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

Optimal. Leaf size=146 \[ -\frac{a^2}{32 d (a \sin (c+d x)+a)^4}+\frac{5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac{a}{16 d (a \sin (c+d x)+a)^3}+\frac{1}{64 d (a-a \sin (c+d x))^2}-\frac{3}{32 d (a \sin (c+d x)+a)^2} \]

[Out]

(15*ArcTanh[Sin[c + d*x]])/(64*a^2*d) + 1/(64*d*(a - a*Sin[c + d*x])^2) - a^2/(32*d*(a + a*Sin[c + d*x])^4) -
a/(16*d*(a + a*Sin[c + d*x])^3) - 3/(32*d*(a + a*Sin[c + d*x])^2) + 5/(64*d*(a^2 - a^2*Sin[c + d*x])) - 5/(32*
d*(a^2 + a^2*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.10989, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2667, 44, 206} \[ -\frac{a^2}{32 d (a \sin (c+d x)+a)^4}+\frac{5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac{a}{16 d (a \sin (c+d x)+a)^3}+\frac{1}{64 d (a-a \sin (c+d x))^2}-\frac{3}{32 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*ArcTanh[Sin[c + d*x]])/(64*a^2*d) + 1/(64*d*(a - a*Sin[c + d*x])^2) - a^2/(32*d*(a + a*Sin[c + d*x])^4) -
a/(16*d*(a + a*Sin[c + d*x])^3) - 3/(32*d*(a + a*Sin[c + d*x])^2) + 5/(64*d*(a^2 - a^2*Sin[c + d*x])) - 5/(32*
d*(a^2 + a^2*Sin[c + d*x]))

Rule 2667

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])

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{1}{(a-x)^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^5 \operatorname{Subst}\left (\int \left (\frac{1}{32 a^5 (a-x)^3}+\frac{5}{64 a^6 (a-x)^2}+\frac{1}{8 a^3 (a+x)^5}+\frac{3}{16 a^4 (a+x)^4}+\frac{3}{16 a^5 (a+x)^3}+\frac{5}{32 a^6 (a+x)^2}+\frac{15}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{1}{64 d (a-a \sin (c+d x))^2}-\frac{a^2}{32 d (a+a \sin (c+d x))^4}-\frac{a}{16 d (a+a \sin (c+d x))^3}-\frac{3}{32 d (a+a \sin (c+d x))^2}+\frac{5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{64 a d}\\ &=\frac{15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}+\frac{1}{64 d (a-a \sin (c+d x))^2}-\frac{a^2}{32 d (a+a \sin (c+d x))^4}-\frac{a}{16 d (a+a \sin (c+d x))^3}-\frac{3}{32 d (a+a \sin (c+d x))^2}+\frac{5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.337791, size = 137, normalized size = 0.94 \[ \frac{(1-\sin (c+d x))^2 (\sin (c+d x)+1)^2 \sec ^4(c+d x) \left (\frac{5}{64 (1-\sin (c+d x))}-\frac{5}{32 (\sin (c+d x)+1)}+\frac{1}{64 (1-\sin (c+d x))^2}-\frac{3}{32 (\sin (c+d x)+1)^2}-\frac{1}{16 (\sin (c+d x)+1)^3}-\frac{1}{32 (\sin (c+d x)+1)^4}+\frac{15}{64} \tanh ^{-1}(\sin (c+d x))\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^4*(1 - Sin[c + d*x])^2*(1 + Sin[c + d*x])^2*((15*ArcTanh[Sin[c + d*x]])/64 + 1/(64*(1 - Sin[c +
d*x])^2) + 5/(64*(1 - Sin[c + d*x])) - 1/(32*(1 + Sin[c + d*x])^4) - 1/(16*(1 + Sin[c + d*x])^3) - 3/(32*(1 +
Sin[c + d*x])^2) - 5/(32*(1 + Sin[c + d*x]))))/(a^2*d)

________________________________________________________________________________________

Maple [A]  time = 0.086, size = 144, normalized size = 1. \begin{align*}{\frac{1}{64\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{5}{64\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{15\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{128\,d{a}^{2}}}-{\frac{1}{32\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{16\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{3}{32\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5}{32\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{15\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{128\,d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+a*sin(d*x+c))^2,x)

[Out]

1/64/d/a^2/(sin(d*x+c)-1)^2-5/64/d/a^2/(sin(d*x+c)-1)-15/128/d/a^2*ln(sin(d*x+c)-1)-1/32/d/a^2/(1+sin(d*x+c))^
4-1/16/d/a^2/(1+sin(d*x+c))^3-3/32/d/a^2/(1+sin(d*x+c))^2-5/32/d/a^2/(1+sin(d*x+c))+15/128*ln(1+sin(d*x+c))/a^
2/d

________________________________________________________________________________________

Maxima [A]  time = 0.977011, size = 225, normalized size = 1.54 \begin{align*} -\frac{\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} + 30 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{3} - 50 \, \sin \left (d x + c\right )^{2} - 17 \, \sin \left (d x + c\right ) + 16\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac{15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{128 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/128*(2*(15*sin(d*x + c)^5 + 30*sin(d*x + c)^4 - 10*sin(d*x + c)^3 - 50*sin(d*x + c)^2 - 17*sin(d*x + c) + 1
6)/(a^2*sin(d*x + c)^6 + 2*a^2*sin(d*x + c)^5 - a^2*sin(d*x + c)^4 - 4*a^2*sin(d*x + c)^3 - a^2*sin(d*x + c)^2
+ 2*a^2*sin(d*x + c) + a^2) - 15*log(sin(d*x + c) + 1)/a^2 + 15*log(sin(d*x + c) - 1)/a^2)/d

________________________________________________________________________________________

Fricas [A]  time = 2.18425, size = 527, normalized size = 3.61 \begin{align*} \frac{60 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} + 15 \,{\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \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 )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 12\right )} \sin \left (d x + c\right ) - 8}{128 \,{\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/128*(60*cos(d*x + c)^4 - 20*cos(d*x + c)^2 + 15*(cos(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d*x
+ c)^4)*log(sin(d*x + c) + 1) - 15*(cos(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d*x + c)^4)*log(-si
n(d*x + c) + 1) + 2*(15*cos(d*x + c)^4 - 20*cos(d*x + c)^2 - 12)*sin(d*x + c) - 8)/(a^2*d*cos(d*x + c)^6 - 2*a
^2*d*cos(d*x + c)^4*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^4)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.18629, size = 170, normalized size = 1.16 \begin{align*} \frac{\frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \,{\left (45 \, \sin \left (d x + c\right )^{2} - 110 \, \sin \left (d x + c\right ) + 69\right )}}{a^{2}{\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac{125 \, \sin \left (d x + c\right )^{4} + 580 \, \sin \left (d x + c\right )^{3} + 1038 \, \sin \left (d x + c\right )^{2} + 868 \, \sin \left (d x + c\right ) + 301}{a^{2}{\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{512 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/512*(60*log(abs(sin(d*x + c) + 1))/a^2 - 60*log(abs(sin(d*x + c) - 1))/a^2 + 2*(45*sin(d*x + c)^2 - 110*sin(
d*x + c) + 69)/(a^2*(sin(d*x + c) - 1)^2) - (125*sin(d*x + c)^4 + 580*sin(d*x + c)^3 + 1038*sin(d*x + c)^2 + 8
68*sin(d*x + c) + 301)/(a^2*(sin(d*x + c) + 1)^4))/d