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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3957, 2917, 2672, 308, 212, 2715, 8} \[ \int (a+a \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a x}{8} \]

[In]

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

[Out]

(3*a*x)/8 + (a*ArcTanh[Sin[c + d*x]])/d - (a*Sin[c + d*x])/d - (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*Sin[
c + d*x]^3)/(3*d) - (a*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

Rule 8

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[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 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 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[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 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)) \sin ^3(c+d x) \tan (c+d x) \, dx \\ & = a \int \sin ^4(c+d x) \, dx+a \int \sin ^3(c+d x) \tan (c+d x) \, dx \\ & = -\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{4} (3 a) \int \sin ^2(c+d x) \, dx+\frac {a \text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} (3 a) \int 1 \, dx+\frac {a \text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {3 a x}{8}-\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}+\frac {a \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(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)

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	\(\frac {\frac {3 a x}{8}-\frac {11 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {137 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}-\frac {71 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {9 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(184\)

[In]

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

[Out]

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*(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+
c)+3/8*d*x+3/8*c))

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

-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

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

1/24*(9*(d*x + c)*a + 24*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(1
5*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

Mupad [B] (verification not implemented)

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

[In]

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

[Out]

(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)