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

Optimal. Leaf size=73 \[ \frac {\tan ^5(c+d x)}{5 a d}+\frac {\tan ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\sec ^3(c+d x)}{3 a d} \]

[Out]

1/3*sec(d*x+c)^3/a/d-1/5*sec(d*x+c)^5/a/d+1/3*tan(d*x+c)^3/a/d+1/5*tan(d*x+c)^5/a/d

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Rubi [A]  time = 0.16, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2839, 2607, 14, 2606} \[ \frac {\tan ^5(c+d x)}{5 a d}+\frac {\tan ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\sec ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sec[c + d*x]^3/(3*a*d) - Sec[c + d*x]^5/(5*a*d) + Tan[c + d*x]^3/(3*a*d) + Tan[c + d*x]^5/(5*a*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 106, normalized size = 1.45 \[ -\frac {\sec ^3(c+d x) (-224 \sin (c+d x)+22 \sin (2 (c+d x))+32 \sin (3 (c+d x))+11 \sin (4 (c+d x))+66 \cos (c+d x)-32 \cos (2 (c+d x))+22 \cos (3 (c+d x))-16 \cos (4 (c+d x))-80)}{960 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/960*(Sec[c + d*x]^3*(-80 + 66*Cos[c + d*x] - 32*Cos[2*(c + d*x)] + 22*Cos[3*(c + d*x)] - 16*Cos[4*(c + d*x)
] - 224*Sin[c + d*x] + 22*Sin[2*(c + d*x)] + 32*Sin[3*(c + d*x)] + 11*Sin[4*(c + d*x)]))/(a*d*(1 + Sin[c + d*x
]))

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fricas [A]  time = 0.45, size = 73, normalized size = 1.00 \[ \frac {2 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 1}{15 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*cos(d*x + c)^4 - cos(d*x + c)^2 - 2*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 1)/(a*d*cos(d*x + c)^3*sin(d*x
+ c) + a*d*cos(d*x + c)^3)

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giac [A]  time = 0.23, size = 109, normalized size = 1.49 \[ -\frac {\frac {5 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {3 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 50 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(5*(3*tan(1/2*d*x + 1/2*c)^2 + 1)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) - 3*(5*tan(1/2*d*x + 1/2*c)^4 + 40*t
an(1/2*d*x + 1/2*c)^3 + 50*tan(1/2*d*x + 1/2*c)^2 + 40*tan(1/2*d*x + 1/2*c) + 9)/(a*(tan(1/2*d*x + 1/2*c) + 1)
^5))/d

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maple [A]  time = 0.38, size = 130, normalized size = 1.78 \[ \frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+64}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/d/a*(-1/48/(tan(1/2*d*x+1/2*c)-1)^3-1/32/(tan(1/2*d*x+1/2*c)-1)^2-1/64/(tan(1/2*d*x+1/2*c)-1)-1/20/(tan(1/2*
d*x+1/2*c)+1)^5+1/8/(tan(1/2*d*x+1/2*c)+1)^4-1/8/(tan(1/2*d*x+1/2*c)+1)^3+1/16/(tan(1/2*d*x+1/2*c)+1)^2+1/64/(
tan(1/2*d*x+1/2*c)+1))

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maxima [B]  time = 0.34, size = 254, normalized size = 3.48 \[ \frac {4 \, {\left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 1\right )}}{15 \, {\left (a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(2*sin(d*x + c)/(cos(d*x + c) + 1) - 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 1)/((a + 2*a*sin(
d*x + c)/(cos(d*x + c) + 1) - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^
3 + 6*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7 - a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d)

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mupad [B]  time = 10.16, size = 99, normalized size = 1.36 \[ -\frac {4\,\left (10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{15\,a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^2 + 4*tan(c/2 + (d*x)/2)^3 + 5*tan(c/2 + (d*x)/2)^4 + 10*tan(
c/2 + (d*x)/2)^5 + 1))/(15*a*d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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