\(\int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [85]

Optimal result

Integrand size = 21, antiderivative size = 87 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7 x}{8 a^2}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}+\frac {2 \sin ^3(c+d x)}{3 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3957, 2948, 2836, 2715, 8, 2713} \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \sin ^3(c+d x)}{3 a^2 d}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a^2 d}+\frac {7 \sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac {7 x}{8 a^2} \]

[In]

Int[Sin[c + d*x]^4/(a + a*Sec[c + d*x])^2,x]

[Out]

(7*x)/(8*a^2) - (2*Sin[c + d*x])/(a^2*d) + (7*Cos[c + d*x]*Sin[c + d*x])/(8*a^2*d) + (Cos[c + d*x]^3*Sin[c + d
*x])/(4*a^2*d) + (2*Sin[c + d*x]^3)/(3*a^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 2836

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 2948

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

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 \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{(-a-a \cos (c+d x))^2} \, dx \\ & = \frac {\int \cos ^2(c+d x) (-a+a \cos (c+d x))^2 \, dx}{a^4} \\ & = \frac {\int \left (a^2 \cos ^2(c+d x)-2 a^2 \cos ^3(c+d x)+a^2 \cos ^4(c+d x)\right ) \, dx}{a^4} \\ & = \frac {\int \cos ^2(c+d x) \, dx}{a^2}+\frac {\int \cos ^4(c+d x) \, dx}{a^2}-\frac {2 \int \cos ^3(c+d x) \, dx}{a^2} \\ & = \frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}+\frac {\int 1 \, dx}{2 a^2}+\frac {3 \int \cos ^2(c+d x) \, dx}{4 a^2}+\frac {2 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d} \\ & = \frac {x}{2 a^2}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}+\frac {2 \sin ^3(c+d x)}{3 a^2 d}+\frac {3 \int 1 \, dx}{8 a^2} \\ & = \frac {7 x}{8 a^2}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}+\frac {2 \sin ^3(c+d x)}{3 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (84 d x-144 \sin (c+d x)+48 \sin (2 (c+d x))-16 \sin (3 (c+d x))+3 \sin (4 (c+d x))+2 \tan \left (\frac {c}{2}\right )\right )}{24 a^2 d (1+\sec (c+d x))^2} \]

[In]

Integrate[Sin[c + d*x]^4/(a + a*Sec[c + d*x])^2,x]

[Out]

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(84*d*x - 144*Sin[c + d*x] + 48*Sin[2*(c + d*x)] - 16*Sin[3*(c + d*x)] + 3*
Sin[4*(c + d*x)] + 2*Tan[c/2]))/(24*a^2*d*(1 + Sec[c + d*x])^2)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.63

method	result	size
parallelrisch	\(\frac {84 d x -144 \sin \left (d x +c \right )-16 \sin \left (3 d x +3 c \right )+3 \sin \left (4 d x +4 c \right )+48 \sin \left (2 d x +2 c \right )}{96 a^{2} d}\)	\(55\)
risch	\(\frac {7 x}{8 a^{2}}-\frac {3 \sin \left (d x +c \right )}{2 a^{2} d}+\frac {\sin \left (4 d x +4 c \right )}{32 a^{2} d}-\frac {\sin \left (3 d x +3 c \right )}{6 a^{2} d}+\frac {\sin \left (2 d x +2 c \right )}{2 a^{2} d}\)	\(73\)
derivativedivides	\(\frac {\frac {8 \left (-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{32}-\frac {83 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{96}-\frac {77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{96}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{2} d}\)	\(89\)
default	\(\frac {\frac {8 \left (-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{32}-\frac {83 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{96}-\frac {77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{96}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{2} d}\)	\(89\)
norman	\(\frac {\frac {7 x}{8 a}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {83 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 a d}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a d}+\frac {7 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}+\frac {21 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 a}+\frac {7 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}+\frac {7 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a}\)	\(169\)

[In]

int(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/96*(84*d*x-144*sin(d*x+c)-16*sin(3*d*x+3*c)+3*sin(4*d*x+4*c)+48*sin(2*d*x+2*c))/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.57 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {21 \, d x + {\left (6 \, \cos \left (d x + c\right )^{3} - 16 \, \cos \left (d x + c\right )^{2} + 21 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right )}{24 \, a^{2} d} \]

[In]

integrate(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(21*d*x + (6*cos(d*x + c)^3 - 16*cos(d*x + c)^2 + 21*cos(d*x + c) - 32)*sin(d*x + c))/(a^2*d)

Sympy [F]

\[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\sin ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(sin(d*x+c)**4/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(sin(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (79) = 158\).

Time = 0.28 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.37 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {83 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {75 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {21 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{12 \, d} \]

[In]

integrate(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*((21*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 83*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5 + 75*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^2 + 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*
a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^8/(cos(
d*x + c) + 1)^8) - 21*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {21 \, {\left (d x + c\right )}}{a^{2}} - \frac {2 \, {\left (75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 83 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 77 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{2}}}{24 \, d} \]

[In]

integrate(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(21*(d*x + c)/a^2 - 2*(75*tan(1/2*d*x + 1/2*c)^7 + 83*tan(1/2*d*x + 1/2*c)^5 + 77*tan(1/2*d*x + 1/2*c)^3
+ 21*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^2))/d

Mupad [B] (verification not implemented)

Time = 17.56 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7\,x}{8\,a^2}-\frac {\frac {25\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {83\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {77\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

[In]

int(sin(c + d*x)^4/(a + a/cos(c + d*x))^2,x)

[Out]

(7*x)/(8*a^2) - ((7*tan(c/2 + (d*x)/2))/4 + (77*tan(c/2 + (d*x)/2)^3)/12 + (83*tan(c/2 + (d*x)/2)^5)/12 + (25*
tan(c/2 + (d*x)/2)^7)/4)/(a^2*d*(tan(c/2 + (d*x)/2)^2 + 1)^4)