3.887 \(\int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=148 \[ -\frac {\sec ^8(c+d x)}{8 a d}+\frac {\sec ^6(c+d x)}{6 a d}-\frac {5 \tanh ^{-1}(\sin (c+d x))}{128 a d}+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 a d}-\frac {\tan (c+d x) \sec ^5(c+d x)}{48 a d}-\frac {5 \tan (c+d x) \sec ^3(c+d x)}{192 a d}-\frac {5 \tan (c+d x) \sec (c+d x)}{128 a d} \]

[Out]

-5/128*arctanh(sin(d*x+c))/a/d+1/6*sec(d*x+c)^6/a/d-1/8*sec(d*x+c)^8/a/d-5/128*sec(d*x+c)*tan(d*x+c)/a/d-5/192
*sec(d*x+c)^3*tan(d*x+c)/a/d-1/48*sec(d*x+c)^5*tan(d*x+c)/a/d+1/8*sec(d*x+c)^7*tan(d*x+c)/a/d

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Rubi [A]  time = 0.20, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2835, 2611, 3768, 3770, 2606, 14} \[ -\frac {\sec ^8(c+d x)}{8 a d}+\frac {\sec ^6(c+d x)}{6 a d}-\frac {5 \tanh ^{-1}(\sin (c+d x))}{128 a d}+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 a d}-\frac {\tan (c+d x) \sec ^5(c+d x)}{48 a d}-\frac {5 \tan (c+d x) \sec ^3(c+d x)}{192 a d}-\frac {5 \tan (c+d x) \sec (c+d x)}{128 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^5*Tan[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

(-5*ArcTanh[Sin[c + d*x]])/(128*a*d) + Sec[c + d*x]^6/(6*a*d) - Sec[c + d*x]^8/(8*a*d) - (5*Sec[c + d*x]*Tan[c
+ d*x])/(128*a*d) - (5*Sec[c + d*x]^3*Tan[c + d*x])/(192*a*d) - (Sec[c + d*x]^5*Tan[c + d*x])/(48*a*d) + (Sec
[c + d*x]^7*Tan[c + d*x])/(8*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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \sec ^7(c+d x) \tan ^2(c+d x) \, dx}{a}-\frac {\int \sec ^6(c+d x) \tan ^3(c+d x) \, dx}{a}\\ &=\frac {\sec ^7(c+d x) \tan (c+d x)}{8 a d}-\frac {\int \sec ^7(c+d x) \, dx}{8 a}-\frac {\operatorname {Subst}\left (\int x^5 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{48 a d}+\frac {\sec ^7(c+d x) \tan (c+d x)}{8 a d}-\frac {5 \int \sec ^5(c+d x) \, dx}{48 a}-\frac {\operatorname {Subst}\left (\int \left (-x^5+x^7\right ) \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac {\sec ^6(c+d x)}{6 a d}-\frac {\sec ^8(c+d x)}{8 a d}-\frac {5 \sec ^3(c+d x) \tan (c+d x)}{192 a d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{48 a d}+\frac {\sec ^7(c+d x) \tan (c+d x)}{8 a d}-\frac {5 \int \sec ^3(c+d x) \, dx}{64 a}\\ &=\frac {\sec ^6(c+d x)}{6 a d}-\frac {\sec ^8(c+d x)}{8 a d}-\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {5 \sec ^3(c+d x) \tan (c+d x)}{192 a d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{48 a d}+\frac {\sec ^7(c+d x) \tan (c+d x)}{8 a d}-\frac {5 \int \sec (c+d x) \, dx}{128 a}\\ &=-\frac {5 \tanh ^{-1}(\sin (c+d x))}{128 a d}+\frac {\sec ^6(c+d x)}{6 a d}-\frac {\sec ^8(c+d x)}{8 a d}-\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {5 \sec ^3(c+d x) \tan (c+d x)}{192 a d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{48 a d}+\frac {\sec ^7(c+d x) \tan (c+d x)}{8 a d}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 92, normalized size = 0.62 \[ -\frac {-\frac {3}{\sin (c+d x)-1}-\frac {12}{\sin (c+d x)+1}-\frac {3}{(\sin (c+d x)-1)^2}-\frac {6}{(\sin (c+d x)+1)^2}+\frac {4}{(\sin (c+d x)-1)^3}+\frac {6}{(\sin (c+d x)+1)^4}+15 \tanh ^{-1}(\sin (c+d x))}{384 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^5*Tan[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

-1/384*(15*ArcTanh[Sin[c + d*x]] + 4/(-1 + Sin[c + d*x])^3 - 3/(-1 + Sin[c + d*x])^2 - 3/(-1 + Sin[c + d*x]) +
6/(1 + Sin[c + d*x])^4 - 6/(1 + Sin[c + d*x])^2 - 12/(1 + Sin[c + d*x]))/(a*d)

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fricas [A]  time = 0.48, size = 167, normalized size = 1.13 \[ \frac {30 \, \cos \left (d x + c\right )^{6} - 10 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{2} - 56\right )} \sin \left (d x + c\right ) + 16}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/768*(30*cos(d*x + c)^6 - 10*cos(d*x + c)^4 - 4*cos(d*x + c)^2 - 15*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x +
c)^6)*log(sin(d*x + c) + 1) + 15*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(15
*cos(d*x + c)^4 + 10*cos(d*x + c)^2 - 56)*sin(d*x + c) + 16)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*x +
c)^6)

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giac [A]  time = 0.28, size = 136, normalized size = 0.92 \[ -\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (55 \, \sin \left (d x + c\right )^{3} - 177 \, \sin \left (d x + c\right )^{2} + 177 \, \sin \left (d x + c\right ) - 39\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 596 \, \sin \left (d x + c\right )^{3} + 1086 \, \sin \left (d x + c\right )^{2} + 884 \, \sin \left (d x + c\right ) + 221}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/3072*(60*log(abs(sin(d*x + c) + 1))/a - 60*log(abs(sin(d*x + c) - 1))/a + 2*(55*sin(d*x + c)^3 - 177*sin(d*
x + c)^2 + 177*sin(d*x + c) - 39)/(a*(sin(d*x + c) - 1)^3) - (125*sin(d*x + c)^4 + 596*sin(d*x + c)^3 + 1086*s
in(d*x + c)^2 + 884*sin(d*x + c) + 221)/(a*(sin(d*x + c) + 1)^4))/d

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maple [A]  time = 0.36, size = 144, normalized size = 0.97 \[ -\frac {1}{96 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {1}{128 a d \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {1}{128 a d \left (\sin \left (d x +c \right )-1\right )}+\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256 a d}-\frac {1}{64 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{64 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{32 a d \left (1+\sin \left (d x +c \right )\right )}-\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

-1/96/a/d/(sin(d*x+c)-1)^3+1/128/a/d/(sin(d*x+c)-1)^2+1/128/a/d/(sin(d*x+c)-1)+5/256/a/d*ln(sin(d*x+c)-1)-1/64
/a/d/(1+sin(d*x+c))^4+1/64/a/d/(1+sin(d*x+c))^2+1/32/a/d/(1+sin(d*x+c))-5/256*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.31, size = 175, normalized size = 1.18 \[ \frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{6} + 15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{4} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )^{2} - 31 \, \sin \left (d x + c\right ) - 16\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/768*(2*(15*sin(d*x + c)^6 + 15*sin(d*x + c)^5 - 40*sin(d*x + c)^4 - 40*sin(d*x + c)^3 + 33*sin(d*x + c)^2 -
31*sin(d*x + c) - 16)/(a*sin(d*x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x + c)^5 - 3*a*sin(d*x + c)^4 + 3*a*sin
(d*x + c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a) - 15*log(sin(d*x + c) + 1)/a + 15*log(sin(d*x + c) - 1)
/a)/d

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mupad [B]  time = 17.12, size = 388, normalized size = 2.62 \[ \frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}+\frac {221\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}+\frac {43\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}+\frac {625\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{48}+\frac {355\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48}+\frac {625\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}+\frac {43\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{96}+\frac {221\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^2/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)

[Out]

((5*tan(c/2 + (d*x)/2))/64 + (5*tan(c/2 + (d*x)/2)^2)/32 + (221*tan(c/2 + (d*x)/2)^3)/96 + (43*tan(c/2 + (d*x)
/2)^4)/96 + (625*tan(c/2 + (d*x)/2)^5)/192 + (35*tan(c/2 + (d*x)/2)^6)/48 + (355*tan(c/2 + (d*x)/2)^7)/48 + (3
5*tan(c/2 + (d*x)/2)^8)/48 + (625*tan(c/2 + (d*x)/2)^9)/192 + (43*tan(c/2 + (d*x)/2)^10)/96 + (221*tan(c/2 + (
d*x)/2)^11)/96 + (5*tan(c/2 + (d*x)/2)^12)/32 + (5*tan(c/2 + (d*x)/2)^13)/64)/(d*(a + 2*a*tan(c/2 + (d*x)/2) -
5*a*tan(c/2 + (d*x)/2)^2 - 12*a*tan(c/2 + (d*x)/2)^3 + 9*a*tan(c/2 + (d*x)/2)^4 + 30*a*tan(c/2 + (d*x)/2)^5 -
5*a*tan(c/2 + (d*x)/2)^6 - 40*a*tan(c/2 + (d*x)/2)^7 - 5*a*tan(c/2 + (d*x)/2)^8 + 30*a*tan(c/2 + (d*x)/2)^9 +
9*a*tan(c/2 + (d*x)/2)^10 - 12*a*tan(c/2 + (d*x)/2)^11 - 5*a*tan(c/2 + (d*x)/2)^12 + 2*a*tan(c/2 + (d*x)/2)^1
3 + a*tan(c/2 + (d*x)/2)^14)) - (5*atanh(tan(c/2 + (d*x)/2)))/(64*a*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7*sin(d*x+c)**2/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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