3.73 \(\int \frac{\sec ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=93 \[ \frac{4 \tan ^3(c+d x)}{21 a^2 d}+\frac{4 \tan (c+d x)}{7 a^2 d}-\frac{\sec ^3(c+d x)}{7 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2} \]

[Out]

-Sec[c + d*x]^3/(7*d*(a + a*Sin[c + d*x])^2) - Sec[c + d*x]^3/(7*d*(a^2 + a^2*Sin[c + d*x])) + (4*Tan[c + d*x]
)/(7*a^2*d) + (4*Tan[c + d*x]^3)/(21*a^2*d)

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Rubi [A]  time = 0.0975248, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2672, 3767} \[ \frac{4 \tan ^3(c+d x)}{21 a^2 d}+\frac{4 \tan (c+d x)}{7 a^2 d}-\frac{\sec ^3(c+d x)}{7 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^2,x]

[Out]

-Sec[c + d*x]^3/(7*d*(a + a*Sin[c + d*x])^2) - Sec[c + d*x]^3/(7*d*(a^2 + a^2*Sin[c + d*x])) + (4*Tan[c + d*x]
)/(7*a^2*d) + (4*Tan[c + d*x]^3)/(21*a^2*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]

Rubi steps

\begin{align*} \int \frac{\sec ^4(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{5 \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{\sec ^3(c+d x)}{7 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{4 \int \sec ^4(c+d x) \, dx}{7 a^2}\\ &=-\frac{\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac{\sec ^3(c+d x)}{7 d \left (a^2+a^2 \sin (c+d x)\right )}-\frac{4 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{7 a^2 d}\\ &=-\frac{\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac{\sec ^3(c+d x)}{7 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{4 \tan (c+d x)}{7 a^2 d}+\frac{4 \tan ^3(c+d x)}{21 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.0634137, size = 78, normalized size = 0.84 \[ -\frac{\left (8 \sin ^5(c+d x)+16 \sin ^4(c+d x)-4 \sin ^3(c+d x)-24 \sin ^2(c+d x)-9 \sin (c+d x)+6\right ) \sec ^3(c+d x)}{21 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^2,x]

[Out]

-(Sec[c + d*x]^3*(6 - 9*Sin[c + d*x] - 24*Sin[c + d*x]^2 - 4*Sin[c + d*x]^3 + 16*Sin[c + d*x]^4 + 8*Sin[c + d*
x]^5))/(21*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.095, size = 158, normalized size = 1.7 \begin{align*} 2\,{\frac{1}{d{a}^{2}} \left ( -1/24\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-3}-1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-2}-3/16\, \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}-2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}+5/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}-{\frac{55}{24\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+{\frac{23}{16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-{\frac{13}{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)^4/(a+a*sin(d*x+c))^2,x)

[Out]

2/d/a^2*(-1/24/(tan(1/2*d*x+1/2*c)-1)^3-1/16/(tan(1/2*d*x+1/2*c)-1)^2-3/16/(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)^6-2/(tan(1/2*d*x+1/2*c)+1)^5+5/2/(tan(1/2*d*x+1/2*c)+1)^4-55/24/(tan
(1/2*d*x+1/2*c)+1)^3+23/16/(tan(1/2*d*x+1/2*c)+1)^2-13/16/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.00202, size = 535, normalized size = 5.75 \begin{align*} -\frac{2 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{76 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{28 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{42 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{56 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{28 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{42 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{21 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 6\right )}}{21 \,{\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/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-2/21*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 76*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 28*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 42*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 56*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 - 28*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 42*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 -
21*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 6)/((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*sin(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 = 2.00504, size = 261, normalized size = 2.81 \begin{align*} \frac{16 \, \cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{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 ) - 2}{21 \,{\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/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/21*(16*cos(d*x + c)^4 - 8*cos(d*x + c)^2 + (8*cos(d*x + c)^4 - 12*cos(d*x + c)^2 - 5)*sin(d*x + c) - 2)/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{4}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**4/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.15149, size = 196, normalized size = 2.11 \begin{align*} -\frac{\frac{7 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{273 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 1155 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2450 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2870 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2037 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 791 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 152}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{7}}}{168 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/168*(7*(9*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 8)/(a^2*(tan(1/2*d*x + 1/2*c) - 1)^3) + (273*t
an(1/2*d*x + 1/2*c)^6 + 1155*tan(1/2*d*x + 1/2*c)^5 + 2450*tan(1/2*d*x + 1/2*c)^4 + 2870*tan(1/2*d*x + 1/2*c)^
3 + 2037*tan(1/2*d*x + 1/2*c)^2 + 791*tan(1/2*d*x + 1/2*c) + 152)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^7))/d