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

Optimal. Leaf size=99 \[ \frac{8 \tan (c+d x)}{35 a^3 d}-\frac{4 \sec (c+d x)}{35 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{4 \sec (c+d x)}{35 a d (a \sin (c+d x)+a)^2}-\frac{\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3} \]

[Out]

-Sec[c + d*x]/(7*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)

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Rubi [A]  time = 0.134848, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2672, 3767, 8} \[ \frac{8 \tan (c+d x)}{35 a^3 d}-\frac{4 \sec (c+d x)}{35 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{4 \sec (c+d x)}{35 a d (a \sin (c+d x)+a)^2}-\frac{\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sec[c + d*x]/(7*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 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 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]

Rule 8

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

Rubi steps

\begin{align*} \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 \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 \operatorname{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]  time = 0.0977653, size = 63, normalized size = 0.64 \[ \frac{\sec (c+d x) (14 \sin (c+d x)-6 \sin (3 (c+d x))-14 \cos (2 (c+d x))+\cos (4 (c+d x)))}{35 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.081, size = 130, normalized size = 1.3 \begin{align*} 2\,{\frac{1}{d{a}^{3}} \left ( -1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-1}-4/7\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-7}+2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-6}-{\frac{19}{5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}+9/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}-{\frac{15}{4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+{\frac{17}{8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-{\frac{15}{16\,\tan \left ( 1/2\,dx+c/2 \right ) +16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^3*(-1/16/(tan(1/2*d*x+1/2*c)-1)-4/7/(tan(1/2*d*x+1/2*c)+1)^7+2/(tan(1/2*d*x+1/2*c)+1)^6-19/5/(tan(1/2*d*
x+1/2*c)+1)^5+9/2/(tan(1/2*d*x+1/2*c)+1)^4-15/4/(tan(1/2*d*x+1/2*c)+1)^3+17/8/(tan(1/2*d*x+1/2*c)+1)^2-15/16/(
tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 0.98741, size = 419, normalized size = 4.23 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.90802, size = 270, normalized size = 2.73 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.16622, size = 161, normalized size = 1.63 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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