\(\int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx\) [863]

Optimal result

Integrand size = 29, antiderivative size = 91 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}-\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {5 a^5}{4 d \left (a^3-a^3 \sin (c+d x)\right )} \] Output:

-7/8*a^2*ln(1-sin(d*x+c))/d-1/8*a^2*ln(1+sin(d*x+c))/d+1/4*a^4/d/(a-a*sin( 
d*x+c))^2-5/4*a^5/d/(a^3-a^3*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.68 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{4 d}-\frac {a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \sec ^2(c+d x)}{d}+\frac {a^2 \sec ^4(c+d x)}{4 d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}-\frac {3 a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {2 a^2 \sec (c+d x) \tan ^3(c+d x)}{d}+\frac {a^2 \tan ^4(c+d x)}{4 d} \] Input:

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

Output:

(3*a^2*ArcTanh[Sin[c + d*x]])/(4*d) - (a^2*Log[Cos[c + d*x]])/d - (a^2*Sec 
[c + d*x]^2)/d + (a^2*Sec[c + d*x]^4)/(4*d) + (3*a^2*Sec[c + d*x]*Tan[c + 
d*x])/(4*d) - (3*a^2*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) + (2*a^2*Sec[c + d 
*x]*Tan[c + d*x]^3)/d + (a^2*Tan[c + d*x]^4)/(4*d)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \tan ^3(c+d x) \sec ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^3 (a \sin (c+d x)+a)^2}{\cos (c+d x)^5}dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {a^5 \int \frac {\sin ^3(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 \int \frac {a^3 \sin ^3(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {a^2 \int \left (\frac {a^2}{2 (a-a \sin (c+d x))^3}-\frac {5 a}{4 (a-a \sin (c+d x))^2}+\frac {7}{8 (a-a \sin (c+d x))}-\frac {1}{8 (\sin (c+d x) a+a)}\right )d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^2 \left (\frac {a^2}{4 (a-a \sin (c+d x))^2}-\frac {5 a}{4 (a-a \sin (c+d x))}-\frac {7}{8} \log (a-a \sin (c+d x))-\frac {1}{8} \log (a \sin (c+d x)+a)\right )}{d}\)

Input:

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

Output:

(a^2*((-7*Log[a - a*Sin[c + d*x]])/8 - Log[a + a*Sin[c + d*x]]/8 + a^2/(4* 
(a - a*Sin[c + d*x])^2) - (5*a)/(4*(a - a*Sin[c + d*x]))))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3315

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, 
 x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege 
rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.37

method	result	size
risch	\(i a^{2} x +\frac {2 i a^{2} c}{d}+\frac {i \left (-8 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 a^{2} {\mathrm e}^{i \left (d x +c \right )}+5 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{2 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}\)	\(125\)
derivativedivides	\(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(137\)
default	\(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(137\)

Input:

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

Output:

I*a^2*x+2*I*a^2/d*c+1/2*I*(-8*I*a^2*exp(2*I*(d*x+c))-5*a^2*exp(I*(d*x+c))+ 
5*a^2*exp(3*I*(d*x+c)))/(exp(I*(d*x+c))-I)^4/d-1/4*a^2/d*ln(exp(I*(d*x+c)) 
+I)-7/4*a^2/d*ln(exp(I*(d*x+c))-I)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.37 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {10 \, a^{2} \sin \left (d x + c\right ) - 8 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^2*tan(d*x+c)^3,x, algorithm="frica 
s")
 

Output:

-1/8*(10*a^2*sin(d*x + c) - 8*a^2 + (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + 
c) - 2*a^2)*log(sin(d*x + c) + 1) + 7*(a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x 
+ c) - 2*a^2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) 
 - 2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \] Input:

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^2*tan(d*x+c)^3,x, algorithm="maxim 
a")
 

Output:

-1/8*(a^2*log(sin(d*x + c) + 1) + 7*a^2*log(sin(d*x + c) - 1) - 2*(5*a^2*s 
in(d*x + c) - 4*a^2)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {1}{8} \, a^{2} {\left (\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} + \frac {7 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {2 \, {\left (5 \, \sin \left (d x + c\right ) - 4\right )}}{d {\left (\sin \left (d x + c\right ) - 1\right )}^{2}}\right )} \] Input:

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

Output:

-1/8*a^2*(log(abs(sin(d*x + c) + 1))/d + 7*log(abs(sin(d*x + c) - 1))/d - 
2*(5*sin(d*x + c) - 4)/(d*(sin(d*x + c) - 1)^2))
 

Mupad [B] (verification not implemented)

Time = 32.49 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.82 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {7\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{4\,d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{4\,d}-\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \] Input:

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

Output:

(a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (7*a^2*log(tan(c/2 + (d*x)/2) - 1) 
)/(4*d) - (a^2*log(tan(c/2 + (d*x)/2) + 1))/(4*d) - ((3*a^2*tan(c/2 + (d*x 
)/2)^3)/2 - 4*a^2*tan(c/2 + (d*x)/2)^2 + (3*a^2*tan(c/2 + (d*x)/2))/2)/(d* 
(6*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 4*tan(c/2 + (d*x)/2)^3 + 
tan(c/2 + (d*x)/2)^4 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.35 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^{2} \left (8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-14 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+28 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-14 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+5 \sin \left (d x +c \right )^{2}-3\right )}{8 d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:

int(sec(d*x+c)^2*(a+a*sin(d*x+c))^2*tan(d*x+c)^3,x)
 

Output:

(a**2*(8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 - 16*log(tan((c + d* 
x)/2)**2 + 1)*sin(c + d*x) + 8*log(tan((c + d*x)/2)**2 + 1) - 14*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**2 + 28*log(tan((c + d*x)/2) - 1)*sin(c + d* 
x) - 14*log(tan((c + d*x)/2) - 1) - 2*log(tan((c + d*x)/2) + 1)*sin(c + d* 
x)**2 + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 2*log(tan((c + d*x)/2) 
+ 1) + 5*sin(c + d*x)**2 - 3))/(8*d*(sin(c + d*x)**2 - 2*sin(c + d*x) + 1) 
)