\(\int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [84]

Optimal result

Integrand size = 21, antiderivative size = 99 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\sec (c+d x)}{7 d (a+a \sin (c+d x))^3}-\frac {4 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}-\frac {4 \sec (c+d x)}{35 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {8 \tan (c+d x)}{35 a^3 d} \]

[Out]

-1/7*sec(d*x+c)/d/(a+a*sin(d*x+c))^3-4/35*sec(d*x+c)/a/d/(a+a*sin(d*x+c))^2-4/35*sec(d*x+c)/d/(a^3+a^3*sin(d*x
+c))+8/35*tan(d*x+c)/a^3/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2751, 3852, 8} \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {8 \tan (c+d x)}{35 a^3 d}-\frac {4 \sec (c+d x)}{35 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {4 \sec (c+d x)}{35 a d (a \sin (c+d x)+a)^2}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3} \]

[In]

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

[Out]

-1/7*Sec[c + d*x]/(d*(a + a*Sin[c + d*x])^3) - (4*Sec[c + d*x])/(35*a*d*(a + a*Sin[c + d*x])^2) - (4*Sec[c + d
*x])/(35*d*(a^3 + a^3*Sin[c + d*x])) + (8*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 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 {4 \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 {4 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}+\frac {12 \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 {4 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}-\frac {4 \sec (c+d x)}{35 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {8 \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 {4 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}-\frac {4 \sec (c+d x)}{35 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac {8 \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 {4 \sec (c+d x)}{35 a d (a+a \sin (c+d x))^2}-\frac {4 \sec (c+d x)}{35 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {8 \tan (c+d x)}{35 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec (c+d x) (-14 \cos (2 (c+d x))+\cos (4 (c+d x))+14 \sin (c+d x)-6 \sin (3 (c+d x)))}{35 a^3 d (1+\sin (c+d x))^3} \]

[In]

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

[Out]

(Sec[c + d*x]*(-14*Cos[2*(c + d*x)] + Cos[4*(c + d*x)] + 14*Sin[c + d*x] - 6*Sin[3*(c + d*x)]))/(35*a^3*d*(1 +
Sin[c + d*x])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {-\frac {96 \,{\mathrm e}^{i \left (d x +c \right )}}{35}-\frac {16 i}{35}+\frac {32 \,{\mathrm e}^{3 i \left (d x +c \right )}}{5}+\frac {32 i {\mathrm e}^{2 i \left (d x +c \right )}}{5}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{3} d}\)	\(74\)
parallelrisch	\(\frac {\frac {26}{35}-2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {22 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {86 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35}}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(126\)
derivativedivides	\(\frac {-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {38}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {15}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {17}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{a^{3} d}\)	\(130\)
default	\(\frac {-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {38}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {15}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {17}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{a^{3} d}\)	\(130\)
norman	\(\frac {-\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {26}{35 a d}-\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {6 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {22 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {86 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35 d a}}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(171\)

[In]

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

[Out]

16/35*(-6*exp(I*(d*x+c))-I+14*exp(3*I*(d*x+c))+14*I*exp(2*I*(d*x+c)))/(exp(I*(d*x+c))+I)^7/(-I+exp(I*(d*x+c)))
/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {8 \, \cos \left (d x + c\right )^{4} - 36 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (6 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) + 15}{35 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/35*(8*cos(d*x + c)^4 - 36*cos(d*x + c)^2 - 4*(6*cos(d*x + c)^2 - 5)*sin(d*x + c) + 15)/(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]

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

[In]

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

[Out]

Integral(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. 310 vs. \(2 (91) = 182\).

Time = 0.21 (sec) , antiderivative size = 310, normalized size of antiderivative = 3.13 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {43 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {105 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {175 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {105 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {35 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 13\right )}}{35 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]

[In]

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

[Out]

-2/35*(43*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 7*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 105*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 175*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 105*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 - 35*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 13)/((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 = 0.32 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {35}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} + \frac {525 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 1960 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4025 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3143 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1176 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 243}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{280 \, d} \]

[In]

integrate(sec(d*x+c)^2/(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)) + (525*tan(1/2*d*x + 1/2*c)^6 + 1960*tan(1/2*d*x + 1/2*c)^5 + 4025
*tan(1/2*d*x + 1/2*c)^4 + 4480*tan(1/2*d*x + 1/2*c)^3 + 3143*tan(1/2*d*x + 1/2*c)^2 + 1176*tan(1/2*d*x + 1/2*c
) + 243)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

Mupad [B] (verification not implemented)

Time = 6.57 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.30 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (13\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+43\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+77\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+7\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-175\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-105\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-35\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\right )}{35\,a^3\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

[In]

int(1/(cos(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

-(2*cos(c/2 + (d*x)/2)*(13*cos(c/2 + (d*x)/2)^7 - 35*sin(c/2 + (d*x)/2)^7 - 105*cos(c/2 + (d*x)/2)*sin(c/2 + (
d*x)/2)^6 + 43*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) - 175*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^5 - 105*c
os(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^4 + 7*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^3 + 77*cos(c/2 + (d*x)/2)
^5*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)