\(\int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx\) [1452]

Optimal result

Integrand size = 25, antiderivative size = 42 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-2 a b x+\frac {2 b^2 \cos (c+d x)}{d}+\frac {\sec (c+d x) (a+b \sin (c+d x))^2}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2940, 12, 2718} \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {\sec (c+d x) (a+b \sin (c+d x))^2}{d}-2 a b x+\frac {2 b^2 \cos (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2718

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

Rule 2940

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])/(f
*g*(p + 1))), x] + Dist[1/(g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(p
+ 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b
*x])

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

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.57 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {\left (2 a^2+3 b^2+b^2 \cos (2 (c+d x))\right ) \sec (c+d x)-2 \left (a^2+b^2+2 a b (c+d x)-2 a b \tan (c+d x)\right )}{2 d} \]

[In]

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

[Out]

((2*a^2 + 3*b^2 + b^2*Cos[2*(c + d*x)])*Sec[c + d*x] - 2*(a^2 + b^2 + 2*a*b*(c + d*x) - 2*a*b*Tan[c + d*x]))/(
2*d)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.71

method	result	size
parallelrisch	\(\frac {b^{2} \cos \left (2 d x +2 c \right )+\left (-4 a b x d +2 a^{2}+4 b^{2}\right ) \cos \left (d x +c \right )+4 a b \sin \left (d x +c \right )+2 a^{2}+3 b^{2}}{2 d \cos \left (d x +c \right )}\)	\(72\)
derivativedivides	\(\frac {\frac {a^{2}}{\cos \left (d x +c \right )}+2 a b \left (\tan \left (d x +c \right )-d x -c \right )+b^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d}\)	\(75\)
default	\(\frac {\frac {a^{2}}{\cos \left (d x +c \right )}+2 a b \left (\tan \left (d x +c \right )-d x -c \right )+b^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d}\)	\(75\)
risch	\(-2 a b x +\frac {b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}+\frac {4 i a b +2 a^{2} {\mathrm e}^{i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(91\)
norman	\(\frac {\frac {2 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2}+4 b^{2}}{d}+2 a b x -\frac {\left (6 a^{2}+4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {8 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+2 a b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(200\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.40 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, a b d x \cos \left (d x + c\right ) - b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}{d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.33 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a b - b^{2} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - \frac {a^{2}}{\cos \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.95 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, {\left ({\left (d x + c\right )} a b + \frac {2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} + 2 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}\right )}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.74 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.93 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,b^2}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )}-2\,a\,b\,x \]

[In]

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

[Out]

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