\(\int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx\) [61]

Optimal result

Integrand size = 21, antiderivative size = 120 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \text {arctanh}(\sin (c+d x))}{16 a d}+\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))} \]

[Out]

5/16*arctanh(sin(d*x+c))/a/d+1/32*a/d/(a-a*sin(d*x+c))^2+1/8/d/(a-a*sin(d*x+c))-1/24*a^2/d/(a+a*sin(d*x+c))^3-
3/32*a/d/(a+a*sin(d*x+c))^2-3/16/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2746, 46, 212} \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {a^2}{24 d (a \sin (c+d x)+a)^3}+\frac {5 \text {arctanh}(\sin (c+d x))}{16 a d}+\frac {a}{32 d (a-a \sin (c+d x))^2}-\frac {3 a}{32 d (a \sin (c+d x)+a)^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {3}{16 d (a \sin (c+d x)+a)} \]

[In]

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

[Out]

(5*ArcTanh[Sin[c + d*x]])/(16*a*d) + a/(32*d*(a - a*Sin[c + d*x])^2) + 1/(8*d*(a - a*Sin[c + d*x])) - a^2/(24*
d*(a + a*Sin[c + d*x])^3) - (3*a)/(32*d*(a + a*Sin[c + d*x])^2) - 3/(16*d*(a + a*Sin[c + d*x]))

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 \text {Subst}\left (\int \left (\frac {1}{16 a^4 (a-x)^3}+\frac {1}{8 a^5 (a-x)^2}+\frac {1}{8 a^3 (a+x)^4}+\frac {3}{16 a^4 (a+x)^3}+\frac {3}{16 a^5 (a+x)^2}+\frac {5}{16 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))}+\frac {5 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 d} \\ & = \frac {5 \text {arctanh}(\sin (c+d x))}{16 a d}+\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.81 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sec ^4(c+d x) \left (-8+25 \sin (c+d x)+25 \sin ^2(c+d x)-15 \sin ^3(c+d x)-15 \sin ^4(c+d x)+15 \text {arctanh}(\sin (c+d x)) (-1+\sin (c+d x))^2 (1+\sin (c+d x))^3\right )}{48 a d (1+\sin (c+d x))} \]

[In]

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

[Out]

(Sec[c + d*x]^4*(-8 + 25*Sin[c + d*x] + 25*Sin[c + d*x]^2 - 15*Sin[c + d*x]^3 - 15*Sin[c + d*x]^4 + 15*ArcTanh
[Sin[c + d*x]]*(-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^3))/(48*a*d*(1 + Sin[c + d*x]))

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}+\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}}{d a}\)	\(91\)
default	\(\frac {-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}+\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}}{d a}\)	\(91\)
risch	\(-\frac {i \left (30 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}+110 i {\mathrm e}^{6 i \left (d x +c \right )}+40 \,{\mathrm e}^{7 i \left (d x +c \right )}-110 i {\mathrm e}^{4 i \left (d x +c \right )}+18 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}+40 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{4} d a}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{16 a d}-\frac {5 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{16 a d}\)	\(185\)
norman	\(\frac {\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {11 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}+\frac {13 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 a d}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 a d}\)	\(239\)
parallelrisch	\(\frac {\left (-120 \cos \left (2 d x +2 c \right )-30 \cos \left (4 d x +4 c \right )-30 \sin \left (d x +c \right )-45 \sin \left (3 d x +3 c \right )-15 \sin \left (5 d x +5 c \right )-90\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (120 \cos \left (2 d x +2 c \right )+30 \cos \left (4 d x +4 c \right )+30 \sin \left (d x +c \right )+45 \sin \left (3 d x +3 c \right )+15 \sin \left (5 d x +5 c \right )+90\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-16 \cos \left (2 d x +2 c \right )-14 \cos \left (4 d x +4 c \right )+236 \sin \left (d x +c \right )+84 \sin \left (3 d x +3 c \right )+8 \sin \left (5 d x +5 c \right )+30}{48 a d \left (\sin \left (5 d x +5 c \right )+3 \sin \left (3 d x +3 c \right )+2 \sin \left (d x +c \right )+2 \cos \left (4 d x +4 c \right )+8 \cos \left (2 d x +2 c \right )+6\right )}\)	\(251\)

[In]

int(sec(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-1/24/(1+sin(d*x+c))^3-3/32/(1+sin(d*x+c))^2-3/16/(1+sin(d*x+c))+5/32*ln(1+sin(d*x+c))+1/32/(sin(d*x+c)
-1)^2-1/8/(sin(d*x+c)-1)-5/32*ln(sin(d*x+c)-1))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 4}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/96*(30*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(sin(d*x +
c) + 1) + 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 10*(3*cos(d*x + c)^2 + 2
)*sin(d*x + c) - 4)/(a*d*cos(d*x + c)^4*sin(d*x + c) + a*d*cos(d*x + c)^4)

Sympy [F]

\[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

integrate(sec(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**5/(sin(c + d*x) + 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 15 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 25 \, \sin \left (d x + c\right ) + 8\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{96 \, d} \]

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/96*(2*(15*sin(d*x + c)^4 + 15*sin(d*x + c)^3 - 25*sin(d*x + c)^2 - 25*sin(d*x + c) + 8)/(a*sin(d*x + c)^5 +
a*sin(d*x + c)^4 - 2*a*sin(d*x + c)^3 - 2*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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {3 \, {\left (15 \, \sin \left (d x + c\right )^{2} - 38 \, \sin \left (d x + c\right ) + 25\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {55 \, \sin \left (d x + c\right )^{3} + 201 \, \sin \left (d x + c\right )^{2} + 255 \, \sin \left (d x + c\right ) + 117}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{192 \, d} \]

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*(30*log(abs(sin(d*x + c) + 1))/a - 30*log(abs(sin(d*x + c) - 1))/a + 3*(15*sin(d*x + c)^2 - 38*sin(d*x +
c) + 25)/(a*(sin(d*x + c) - 1)^2) - (55*sin(d*x + c)^3 + 201*sin(d*x + c)^2 + 255*sin(d*x + c) + 117)/(a*(sin
(d*x + c) + 1)^3))/d

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{16\,a\,d}-\frac {\frac {5\,{\sin \left (c+d\,x\right )}^4}{16}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{16}-\frac {25\,{\sin \left (c+d\,x\right )}^2}{48}-\frac {25\,\sin \left (c+d\,x\right )}{48}+\frac {1}{6}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,a\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]

[In]

int(1/(cos(c + d*x)^5*(a + a*sin(c + d*x))),x)

[Out]

(5*atanh(sin(c + d*x)))/(16*a*d) - ((5*sin(c + d*x)^3)/16 - (25*sin(c + d*x)^2)/48 - (25*sin(c + d*x))/48 + (5
*sin(c + d*x)^4)/16 + 1/6)/(d*(a + a*sin(c + d*x) - 2*a*sin(c + d*x)^2 - 2*a*sin(c + d*x)^3 + a*sin(c + d*x)^4
+ a*sin(c + d*x)^5))