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

Optimal. Leaf size=91 \[ \frac {2 \tan ^7(c+d x)}{7 a^2 d}+\frac {3 \tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {2 \sec ^5(c+d x)}{5 a^2 d} \]

[Out]

2/5*sec(d*x+c)^5/a^2/d-2/7*sec(d*x+c)^7/a^2/d+1/3*tan(d*x+c)^3/a^2/d+3/5*tan(d*x+c)^5/a^2/d+2/7*tan(d*x+c)^7/a
^2/d

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Rubi [A]  time = 0.31, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2875, 2873, 2607, 270, 2606, 14} \[ \frac {2 \tan ^7(c+d x)}{7 a^2 d}+\frac {3 \tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {2 \sec ^5(c+d x)}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sec[c + d*x]^5)/(5*a^2*d) - (2*Sec[c + d*x]^7)/(7*a^2*d) + Tan[c + d*x]^3/(3*a^2*d) + (3*Tan[c + d*x]^5)/(5
*a^2*d) + (2*Tan[c + d*x]^7)/(7*a^2*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 2873

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

Rule 2875

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

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \sec ^6(c+d x) (a-a \sin (c+d x))^2 \tan ^2(c+d x) \, dx}{a^4}\\ &=\frac {\int \left (a^2 \sec ^6(c+d x) \tan ^2(c+d x)-2 a^2 \sec ^5(c+d x) \tan ^3(c+d x)+a^2 \sec ^4(c+d x) \tan ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \sec ^6(c+d x) \tan ^2(c+d x) \, dx}{a^2}+\frac {\int \sec ^4(c+d x) \tan ^4(c+d x) \, dx}{a^2}-\frac {2 \int \sec ^5(c+d x) \tan ^3(c+d x) \, dx}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {\operatorname {Subst}\left (\int x^2 \left (1+x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {2 \operatorname {Subst}\left (\int x^4 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {\operatorname {Subst}\left (\int \left (x^2+2 x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {2 \operatorname {Subst}\left (\int \left (-x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {3 \tan ^5(c+d x)}{5 a^2 d}+\frac {2 \tan ^7(c+d x)}{7 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 126, normalized size = 1.38 \[ \frac {\sec ^3(c+d x) (3136 \sin (c+d x)-408 \sin (2 (c+d x))-48 \sin (3 (c+d x))-204 \sin (4 (c+d x))+16 \sin (5 (c+d x))-714 \cos (c+d x)+128 \cos (2 (c+d x))-153 \cos (3 (c+d x))+64 \cos (4 (c+d x))+51 \cos (5 (c+d x))+1344)}{13440 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(1344 - 714*Cos[c + d*x] + 128*Cos[2*(c + d*x)] - 153*Cos[3*(c + d*x)] + 64*Cos[4*(c + d*x)] +
51*Cos[5*(c + d*x)] + 3136*Sin[c + d*x] - 408*Sin[2*(c + d*x)] - 48*Sin[3*(c + d*x)] - 204*Sin[4*(c + d*x)] +
16*Sin[5*(c + d*x)]))/(13440*a^2*d*(1 + Sin[c + d*x])^2)

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fricas [A]  time = 0.44, size = 103, normalized size = 1.13 \[ -\frac {4 \, \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + {\left (2 \, \cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} + 25\right )} \sin \left (d x + c\right ) + 10}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

-1/105*(4*cos(d*x + c)^4 - 2*cos(d*x + c)^2 + (2*cos(d*x + c)^4 - 3*cos(d*x + c)^2 + 25)*sin(d*x + c) + 10)/(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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giac [A]  time = 0.26, size = 146, normalized size = 1.60 \[ -\frac {\frac {35 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 945 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1820 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1617 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 749 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 122}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

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))^2,x, algorithm="giac")

[Out]

-1/840*(35*(3*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) + 2)/(a^2*(tan(1/2*d*x + 1/2*c) - 1)^3) - (105*t
an(1/2*d*x + 1/2*c)^6 + 945*tan(1/2*d*x + 1/2*c)^5 + 1820*tan(1/2*d*x + 1/2*c)^4 + 2450*tan(1/2*d*x + 1/2*c)^3
+ 1617*tan(1/2*d*x + 1/2*c)^2 + 749*tan(1/2*d*x + 1/2*c) + 122)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

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maple [A]  time = 0.50, size = 160, normalized size = 1.76 \[ \frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {16}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {19}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+64}}{a^{2} d} \]

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))^2,x)

[Out]

8/d/a^2*(-1/96/(tan(1/2*d*x+1/2*c)-1)^3-1/64/(tan(1/2*d*x+1/2*c)-1)^2-1/64/(tan(1/2*d*x+1/2*c)-1)-1/14/(tan(1/
2*d*x+1/2*c)+1)^7+1/4/(tan(1/2*d*x+1/2*c)+1)^6-2/5/(tan(1/2*d*x+1/2*c)+1)^5+3/8/(tan(1/2*d*x+1/2*c)+1)^4-19/96
/(tan(1/2*d*x+1/2*c)+1)^3+3/64/(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 = 356, normalized size = 3.91 \[ \frac {8 \, {\left (\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {11 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {7 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {42 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {35 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {35 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 3\right )}}{105 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} d} \]

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))^2,x, algorithm="maxima")

[Out]

8/105*(12*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 11*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 7*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 42*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 35*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 + 35*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3)/((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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mupad [B]  time = 14.34, size = 231, normalized size = 2.54 \[ \frac {\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{35}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+\frac {24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}+\frac {88\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{105}-\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}}{a^2\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((8*cos(c/2 + (d*x)/2)^10)/35 + (32*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2))/35 + (8*cos(c/2 + (d*x)/2)^3*sin(
c/2 + (d*x)/2)^7)/3 + (8*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6)/3 + (16*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*
x)/2)^5)/5 - (8*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4)/15 + (88*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^3)
/105 + (24*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2)/35)/(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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