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]

-1/7*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^2-1/7*sec(d*x+c)^3/d/(a^2+a^2*sin(d*x+c))+4/7*tan(d*x+c)/a^2/d+4/21*tan(d
*x+c)^3/a^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = 0.06, 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]

-1/21*(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))/(a^2*d*(1 + Sin[c + d*x])^2)

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 103, normalized size = 1.11 \[ \frac {16 \, \cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + {\left (8 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 2}{21 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.78, size = 145, normalized size = 1.56 \[ -\frac {\frac {7 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {273 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 1155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2870 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2037 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 791 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 152}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{168 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 0.23, size = 158, normalized size = 1.70 \[ \frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {5}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {55}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {23}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {13}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a^{2} d} \]

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))

________________________________________________________________________________________

maxima [B]  time = 0.78, size = 396, normalized size = 4.26 \[ -\frac {2 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {76 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {28 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {42 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {56 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {28 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {42 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {21 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 6\right )}}{21 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} d} \]

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)

________________________________________________________________________________________

mupad [B]  time = 5.18, size = 276, normalized size = 2.97 \[ \frac {2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+76\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+28\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-42\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-56\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+28\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+42\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+21\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\right )}{21\,a^2\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^2),x)

[Out]

(2*cos(c/2 + (d*x)/2)*(21*sin(c/2 + (d*x)/2)^9 - 6*cos(c/2 + (d*x)/2)^9 + 42*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x
)/2)^8 - 3*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2) + 28*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^7 - 56*cos(c/2
+ (d*x)/2)^3*sin(c/2 + (d*x)/2)^6 - 42*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^5 + 28*cos(c/2 + (d*x)/2)^5*si
n(c/2 + (d*x)/2)^4 + 76*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^3 + 24*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)
^2))/(21*a^2*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^7)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

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

________________________________________________________________________________________