3.1.12 \(\int (a+a \sec (c+d x)) \sin ^4(c+d x) \, dx\) [12]

3.1.12.1 Optimal result
3.1.12.2 Mathematica [A] (verified)
3.1.12.3 Rubi [A] (verified)
3.1.12.4 Maple [A] (verified)
3.1.12.5 Fricas [A] (verification not implemented)
3.1.12.6 Sympy [F]
3.1.12.7 Maxima [A] (verification not implemented)
3.1.12.8 Giac [A] (verification not implemented)
3.1.12.9 Mupad [B] (verification not implemented)

3.1.12.1 Optimal result

Integrand size = 19, antiderivative size = 89 \[ \int (a+a \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {3 a x}{8}+\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 d} \]

output
3/8*a*x+a*arctanh(sin(d*x+c))/d-a*sin(d*x+c)/d-3/8*a*cos(d*x+c)*sin(d*x+c) 
/d-1/3*a*sin(d*x+c)^3/d-1/4*a*cos(d*x+c)*sin(d*x+c)^3/d
 
3.1.12.2 Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int (a+a \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {3 a (c+d x)}{8 d}+\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin (2 (c+d x))}{4 d}+\frac {a \sin (4 (c+d x))}{32 d} \]

input
Integrate[(a + a*Sec[c + d*x])*Sin[c + d*x]^4,x]
 
output
(3*a*(c + d*x))/(8*d) + (a*ArcTanh[Sin[c + d*x]])/d - (a*Sin[c + d*x])/d - 
 (a*Sin[c + d*x]^3)/(3*d) - (a*Sin[2*(c + d*x)])/(4*d) + (a*Sin[4*(c + d*x 
)])/(32*d)
 
3.1.12.3 Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.737, Rules used = {3042, 4360, 25, 25, 3042, 3317, 3042, 3072, 254, 2009, 3115, 3042, 3115, 24}

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.

\(\displaystyle \int \sin ^4(c+d x) (a \sec (c+d x)+a) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int -\left (\sin ^3(c+d x) \tan (c+d x) (a (-\cos (c+d x))-a)\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -\left ((\cos (c+d x) a+a) \sin ^3(c+d x) \tan (c+d x)\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \sin ^3(c+d x) \tan (c+d x) (a \cos (c+d x)+a)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3317

\(\displaystyle a \int \sin ^4(c+d x)dx+a \int \sin ^3(c+d x) \tan (c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \sin (c+d x)^4dx+a \int \sin (c+d x)^3 \tan (c+d x)dx\)

\(\Big \downarrow \) 3072

\(\displaystyle \frac {a \int \frac {\sin ^4(c+d x)}{1-\sin ^2(c+d x)}d\sin (c+d x)}{d}+a \int \sin (c+d x)^4dx\)

\(\Big \downarrow \) 254

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

\(\Big \downarrow \) 2009

\(\displaystyle a \int \sin (c+d x)^4dx+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {3}{4} \int \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {3}{4} \int \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+a \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )\)

input
Int[(a + a*Sec[c + d*x])*Sin[c + d*x]^4,x]
 
output
(a*(ArcTanh[Sin[c + d*x]] - Sin[c + d*x] - Sin[c + d*x]^3/3))/d + a*(-1/4* 
(Cos[c + d*x]*Sin[c + d*x]^3)/d + (3*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/(2 
*d)))/4)
 

3.1.12.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 254
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, 
 a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
 

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 3072
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   Subst[Int[ 
(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x 
]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
 

rule 3115
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] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3317
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 

rule 4360
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 
3.1.12.4 Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85

method result size
derivativedivides \(\frac {a \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) \(76\)
default \(\frac {a \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) \(76\)
parts \(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {a \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) \(78\)
parallelrisch \(-\frac {a \left (-36 d x +120 \sin \left (d x +c \right )-96 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+96 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+24 \sin \left (2 d x +2 c \right )-3 \sin \left (4 d x +4 c \right )-8 \sin \left (3 d x +3 c \right )\right )}{96 d}\) \(81\)
risch \(\frac {3 a x}{8}+\frac {5 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {5 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a \sin \left (4 d x +4 c \right )}{32 d}+\frac {a \sin \left (3 d x +3 c \right )}{12 d}-\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) \(120\)
norman \(\frac {\frac {3 a x}{8}-\frac {11 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {137 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}-\frac {71 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {9 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) \(184\)

input
int((a+a*sec(d*x+c))*sin(d*x+c)^4,x,method=_RETURNVERBOSE)
 
output
1/d*(a*(-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+a*(-1/4*(s 
in(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+c)+3/8*d*x+3/8*c))
 
3.1.12.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int (a+a \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {9 \, a d x + 12 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, a \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, a \cos \left (d x + c\right )^{3} + 8 \, a \cos \left (d x + c\right )^{2} - 15 \, a \cos \left (d x + c\right ) - 32 \, a\right )} \sin \left (d x + c\right )}{24 \, d} \]

input
integrate((a+a*sec(d*x+c))*sin(d*x+c)^4,x, algorithm="fricas")
 
output
1/24*(9*a*d*x + 12*a*log(sin(d*x + c) + 1) - 12*a*log(-sin(d*x + c) + 1) + 
 (6*a*cos(d*x + c)^3 + 8*a*cos(d*x + c)^2 - 15*a*cos(d*x + c) - 32*a)*sin( 
d*x + c))/d
 
3.1.12.6 Sympy [F]

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

input
integrate((a+a*sec(d*x+c))*sin(d*x+c)**4,x)
 
output
a*(Integral(sin(c + d*x)**4*sec(c + d*x), x) + Integral(sin(c + d*x)**4, x 
))
 
3.1.12.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91 \[ \int (a+a \sec (c+d x)) \sin ^4(c+d x) \, dx=-\frac {16 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{96 \, d} \]

input
integrate((a+a*sec(d*x+c))*sin(d*x+c)^4,x, algorithm="maxima")
 
output
-1/96*(16*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) 
 - 1) + 6*sin(d*x + c))*a - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2* 
d*x + 2*c))*a)/d
 
3.1.12.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.33 \[ \int (a+a \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {9 \, {\left (d x + c\right )} a + 24 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 71 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 137 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

input
integrate((a+a*sec(d*x+c))*sin(d*x+c)^4,x, algorithm="giac")
 
output
1/24*(9*(d*x + c)*a + 24*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*a*log(a 
bs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*a*tan(1/2*d*x + 1/2*c)^7 + 71*a*tan( 
1/2*d*x + 1/2*c)^5 + 137*a*tan(1/2*d*x + 1/2*c)^3 + 33*a*tan(1/2*d*x + 1/2 
*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
 
3.1.12.9 Mupad [B] (verification not implemented)

Time = 13.45 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int (a+a \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {3\,a\,x}{8}+\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {a\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}-\frac {5\,a\,\sin \left (c+d\,x\right )}{4\,d} \]

input
int(sin(c + d*x)^4*(a + a/cos(c + d*x)),x)
 
output
(3*a*x)/8 + (2*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a*sin( 
2*c + 2*d*x))/(4*d) + (a*sin(3*c + 3*d*x))/(12*d) + (a*sin(4*c + 4*d*x))/( 
32*d) - (5*a*sin(c + d*x))/(4*d)