(c)*sec(c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {{\left (b \sec \left (d x + c\right ) + a \sin \left (d x + c\right )\right )}^{2}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) - \frac {a^{2}}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 26.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.35 \[ \int (b \sec (c+d x)+a \sin (c+d x)) (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=-\frac {\frac {a^2\,\left (2\,{\sin \left (2\,c+2\,d\,x\right )}^2-1\right )}{16}+\frac {a^2}{16}+\frac {b^2}{2}+\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \]

[In]

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

[Out]

-((a^2*(2*sin(2*c + 2*d*x)^2 - 1))/16 + a^2/16 + b^2/2 + (a*b*sin(2*c + 2*d*x))/2)/(d*(sin(c + d*x)^2 - 1))