\(\int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [792]

Optimal result

Integrand size = 25, antiderivative size = 99 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec (c+d x)}{7 d (a+a \sin (c+d x))^3}-\frac {3 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}-\frac {3 \sec (c+d x)}{35 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {6 \tan (c+d x)}{35 a^3 d} \]

[Out]

1/7*sec(d*x+c)/d/(a+a*sin(d*x+c))^3-3/35*sec(d*x+c)/a/d/(a+a*sin(d*x+c))^2-3/35*sec(d*x+c)/d/(a^3+a^3*sin(d*x+
c))+6/35*tan(d*x+c)/a^3/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2938, 2751, 3852, 8} \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {6 \tan (c+d x)}{35 a^3 d}-\frac {3 \sec (c+d x)}{35 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {3 \sec (c+d x)}{35 a d (a \sin (c+d x)+a)^2}+\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3} \]

[In]

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

[Out]

Sec[c + d*x]/(7*d*(a + a*Sin[c + d*x])^3) - (3*Sec[c + d*x])/(35*a*d*(a + a*Sin[c + d*x])^2) - (3*Sec[c + d*x]
)/(35*d*(a^3 + a^3*Sin[c + d*x])) + (6*Tan[c + d*x])/(35*a^3*d)

Rule 8

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

Rule 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2938

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 3852

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*} \text {integral}& = \frac {\sec (c+d x)}{7 d (a+a \sin (c+d x))^3}+\frac {3 \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx}{7 a} \\ & = \frac {\sec (c+d x)}{7 d (a+a \sin (c+d x))^3}-\frac {3 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}+\frac {9 \int \frac {\sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{35 a^2} \\ & = \frac {\sec (c+d x)}{7 d (a+a \sin (c+d x))^3}-\frac {3 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}-\frac {3 \sec (c+d x)}{35 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {6 \int \sec ^2(c+d x) \, dx}{35 a^3} \\ & = \frac {\sec (c+d x)}{7 d (a+a \sin (c+d x))^3}-\frac {3 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}-\frac {3 \sec (c+d x)}{35 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac {6 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{35 a^3 d} \\ & = \frac {\sec (c+d x)}{7 d (a+a \sin (c+d x))^3}-\frac {3 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}-\frac {3 \sec (c+d x)}{35 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {6 \tan (c+d x)}{35 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec (c+d x) (560+182 \cos (c+d x)-672 \cos (2 (c+d x))-78 \cos (3 (c+d x))+48 \cos (4 (c+d x))+672 \sin (c+d x)+182 \sin (2 (c+d x))-288 \sin (3 (c+d x))-13 \sin (4 (c+d x)))}{2240 a^3 d (1+\sin (c+d x))^3} \]

[In]

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

[Out]

(Sec[c + d*x]*(560 + 182*Cos[c + d*x] - 672*Cos[2*(c + d*x)] - 78*Cos[3*(c + d*x)] + 48*Cos[4*(c + d*x)] + 672
*Sin[c + d*x] + 182*Sin[2*(c + d*x)] - 288*Sin[3*(c + d*x)] - 13*Sin[4*(c + d*x)]))/(2240*a^3*d*(1 + Sin[c + d
*x])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.87

method	result	size
risch	\(-\frac {4 i \left (-42 \,{\mathrm e}^{2 i \left (d x +c \right )}-18 i {\mathrm e}^{i \left (d x +c \right )}+3+35 \,{\mathrm e}^{4 i \left (d x +c \right )}+42 i {\mathrm e}^{3 i \left (d x +c \right )}\right )}{35 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d \,a^{3}}\)	\(86\)
parallelrisch	\(\frac {\frac {2}{35}-2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35}}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(113\)
derivativedivides	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {34}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {7}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {9}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+32}}{d \,a^{3}}\)	\(130\)
default	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {34}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {7}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {9}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+32}}{d \,a^{3}}\)	\(130\)
norman	\(\frac {-\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2}{35 a d}-\frac {4 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35 d a}-\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(152\)

[In]

int(sec(d*x+c)^2*sin(d*x+c)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-4/35*I*(-42*exp(2*I*(d*x+c))-18*I*exp(I*(d*x+c))+3+35*exp(4*I*(d*x+c))+42*I*exp(3*I*(d*x+c)))/(exp(I*(d*x+c))
-I)/(exp(I*(d*x+c))+I)^7/d/a^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.07 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{4} - 27 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (6 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) + 20}{35 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/35*(6*cos(d*x + c)^4 - 27*cos(d*x + c)^2 - 3*(6*cos(d*x + c)^2 - 5)*sin(d*x + c) + 20)/(3*a^3*d*cos(d*x + c
)^3 - 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c))*sin(d*x + c))

Sympy [F]

\[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)/(a+a*sin(d*x+c))**3,x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 290 vs. \(2 (91) = 182\).

Time = 0.22 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.93 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {21 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {56 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {105 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {70 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {35 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}{35 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-2/35*(6*sin(d*x + c)/(cos(d*x + c) + 1) - 21*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 56*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 105*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 70*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 35*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 + 1)/((a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 14*a^3*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 14*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*a^
3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a^3*sin(d*x + c)^8/(cos(d*
x + c) + 1)^8)*d)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.21 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {35}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 280 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 665 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 791 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 392 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 51}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{280 \, d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/280*(35/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) - (35*tan(1/2*d*x + 1/2*c)^6 - 280*tan(1/2*d*x + 1/2*c)^5 - 665*ta
n(1/2*d*x + 1/2*c)^4 - 1120*tan(1/2*d*x + 1/2*c)^3 - 791*tan(1/2*d*x + 1/2*c)^2 - 392*tan(1/2*d*x + 1/2*c) - 5
1)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

Mupad [B] (verification not implemented)

Time = 11.31 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.08 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+21\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+56\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+70\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+35\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\right )}{35\,a^3\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

[In]

int(sin(c + d*x)/(cos(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

(2*cos(c/2 + (d*x)/2)^2*(35*sin(c/2 + (d*x)/2)^6 - cos(c/2 + (d*x)/2)^6 + 70*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x
)/2)^5 - 6*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2) + 105*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^4 + 56*cos(c/
2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^3 + 21*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^2))/(35*a^3*d*(cos(c/2 + (d*x
)/2) - sin(c/2 + (d*x)/2))*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^7)