\(\int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx\) [50]

Optimal result

Integrand size = 21, antiderivative size = 138 \[ \int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx=-\frac {33 a^3 x}{8}+\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {2 a^3 \sin (c+d x)}{d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

-33/8*a^3*x+3/2*a^3*arctanh(sin(d*x+c))/d-2*a^3*sin(d*x+c)/d+7/8*a^3*cos(d*x+c)*sin(d*x+c)/d+1/4*a^3*cos(d*x+c
)^3*sin(d*x+c)/d-a^3*sin(d*x+c)^3/d+3*a^3*tan(d*x+c)/d+1/2*a^3*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3957, 2951, 2717, 2715, 8, 2713, 3855, 3852, 3853} \[ \int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {a^3 \sin ^3(c+d x)}{d}-\frac {2 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac {33 a^3 x}{8} \]

[In]

Int[(a + a*Sec[c + d*x])^3*Sin[c + d*x]^4,x]

[Out]

(-33*a^3*x)/8 + (3*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (2*a^3*Sin[c + d*x])/d + (7*a^3*Cos[c + d*x]*Sin[c + d*x
])/(8*d) + (a^3*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (a^3*Sin[c + d*x]^3)/d + (3*a^3*Tan[c + d*x])/d + (a^3*Se
c[c + d*x]*Tan[c + d*x])/(2*d)

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2951

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

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

Rule 3855

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\int (-a-a \cos (c+d x))^3 \sin (c+d x) \tan ^3(c+d x) \, dx \\ & = -\frac {\int \left (5 a^7+5 a^7 \cos (c+d x)-a^7 \cos ^2(c+d x)-3 a^7 \cos ^3(c+d x)-a^7 \cos ^4(c+d x)-a^7 \sec (c+d x)-3 a^7 \sec ^2(c+d x)-a^7 \sec ^3(c+d x)\right ) \, dx}{a^4} \\ & = -5 a^3 x+a^3 \int \cos ^2(c+d x) \, dx+a^3 \int \cos ^4(c+d x) \, dx+a^3 \int \sec (c+d x) \, dx+a^3 \int \sec ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^3(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \, dx-\left (5 a^3\right ) \int \cos (c+d x) \, dx \\ & = -5 a^3 x+\frac {a^3 \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^3 \sin (c+d x)}{d}+\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} a^3 \int 1 \, dx+\frac {1}{2} a^3 \int \sec (c+d x) \, dx+\frac {1}{4} \left (3 a^3\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (3 a^3\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}-\frac {\left (3 a^3\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d} \\ & = -\frac {9 a^3 x}{2}+\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {2 a^3 \sin (c+d x)}{d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{8} \left (3 a^3\right ) \int 1 \, dx \\ & = -\frac {33 a^3 x}{8}+\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {2 a^3 \sin (c+d x)}{d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\frac {a^3 \sec ^2(c+d x) \left (-264 c-264 d x+192 \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)-264 (c+d x) \cos (2 (c+d x))-16 \sin (c+d x)+225 \sin (2 (c+d x))-72 \sin (3 (c+d x))+18 \sin (4 (c+d x))+8 \sin (5 (c+d x))+\sin (6 (c+d x))\right )}{128 d} \]

[In]

Integrate[(a + a*Sec[c + d*x])^3*Sin[c + d*x]^4,x]

[Out]

(a^3*Sec[c + d*x]^2*(-264*c - 264*d*x + 192*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 - 264*(c + d*x)*Cos[2*(c + d*
x)] - 16*Sin[c + d*x] + 225*Sin[2*(c + d*x)] - 72*Sin[3*(c + d*x)] + 18*Sin[4*(c + d*x)] + 8*Sin[5*(c + d*x)]
+ Sin[6*(c + d*x)]))/(128*d)

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(\frac {\left (\left (-12 \cos \left (2 d x +2 c \right )-12\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (12 \cos \left (2 d x +2 c \right )+12\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-33 d x \cos \left (2 d x +2 c \right )-33 d x +\sin \left (5 d x +5 c \right )+\frac {\sin \left (6 d x +6 c \right )}{8}-2 \sin \left (d x +c \right )+\frac {225 \sin \left (2 d x +2 c \right )}{8}-9 \sin \left (3 d x +3 c \right )+\frac {9 \sin \left (4 d x +4 c \right )}{4}\right ) a^{3}}{8 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(153\)
derivativedivides	\(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(192\)
default	\(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(192\)
parts	\(\frac {a^{3} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\)	\(200\)
risch	\(-\frac {33 a^{3} x}{8}-\frac {i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{8 d}-\frac {i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{4 d}+\frac {11 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {11 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}+\frac {i a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{8 d}-\frac {i a^{3} \left ({\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-6\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a^{3} \sin \left (4 d x +4 c \right )}{32 d}\)	\(230\)
norman	\(\frac {-\frac {33 a^{3} x}{8}-\frac {33 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4}+\frac {33 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {33 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {33 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}-\frac {33 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4}-\frac {33 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {21 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {27 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {79 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {25 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}-\frac {83 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {45 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(294\)

[In]

int((a+a*sec(d*x+c))^3*sin(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

1/8*((-12*cos(2*d*x+2*c)-12)*ln(tan(1/2*d*x+1/2*c)-1)+(12*cos(2*d*x+2*c)+12)*ln(tan(1/2*d*x+1/2*c)+1)-33*d*x*c
os(2*d*x+2*c)-33*d*x+sin(5*d*x+5*c)+1/8*sin(6*d*x+6*c)-2*sin(d*x+c)+225/8*sin(2*d*x+2*c)-9*sin(3*d*x+3*c)+9/4*
sin(4*d*x+4*c))*a^3/d/(1+cos(2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.10 \[ \int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx=-\frac {33 \, a^{3} d x \cos \left (d x + c\right )^{2} - 6 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} \cos \left (d x + c\right )^{5} + 8 \, a^{3} \cos \left (d x + c\right )^{4} + 7 \, a^{3} \cos \left (d x + c\right )^{3} - 24 \, a^{3} \cos \left (d x + c\right )^{2} + 24 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="fricas")

[Out]

-1/8*(33*a^3*d*x*cos(d*x + c)^2 - 6*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) + 6*a^3*cos(d*x + c)^2*log(-sin(d
*x + c) + 1) - (2*a^3*cos(d*x + c)^5 + 8*a^3*cos(d*x + c)^4 + 7*a^3*cos(d*x + c)^3 - 24*a^3*cos(d*x + c)^2 + 2
4*a^3*cos(d*x + c) + 4*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.32 \[ \int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx=-\frac {16 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 48 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{3} + 8 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )}}{32 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="maxima")

[Out]

-1/32*(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^3 + 48*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) -
2*tan(d*x + c))*a^3 + 8*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c
) - 1) - 4*sin(d*x + c)))/d

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.30 \[ \int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx=-\frac {33 \, {\left (d x + c\right )} a^{3} - 12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {8 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {2 \, {\left (25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 81 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 79 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^4,x, algorithm="giac")

[Out]

-1/8*(33*(d*x + c)*a^3 - 12*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1))
+ 8*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(25*a^3*ta
n(1/2*d*x + 1/2*c)^7 + 81*a^3*tan(1/2*d*x + 1/2*c)^5 + 79*a^3*tan(1/2*d*x + 1/2*c)^3 + 7*a^3*tan(1/2*d*x + 1/2
*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

Mupad [B] (verification not implemented)

Time = 14.38 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.48 \[ \int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx=\frac {3\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {33\,a^3\,x}{8}+\frac {-\frac {45\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{4}-\frac {83\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {25\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+\frac {79\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {27\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {21\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int(sin(c + d*x)^4*(a + a/cos(c + d*x))^3,x)

[Out]

(3*a^3*atanh(tan(c/2 + (d*x)/2)))/d - (33*a^3*x)/8 + ((27*a^3*tan(c/2 + (d*x)/2)^3)/4 + (79*a^3*tan(c/2 + (d*x
)/2)^5)/2 + (25*a^3*tan(c/2 + (d*x)/2)^7)/2 - (83*a^3*tan(c/2 + (d*x)/2)^9)/4 - (45*a^3*tan(c/2 + (d*x)/2)^11)
/4 + (21*a^3*tan(c/2 + (d*x)/2))/4)/(d*(2*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^6
- tan(c/2 + (d*x)/2)^8 + 2*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))