3.835 \(\int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=93 \[ \frac {8 \tan ^3(c+d x)}{105 a^2 d}+\frac {8 \tan (c+d x)}{35 a^2 d}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2} \]

[Out]

1/7*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^2-2/35*sec(d*x+c)^3/d/(a^2+a^2*sin(d*x+c))+8/35*tan(d*x+c)/a^2/d+8/105*tan
(d*x+c)^3/a^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2859, 2672, 3767} \[ \frac {8 \tan ^3(c+d x)}{105 a^2 d}+\frac {8 \tan (c+d x)}{35 a^2 d}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*Tan[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

Sec[c + d*x]^3/(7*d*(a + a*Sin[c + d*x])^2) - (2*Sec[c + d*x]^3)/(35*d*(a^2 + a^2*Sin[c + d*x])) + (8*Tan[c +
d*x])/(35*a^2*d) + (8*Tan[c + d*x]^3)/(105*a^2*d)

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 3767

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]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}+\frac {2 \int \frac {\sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx}{7 a}\\ &=\frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {8 \int \sec ^4(c+d x) \, dx}{35 a^2}\\ &=\frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2+a^2 \sin (c+d x)\right )}-\frac {8 \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{35 a^2 d}\\ &=\frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {8 \tan (c+d x)}{35 a^2 d}+\frac {8 \tan ^3(c+d x)}{105 a^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.30, size = 134, normalized size = 1.44 \[ -\frac {\sec ^3(c+d x) \left (-56 \sin (c+d x)+3 \sin (2 (c+d x))-12 \sin (3 (c+d x))+\frac {3}{2} \sin (4 (c+d x))+4 \sin (5 (c+d x))+\frac {21}{4} \cos (c+d x)+32 \cos (2 (c+d x))+\frac {9}{8} \cos (3 (c+d x))+16 \cos (4 (c+d x))-\frac {3}{8} \cos (5 (c+d x))-84\right )}{420 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*Tan[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/420*(Sec[c + d*x]^3*(-84 + (21*Cos[c + d*x])/4 + 32*Cos[2*(c + d*x)] + (9*Cos[3*(c + d*x)])/8 + 16*Cos[4*(c
+ d*x)] - (3*Cos[5*(c + d*x)])/8 - 56*Sin[c + d*x] + 3*Sin[2*(c + d*x)] - 12*Sin[3*(c + d*x)] + (3*Sin[4*(c +
d*x)])/2 + 4*Sin[5*(c + d*x)]))/(a^2*d*(1 + Sin[c + d*x])^2)

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 104, normalized size = 1.12 \[ \frac {32 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 25}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(32*cos(d*x + c)^4 - 16*cos(d*x + c)^2 + 2*(8*cos(d*x + c)^4 - 12*cos(d*x + c)^2 - 5)*sin(d*x + c) - 25)
/(a^2*d*cos(d*x + c)^5 - 2*a^2*d*cos(d*x + c)^3*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^3)

________________________________________________________________________________________

giac [A]  time = 0.26, size = 146, normalized size = 1.57 \[ -\frac {\frac {35 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 175 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 910 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 756 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 427 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 31}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(35*(6*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 5)/(a^2*(tan(1/2*d*x + 1/2*c) - 1)^3) - (210*t
an(1/2*d*x + 1/2*c)^6 + 105*tan(1/2*d*x + 1/2*c)^5 - 175*tan(1/2*d*x + 1/2*c)^4 - 910*tan(1/2*d*x + 1/2*c)^3 -
756*tan(1/2*d*x + 1/2*c)^2 - 427*tan(1/2*d*x + 1/2*c) - 31)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

________________________________________________________________________________________

maple [A]  time = 0.46, size = 160, normalized size = 1.72 \[ \frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {35}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {11}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16}}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/d/a^2*(-1/48/(tan(1/2*d*x+1/2*c)-1)^3-1/32/(tan(1/2*d*x+1/2*c)-1)^2-1/16/(tan(1/2*d*x+1/2*c)-1)+1/7/(tan(1/2
*d*x+1/2*c)+1)^7-1/2/(tan(1/2*d*x+1/2*c)+1)^6+9/10/(tan(1/2*d*x+1/2*c)+1)^5-1/(tan(1/2*d*x+1/2*c)+1)^4+35/48/(
tan(1/2*d*x+1/2*c)+1)^3-11/32/(tan(1/2*d*x+1/2*c)+1)^2+1/16/(tan(1/2*d*x+1/2*c)+1))

________________________________________________________________________________________

maxima [B]  time = 0.33, size = 376, normalized size = 4.04 \[ \frac {2 \, {\left (\frac {36 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {132 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {68 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {14 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {84 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {140 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {140 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {105 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 9\right )}}{105 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(36*sin(d*x + c)/(cos(d*x + c) + 1) + 132*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 68*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 14*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 84*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 140*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + 140*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 105*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8 + 9)/((a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*a^2*sin(
d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14*a^2*sin(d*x + c)^6/(cos(d*x
+ c) + 1)^6 + 8*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 4*a^2*si
n(d*x + c)^9/(cos(d*x + c) + 1)^9 - a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10)*d)

________________________________________________________________________________________

mupad [B]  time = 14.35, size = 254, normalized size = 2.73 \[ \frac {2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (9\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+36\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+132\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+68\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+14\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-84\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+140\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+140\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+105\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\right )}{105\,a^2\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*cos(c/2 + (d*x)/2)^2*(9*cos(c/2 + (d*x)/2)^8 + 105*sin(c/2 + (d*x)/2)^8 + 140*cos(c/2 + (d*x)/2)*sin(c/2 +
(d*x)/2)^7 + 36*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2) + 140*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^6 - 84*c
os(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^5 + 14*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^4 + 68*cos(c/2 + (d*x)/2
)^5*sin(c/2 + (d*x)/2)^3 + 132*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^2))/(105*a^2*d*(cos(c/2 + (d*x)/2) - si
n(c/2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^7)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________