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]

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)

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Rubi [A]  time = 0.11545, antiderivative size = 93, normalized size of antiderivative = 1., 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 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]

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 = 0.301531, 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]

-(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)]))/(420*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.105, size = 160, normalized size = 1.7 \begin{align*} 4\,{\frac{1}{d{a}^{2}} \left ( -1/48\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-3}-1/32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-2}-1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-1}+1/7\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-7}-1/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-6}+{\frac{9}{10\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}+{\frac{35}{48\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}-{\frac{11}{32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+1/16\, \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)^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))

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Maxima [B]  time = 1.16389, size = 508, normalized size = 5.46 \begin{align*} \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} \end{align*}

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)

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Fricas [A]  time = 1.61269, size = 267, normalized size = 2.87 \begin{align*} \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 )}} \end{align*}

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.19218, size = 197, normalized size = 2.12 \begin{align*} -\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} \end{align*}

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