3.102 \(\int \frac{\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=108 \[ \frac{\sin ^3(c+d x)}{a^3 d}-\frac{7 \sin (c+d x)}{a^3 d}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a^3 d}+\frac{19 \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{4 \sin (c+d x)}{a^3 d (\cos (c+d x)+1)}+\frac{51 x}{8 a^3} \]

[Out]

(51*x)/(8*a^3) - (7*Sin[c + d*x])/(a^3*d) + (19*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d) + (Cos[c + d*x]^3*Sin[c +
d*x])/(4*a^3*d) - (4*Sin[c + d*x])/(a^3*d*(1 + Cos[c + d*x])) + Sin[c + d*x]^3/(a^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.318477, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2875, 2872, 2648, 2637, 2635, 8, 2633} \[ \frac{\sin ^3(c+d x)}{a^3 d}-\frac{7 \sin (c+d x)}{a^3 d}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a^3 d}+\frac{19 \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{4 \sin (c+d x)}{a^3 d (\cos (c+d x)+1)}+\frac{51 x}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(51*x)/(8*a^3) - (7*Sin[c + d*x])/(a^3*d) + (19*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d) + (Cos[c + d*x]^3*Sin[c +
d*x])/(4*a^3*d) - (4*Sin[c + d*x])/(a^3*d*(1 + Cos[c + d*x])) + Sin[c + d*x]^3/(a^3*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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]

Rule 2875

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

Rule 2872

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 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2637

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

Rule 2635

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] && Integer
Q[2*n]

Rule 8

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

Rule 2633

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]

Rubi steps

\begin{align*} \int \frac{\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\int \frac{\cos ^3(c+d x) \sin ^4(c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=-\frac{\int \cos (c+d x) (-a+a \cos (c+d x))^3 \cot ^2(c+d x) \, dx}{a^6}\\ &=\frac{\int \left (4 a+\frac{4 a}{-1-\cos (c+d x)}-4 a \cos (c+d x)+4 a \cos ^2(c+d x)-3 a \cos ^3(c+d x)+a \cos ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac{4 x}{a^3}+\frac{\int \cos ^4(c+d x) \, dx}{a^3}-\frac{3 \int \cos ^3(c+d x) \, dx}{a^3}+\frac{4 \int \frac{1}{-1-\cos (c+d x)} \, dx}{a^3}-\frac{4 \int \cos (c+d x) \, dx}{a^3}+\frac{4 \int \cos ^2(c+d x) \, dx}{a^3}\\ &=\frac{4 x}{a^3}-\frac{4 \sin (c+d x)}{a^3 d}+\frac{2 \cos (c+d x) \sin (c+d x)}{a^3 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}-\frac{4 \sin (c+d x)}{a^3 d (1+\cos (c+d x))}+\frac{3 \int \cos ^2(c+d x) \, dx}{4 a^3}+\frac{2 \int 1 \, dx}{a^3}+\frac{3 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^3 d}\\ &=\frac{6 x}{a^3}-\frac{7 \sin (c+d x)}{a^3 d}+\frac{19 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}-\frac{4 \sin (c+d x)}{a^3 d (1+\cos (c+d x))}+\frac{\sin ^3(c+d x)}{a^3 d}+\frac{3 \int 1 \, dx}{8 a^3}\\ &=\frac{51 x}{8 a^3}-\frac{7 \sin (c+d x)}{a^3 d}+\frac{19 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}-\frac{4 \sin (c+d x)}{a^3 d (1+\cos (c+d x))}+\frac{\sin ^3(c+d x)}{a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.645708, size = 173, normalized size = 1.6 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (-997 \sin \left (c+\frac{d x}{2}\right )-800 \sin \left (c+\frac{3 d x}{2}\right )-800 \sin \left (2 c+\frac{3 d x}{2}\right )+160 \sin \left (2 c+\frac{5 d x}{2}\right )+160 \sin \left (3 c+\frac{5 d x}{2}\right )-35 \sin \left (3 c+\frac{7 d x}{2}\right )-35 \sin \left (4 c+\frac{7 d x}{2}\right )+5 \sin \left (4 c+\frac{9 d x}{2}\right )+5 \sin \left (5 c+\frac{9 d x}{2}\right )+2040 d x \cos \left (c+\frac{d x}{2}\right )-3563 \sin \left (\frac{d x}{2}\right )+2040 d x \cos \left (\frac{d x}{2}\right )\right )}{640 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]*(2040*d*x*Cos[(d*x)/2] + 2040*d*x*Cos[c + (d*x)/2] - 3563*Sin[(d*x)/2] - 997*Sin[c
+ (d*x)/2] - 800*Sin[c + (3*d*x)/2] - 800*Sin[2*c + (3*d*x)/2] + 160*Sin[2*c + (5*d*x)/2] + 160*Sin[3*c + (5*d
*x)/2] - 35*Sin[3*c + (7*d*x)/2] - 35*Sin[4*c + (7*d*x)/2] + 5*Sin[4*c + (9*d*x)/2] + 5*Sin[5*c + (9*d*x)/2]))
/(640*a^3*d)

________________________________________________________________________________________

Maple [A]  time = 0.107, size = 171, normalized size = 1.6 \begin{align*} -4\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3}}}-{\frac{77}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-{\frac{149}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-{\frac{123}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-{\frac{35}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+{\frac{51}{4\,d{a}^{3}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/d/a^3*tan(1/2*d*x+1/2*c)-77/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7-149/4/d/a^3/(1+tan(1/2*
d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-123/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3-35/4/d/a^3/(1
+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+51/4/d/a^3*arctan(tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [B]  time = 1.51965, size = 306, normalized size = 2.83 \begin{align*} -\frac{\frac{\frac{35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{123 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{149 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{77 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{3} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{51 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac{16 \, \sin \left (d x + c\right )}{a^{3}{\left (\cos \left (d x + c\right ) + 1\right )}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((35*sin(d*x + c)/(cos(d*x + c) + 1) + 123*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 149*sin(d*x + c)^5/(cos(
d*x + c) + 1)^5 + 77*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^3 + 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6
*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^8/(cos
(d*x + c) + 1)^8) - 51*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + 16*sin(d*x + c)/(a^3*(cos(d*x + c) + 1)))
/d

________________________________________________________________________________________

Fricas [A]  time = 1.70069, size = 217, normalized size = 2.01 \begin{align*} \frac{51 \, d x \cos \left (d x + c\right ) + 51 \, d x +{\left (2 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{3} + 11 \, \cos \left (d x + c\right )^{2} - 29 \, \cos \left (d x + c\right ) - 80\right )} \sin \left (d x + c\right )}{8 \,{\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(51*d*x*cos(d*x + c) + 51*d*x + (2*cos(d*x + c)^4 - 6*cos(d*x + c)^3 + 11*cos(d*x + c)^2 - 29*cos(d*x + c)
- 80)*sin(d*x + c))/(a^3*d*cos(d*x + c) + a^3*d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sin ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

________________________________________________________________________________________

Giac [A]  time = 1.35869, size = 136, normalized size = 1.26 \begin{align*} \frac{\frac{51 \,{\left (d x + c\right )}}{a^{3}} - \frac{32 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{2 \,{\left (77 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 149 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 123 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 35 \, \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^{3}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(51*(d*x + c)/a^3 - 32*tan(1/2*d*x + 1/2*c)/a^3 - 2*(77*tan(1/2*d*x + 1/2*c)^7 + 149*tan(1/2*d*x + 1/2*c)^
5 + 123*tan(1/2*d*x + 1/2*c)^3 + 35*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^3))/d