3.86 \(\int \frac{\sec ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=123 \[ \frac{8 \tan ^3(c+d x)}{63 a^3 d}+\frac{8 \tan (c+d x)}{21 a^3 d}-\frac{2 \sec ^3(c+d x)}{21 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{2 \sec ^3(c+d x)}{21 a d (a \sin (c+d x)+a)^2}-\frac{\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3} \]

[Out]

-Sec[c + d*x]^3/(9*d*(a + a*Sin[c + d*x])^3) - (2*Sec[c + d*x]^3)/(21*a*d*(a + a*Sin[c + d*x])^2) - (2*Sec[c +
d*x]^3)/(21*d*(a^3 + a^3*Sin[c + d*x])) + (8*Tan[c + d*x])/(21*a^3*d) + (8*Tan[c + d*x]^3)/(63*a^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.145029, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2672, 3767} \[ \frac{8 \tan ^3(c+d x)}{63 a^3 d}+\frac{8 \tan (c+d x)}{21 a^3 d}-\frac{2 \sec ^3(c+d x)}{21 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{2 \sec ^3(c+d x)}{21 a d (a \sin (c+d x)+a)^2}-\frac{\sec ^3(c+d x)}{9 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

-Sec[c + d*x]^3/(9*d*(a + a*Sin[c + d*x])^3) - (2*Sec[c + d*x]^3)/(21*a*d*(a + a*Sin[c + d*x])^2) - (2*Sec[c +
d*x]^3)/(21*d*(a^3 + a^3*Sin[c + d*x])) + (8*Tan[c + d*x])/(21*a^3*d) + (8*Tan[c + d*x]^3)/(63*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]

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=-\frac{\sec ^3(c+d x)}{9 d (a+a \sin (c+d x))^3}+\frac{2 \int \frac{\sec ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx}{3 a}\\ &=-\frac{\sec ^3(c+d x)}{9 d (a+a \sin (c+d x))^3}-\frac{2 \sec ^3(c+d x)}{21 a d (a+a \sin (c+d x))^2}+\frac{10 \int \frac{\sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx}{21 a^2}\\ &=-\frac{\sec ^3(c+d x)}{9 d (a+a \sin (c+d x))^3}-\frac{2 \sec ^3(c+d x)}{21 a d (a+a \sin (c+d x))^2}-\frac{2 \sec ^3(c+d x)}{21 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac{8 \int \sec ^4(c+d x) \, dx}{21 a^3}\\ &=-\frac{\sec ^3(c+d x)}{9 d (a+a \sin (c+d x))^3}-\frac{2 \sec ^3(c+d x)}{21 a d (a+a \sin (c+d x))^2}-\frac{2 \sec ^3(c+d x)}{21 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac{8 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{21 a^3 d}\\ &=-\frac{\sec ^3(c+d x)}{9 d (a+a \sin (c+d x))^3}-\frac{2 \sec ^3(c+d x)}{21 a d (a+a \sin (c+d x))^2}-\frac{2 \sec ^3(c+d x)}{21 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac{8 \tan (c+d x)}{21 a^3 d}+\frac{8 \tan ^3(c+d x)}{63 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.113553, size = 85, normalized size = 0.69 \[ \frac{\sec ^3(c+d x) (36 \sin (c+d x)+2 \sin (3 (c+d x))-6 \sin (5 (c+d x))-27 \cos (2 (c+d x))-12 \cos (4 (c+d x))+\cos (6 (c+d x)))}{126 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(-27*Cos[2*(c + d*x)] - 12*Cos[4*(c + d*x)] + Cos[6*(c + d*x)] + 36*Sin[c + d*x] + 2*Sin[3*(c
+ d*x)] - 6*Sin[5*(c + d*x)]))/(126*a^3*d*(1 + Sin[c + d*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.109, size = 190, normalized size = 1.5 \begin{align*} 2\,{\frac{1}{d{a}^{3}} \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}-{\frac{7}{64\,\tan \left ( 1/2\,dx+c/2 \right ) -64}}-4/9\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-9}+2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-8}-{\frac{34}{7\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{7}}}+{\frac{23}{3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}-{\frac{35}{4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}+{\frac{59}{8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}-{\frac{19}{4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+9/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}-{\frac{57}{64\,\tan \left ( 1/2\,dx+c/2 \right ) +64}} \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))^3,x)

[Out]

2/d/a^3*(-1/48/(tan(1/2*d*x+1/2*c)-1)^3-1/32/(tan(1/2*d*x+1/2*c)-1)^2-7/64/(tan(1/2*d*x+1/2*c)-1)-4/9/(tan(1/2
*d*x+1/2*c)+1)^9+2/(tan(1/2*d*x+1/2*c)+1)^8-34/7/(tan(1/2*d*x+1/2*c)+1)^7+23/3/(tan(1/2*d*x+1/2*c)+1)^6-35/4/(
tan(1/2*d*x+1/2*c)+1)^5+59/8/(tan(1/2*d*x+1/2*c)+1)^4-19/4/(tan(1/2*d*x+1/2*c)+1)^3+9/4/(tan(1/2*d*x+1/2*c)+1)
^2-57/64/(tan(1/2*d*x+1/2*c)+1))

________________________________________________________________________________________

Maxima [B]  time = 1.02829, size = 651, normalized size = 5.29 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="maxima")

[Out]

-2/63*(51*sin(d*x + c)/(cos(d*x + c) + 1) + 39*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 235*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 450*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 306*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 294*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 378*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 63*sin(d*x + c)^8/(cos(d*x + c) + 1
)^8 - 273*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 189*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 63*sin(d*x + c)^11
/(cos(d*x + c) + 1)^11 + 19)/((a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 2*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 27*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*a^3*si
n(d*x + c)^5/(cos(d*x + c) + 1)^5 + 36*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 27*a^3*sin(d*x + c)^8/(cos(d*
x + c) + 1)^8 - 2*a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 12*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 6*a
^3*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*d)

________________________________________________________________________________________

Fricas [A]  time = 1.70969, size = 331, normalized size = 2.69 \begin{align*} -\frac{16 \, \cos \left (d x + c\right )^{6} - 72 \, \cos \left (d x + c\right )^{4} + 30 \, \cos \left (d x + c\right )^{2} - 2 \,{\left (24 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 7\right )} \sin \left (d x + c\right ) + 7}{63 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} +{\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\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)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/63*(16*cos(d*x + c)^6 - 72*cos(d*x + c)^4 + 30*cos(d*x + c)^2 - 2*(24*cos(d*x + c)^4 - 20*cos(d*x + c)^2 -
7)*sin(d*x + c) + 7)/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*
x + c)^3)*sin(d*x + c))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{4}{\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)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**4/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

________________________________________________________________________________________

Giac [A]  time = 1.18373, size = 231, normalized size = 1.88 \begin{align*} -\frac{\frac{21 \,{\left (21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 36 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 19\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{3591 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 19656 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 56196 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 95760 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 107730 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 79464 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 38484 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 10944 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1615}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{9}}}{2016 \, 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))^3,x, algorithm="giac")

[Out]

-1/2016*(21*(21*tan(1/2*d*x + 1/2*c)^2 - 36*tan(1/2*d*x + 1/2*c) + 19)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) + (3
591*tan(1/2*d*x + 1/2*c)^8 + 19656*tan(1/2*d*x + 1/2*c)^7 + 56196*tan(1/2*d*x + 1/2*c)^6 + 95760*tan(1/2*d*x +
1/2*c)^5 + 107730*tan(1/2*d*x + 1/2*c)^4 + 79464*tan(1/2*d*x + 1/2*c)^3 + 38484*tan(1/2*d*x + 1/2*c)^2 + 1094
4*tan(1/2*d*x + 1/2*c) + 1615)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^9))/d