\(\int \frac {\sin ^4(c+d x)}{a+a \sec (c+d x)} \, dx\) [67]

Optimal result

Integrand size = 21, antiderivative size = 73 \[ \int \frac {\sin ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {x}{8 a}-\frac {\cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac {\sin ^3(c+d x)}{3 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3957, 2918, 2644, 30, 2648, 2715, 8} \[ \int \frac {\sin ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\sin (c+d x) \cos (c+d x)}{8 a d}-\frac {x}{8 a} \]

[In]

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

[Out]

-1/8*x/a - (Cos[c + d*x]*Sin[c + d*x])/(8*a*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d) + Sin[c + d*x]^3/(3*a*d
)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
+ f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*n]

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 2918

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.14 \[ \int \frac {\sin ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (24 \sin (c+d x)-8 \sin (3 (c+d x))+3 \left (4 c-4 d x+\sin (4 (c+d x))-4 \tan \left (\frac {c}{2}\right )\right )\right )}{48 a d (1+\sec (c+d x))} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(24*Sin[c + d*x] - 8*Sin[3*(c + d*x)] + 3*(4*c - 4*d*x + Sin[4*(c + d*x)] - 4
*Tan[c/2])))/(48*a*d*(1 + Sec[c + d*x]))

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.60

method	result	size
parallelrisch	\(\frac {-12 d x +24 \sin \left (d x +c \right )-8 \sin \left (3 d x +3 c \right )+3 \sin \left (4 d x +4 c \right )}{96 d a}\)	\(44\)
risch	\(-\frac {x}{8 a}+\frac {\sin \left (d x +c \right )}{4 a d}+\frac {\sin \left (4 d x +4 c \right )}{32 d a}-\frac {\sin \left (3 d x +3 c \right )}{12 d a}\)	\(56\)
derivativedivides	\(\frac {-\frac {16 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{64}-\frac {53 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{192}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{192}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\)	\(90\)
default	\(\frac {-\frac {16 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{64}-\frac {53 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{192}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{192}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\)	\(90\)
norman	\(\frac {-\frac {x}{8 a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {53 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a d}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(166\)

[In]

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

[Out]

1/96*(-12*d*x+24*sin(d*x+c)-8*sin(3*d*x+3*c)+3*sin(4*d*x+4*c))/d/a

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {3 \, d x - {\left (6 \, \cos \left (d x + c\right )^{3} - 8 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{24 \, a d} \]

[In]

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

[Out]

-1/24*(3*d*x - (6*cos(d*x + c)^3 - 8*cos(d*x + c)^2 - 3*cos(d*x + c) + 8)*sin(d*x + c))/(a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (65) = 130\).

Time = 0.29 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.68 \[ \int \frac {\sin ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {11 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {53 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a + \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{12 \, d} \]

[In]

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

[Out]

1/12*((3*sin(d*x + c)/(cos(d*x + c) + 1) + 11*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 53*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a*sin(d
*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.19 \[ \int \frac {\sin ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 53 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 11 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \]

[In]

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

[Out]

-1/24*(3*(d*x + c)/a + 2*(3*tan(1/2*d*x + 1/2*c)^7 - 53*tan(1/2*d*x + 1/2*c)^5 - 11*tan(1/2*d*x + 1/2*c)^3 - 3
*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d

Mupad [B] (verification not implemented)

Time = 14.00 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \frac {\sin ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )}{4\,a\,d}-\frac {x}{8\,a}-\frac {\sin \left (3\,c+3\,d\,x\right )}{12\,a\,d}+\frac {\sin \left (4\,c+4\,d\,x\right )}{32\,a\,d} \]

[In]

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

[Out]

sin(c + d*x)/(4*a*d) - x/(8*a) - sin(3*c + 3*d*x)/(12*a*d) + sin(4*c + 4*d*x)/(32*a*d)