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

Optimal. Leaf size=99 \[ \frac{6 \tan (c+d x)}{35 a^3 d}-\frac{3 \sec (c+d x)}{35 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{3 \sec (c+d x)}{35 a d (a \sin (c+d x)+a)^2}+\frac{\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3} \]

[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)

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Rubi [A]  time = 0.141934, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2859, 2672, 3767, 8} \[ \frac{6 \tan (c+d x)}{35 a^3 d}-\frac{3 \sec (c+d x)}{35 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{3 \sec (c+d x)}{35 a d (a \sin (c+d x)+a)^2}+\frac{\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[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 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 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 (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 \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 \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{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]  time = 0.329024, size = 104, normalized size = 1.05 \[ \frac{\sec (c+d x) (672 \sin (c+d x)+182 \sin (2 (c+d x))-288 \sin (3 (c+d x))-13 \sin (4 (c+d x))+182 \cos (c+d x)-672 \cos (2 (c+d x))-78 \cos (3 (c+d x))+48 \cos (4 (c+d x))+560)}{2240 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.096, size = 130, normalized size = 1.3 \begin{align*} 4\,{\frac{1}{d{a}^{3}} \left ( -1/32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-1}+2/7\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-7}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-6}+{\frac{17}{10\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}-7/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}+{\frac{9}{8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}-{\frac{7}{16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+1/32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/d/a^3*(-1/32/(tan(1/2*d*x+1/2*c)-1)+2/7/(tan(1/2*d*x+1/2*c)+1)^7-1/(tan(1/2*d*x+1/2*c)+1)^6+17/10/(tan(1/2*d
*x+1/2*c)+1)^5-7/4/(tan(1/2*d*x+1/2*c)+1)^4+9/8/(tan(1/2*d*x+1/2*c)+1)^3-7/16/(tan(1/2*d*x+1/2*c)+1)^2+1/32/(t
an(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.03065, size = 392, normalized size = 3.96 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.08124, size = 270, normalized size = 2.73 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.31122, size = 162, normalized size = 1.64 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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