\(\int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx\) [760]

Optimal result

Integrand size = 21, antiderivative size = 71 \[ \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {5 a^2 x}{2}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

-5/2*a^2*x+2*a^2*cos(d*x+c)/d+2*a^2*cos(d*x+c)/d/(1-sin(d*x+c))+1/2*a^2*cos(d*x+c)*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 71, 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 = {2788, 2727, 2718, 2715, 8} \[ \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {5 a^2 x}{2} \]

[In]

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

[Out]

(-5*a^2*x)/2 + (2*a^2*Cos[c + d*x])/d + (2*a^2*Cos[c + d*x])/(d*(1 - Sin[c + d*x])) + (a^2*Cos[c + d*x]*Sin[c
+ d*x])/(2*d)

Rule 8

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

Rule 2715

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

Rule 2718

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

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rubi steps \begin{align*} \text {integral}& = a^2 \int \left (-2-\frac {2}{-1+\sin (c+d x)}-2 \sin (c+d x)-\sin ^2(c+d x)\right ) \, dx \\ & = -2 a^2 x-a^2 \int \sin ^2(c+d x) \, dx-\left (2 a^2\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx-\left (2 a^2\right ) \int \sin (c+d x) \, dx \\ & = -2 a^2 x+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} a^2 \int 1 \, dx \\ & = -\frac {5 a^2 x}{2}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.69 \[ \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {a^2 \sec (c+d x) (1+\sin (c+d x))^{5/2} \left (-10 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (-8+3 \sin (c+d x)+\sin ^2(c+d x)\right )\right )}{2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

[In]

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

[Out]

-1/2*(a^2*Sec[c + d*x]*(1 + Sin[c + d*x])^(5/2)*(-10*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d
*x]] + Sqrt[1 + Sin[c + d*x]]*(-8 + 3*Sin[c + d*x] + Sin[c + d*x]^2)))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
)^4)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92

method	result	size
parallelrisch	\(\frac {a^{2} \left (-20 d x \cos \left (d x +c \right )+\sin \left (3 d x +3 c \right )+8 \cos \left (2 d x +2 c \right )+17 \sin \left (d x +c \right )+32 \cos \left (d x +c \right )+24\right )}{8 d \cos \left (d x +c \right )}\)	\(65\)
risch	\(-\frac {5 a^{2} x}{2}+\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {4 a^{2}}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{4 d}\)	\(79\)
derivativedivides	\(\frac {a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+2 a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(117\)
default	\(\frac {a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+2 a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(117\)
norman	\(\frac {\frac {5 a^{2} x}{2}-\frac {8 a^{2}}{d}-\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {6 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {5 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {5 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(172\)

[In]

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

[Out]

1/8/d*a^2*(-20*d*x*cos(d*x+c)+sin(3*d*x+3*c)+8*cos(2*d*x+2*c)+17*sin(d*x+c)+32*cos(d*x+c)+24)/cos(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.76 \[ \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^{2} \cos \left (d x + c\right )^{3} - 5 \, a^{2} d x + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{2} - {\left (5 \, a^{2} d x - 7 \, a^{2}\right )} \cos \left (d x + c\right ) + {\left (5 \, a^{2} d x + a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

1/2*(a^2*cos(d*x + c)^3 - 5*a^2*d*x + 4*a^2*cos(d*x + c)^2 + 4*a^2 - (5*a^2*d*x - 7*a^2)*cos(d*x + c) + (5*a^2
*d*x + a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c) + 4*a^2)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) + d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.18 \[ \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{2} + 2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 4 \, a^{2} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.44 \[ \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {5 \, {\left (d x + c\right )} a^{2} + \frac {8 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

-1/2*(5*(d*x + c)*a^2 + 8*a^2/(tan(1/2*d*x + 1/2*c) - 1) + 2*(a^2*tan(1/2*d*x + 1/2*c)^3 - 4*a^2*tan(1/2*d*x +
1/2*c)^2 - a^2*tan(1/2*d*x + 1/2*c) - 4*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d

Mupad [B] (verification not implemented)

Time = 13.23 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.58 \[ \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {5\,a^2\,x}{2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2\,\left (5\,d\,x-6\right )}{2}-\frac {5\,a^2\,d\,x}{2}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (5\,d\,x-10\right )}{2}-\frac {5\,a^2\,d\,x}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^2\,\left (10\,d\,x-10\right )}{2}-5\,a^2\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (10\,d\,x-22\right )}{2}-5\,a^2\,d\,x\right )-\frac {a^2\,\left (5\,d\,x-16\right )}{2}+\frac {5\,a^2\,d\,x}{2}}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

[In]

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

[Out]

- (5*a^2*x)/2 - (tan(c/2 + (d*x)/2)*((a^2*(5*d*x - 6))/2 - (5*a^2*d*x)/2) - tan(c/2 + (d*x)/2)^4*((a^2*(5*d*x
- 10))/2 - (5*a^2*d*x)/2) + tan(c/2 + (d*x)/2)^3*((a^2*(10*d*x - 10))/2 - 5*a^2*d*x) - tan(c/2 + (d*x)/2)^2*((
a^2*(10*d*x - 22))/2 - 5*a^2*d*x) - (a^2*(5*d*x - 16))/2 + (5*a^2*d*x)/2)/(d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2
+ (d*x)/2)^2 + 1)^2)