3.87 \(\int \frac{\sec ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=171 \[ -\frac{a^2}{40 d (a \sin (c+d x)+a)^5}+\frac{3}{64 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac{15}{128 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac{21 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}-\frac{3 a}{64 d (a \sin (c+d x)+a)^4}-\frac{1}{16 d (a \sin (c+d x)+a)^3}+\frac{1}{128 a d (a-a \sin (c+d x))^2}-\frac{5}{64 a d (a \sin (c+d x)+a)^2} \]

[Out]

(21*ArcTanh[Sin[c + d*x]])/(128*a^3*d) + 1/(128*a*d*(a - a*Sin[c + d*x])^2) - a^2/(40*d*(a + a*Sin[c + d*x])^5
) - (3*a)/(64*d*(a + a*Sin[c + d*x])^4) - 1/(16*d*(a + a*Sin[c + d*x])^3) - 5/(64*a*d*(a + a*Sin[c + d*x])^2)
+ 3/(64*d*(a^3 - a^3*Sin[c + d*x])) - 15/(128*d*(a^3 + a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.128418, antiderivative size = 171, 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} \[ -\frac{a^2}{40 d (a \sin (c+d x)+a)^5}+\frac{3}{64 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac{15}{128 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac{21 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}-\frac{3 a}{64 d (a \sin (c+d x)+a)^4}-\frac{1}{16 d (a \sin (c+d x)+a)^3}+\frac{1}{128 a d (a-a \sin (c+d x))^2}-\frac{5}{64 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]

[Out]

(21*ArcTanh[Sin[c + d*x]])/(128*a^3*d) + 1/(128*a*d*(a - a*Sin[c + d*x])^2) - a^2/(40*d*(a + a*Sin[c + d*x])^5
) - (3*a)/(64*d*(a + a*Sin[c + d*x])^4) - 1/(16*d*(a + a*Sin[c + d*x])^3) - 5/(64*a*d*(a + a*Sin[c + d*x])^2)
+ 3/(64*d*(a^3 - a^3*Sin[c + d*x])) - 15/(128*d*(a^3 + a^3*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))^3} \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{1}{(a-x)^3 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^5 \operatorname{Subst}\left (\int \left (\frac{1}{64 a^6 (a-x)^3}+\frac{3}{64 a^7 (a-x)^2}+\frac{1}{8 a^3 (a+x)^6}+\frac{3}{16 a^4 (a+x)^5}+\frac{3}{16 a^5 (a+x)^4}+\frac{5}{32 a^6 (a+x)^3}+\frac{15}{128 a^7 (a+x)^2}+\frac{21}{128 a^7 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{1}{128 a d (a-a \sin (c+d x))^2}-\frac{a^2}{40 d (a+a \sin (c+d x))^5}-\frac{3 a}{64 d (a+a \sin (c+d x))^4}-\frac{1}{16 d (a+a \sin (c+d x))^3}-\frac{5}{64 a d (a+a \sin (c+d x))^2}+\frac{3}{64 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac{15}{128 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac{21 \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 a^2 d}\\ &=\frac{21 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}+\frac{1}{128 a d (a-a \sin (c+d x))^2}-\frac{a^2}{40 d (a+a \sin (c+d x))^5}-\frac{3 a}{64 d (a+a \sin (c+d x))^4}-\frac{1}{16 d (a+a \sin (c+d x))^3}-\frac{5}{64 a d (a+a \sin (c+d x))^2}+\frac{3}{64 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac{15}{128 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.516994, size = 145, normalized size = 0.85 \[ \frac{\sec ^4(c+d x) \left (-105 \sin ^6(c+d x)-315 \sin ^5(c+d x)-140 \sin ^4(c+d x)+420 \sin ^3(c+d x)+469 \sin ^2(c+d x)+7 \sin (c+d x)+105 \tanh ^{-1}(\sin (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^{10}-176\right )}{640 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^4*(-176 + 105*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^10 + 7*Sin[c + d*x] + 469*Sin[c + d*x]^2 + 420*Sin[c + d*x]^3 - 140*Sin[c + d*x]^4 - 315*Sin
[c + d*x]^5 - 105*Sin[c + d*x]^6))/(640*a^3*d*(1 + Sin[c + d*x])^3)

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Maple [A]  time = 0.112, size = 162, normalized size = 1. \begin{align*}{\frac{1}{128\,d{a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{3}{64\,d{a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{21\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{256\,d{a}^{3}}}-{\frac{1}{40\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{3}{64\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{16\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5}{64\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{15}{128\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{21\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{256\,d{a}^{3}}} \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))^3,x)

[Out]

1/128/d/a^3/(sin(d*x+c)-1)^2-3/64/d/a^3/(sin(d*x+c)-1)-21/256/d/a^3*ln(sin(d*x+c)-1)-1/40/d/a^3/(1+sin(d*x+c))
^5-3/64/d/a^3/(1+sin(d*x+c))^4-1/16/d/a^3/(1+sin(d*x+c))^3-5/64/d/a^3/(1+sin(d*x+c))^2-15/128/d/a^3/(1+sin(d*x
+c))+21/256*ln(1+sin(d*x+c))/a^3/d

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Maxima [A]  time = 0.961971, size = 254, normalized size = 1.49 \begin{align*} -\frac{\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{6} + 315 \, \sin \left (d x + c\right )^{5} + 140 \, \sin \left (d x + c\right )^{4} - 420 \, \sin \left (d x + c\right )^{3} - 469 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) + 176\right )}}{a^{3} \sin \left (d x + c\right )^{7} + 3 \, a^{3} \sin \left (d x + c\right )^{6} + a^{3} \sin \left (d x + c\right )^{5} - 5 \, a^{3} \sin \left (d x + c\right )^{4} - 5 \, a^{3} \sin \left (d x + c\right )^{3} + a^{3} \sin \left (d x + c\right )^{2} + 3 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac{105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{1280 \, 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))^3,x, algorithm="maxima")

[Out]

-1/1280*(2*(105*sin(d*x + c)^6 + 315*sin(d*x + c)^5 + 140*sin(d*x + c)^4 - 420*sin(d*x + c)^3 - 469*sin(d*x +
c)^2 - 7*sin(d*x + c) + 176)/(a^3*sin(d*x + c)^7 + 3*a^3*sin(d*x + c)^6 + a^3*sin(d*x + c)^5 - 5*a^3*sin(d*x +
c)^4 - 5*a^3*sin(d*x + c)^3 + a^3*sin(d*x + c)^2 + 3*a^3*sin(d*x + c) + a^3) - 105*log(sin(d*x + c) + 1)/a^3
+ 105*log(sin(d*x + c) - 1)/a^3)/d

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Fricas [A]  time = 1.91453, size = 659, normalized size = 3.85 \begin{align*} -\frac{210 \, \cos \left (d x + c\right )^{6} - 910 \, \cos \left (d x + c\right )^{4} + 252 \, \cos \left (d x + c\right )^{2} - 105 \,{\left (3 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} +{\left (\cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \,{\left (3 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} +{\left (\cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 14 \,{\left (45 \, \cos \left (d x + c\right )^{4} - 30 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 96}{1280 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{6} - 4 \, a^{3} d \cos \left (d x + c\right )^{4} +{\left (a^{3} d \cos \left (d x + c\right )^{6} - 4 \, a^{3} d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\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))^3,x, algorithm="fricas")

[Out]

-1/1280*(210*cos(d*x + c)^6 - 910*cos(d*x + c)^4 + 252*cos(d*x + c)^2 - 105*(3*cos(d*x + c)^6 - 4*cos(d*x + c)
^4 + (cos(d*x + c)^6 - 4*cos(d*x + c)^4)*sin(d*x + c))*log(sin(d*x + c) + 1) + 105*(3*cos(d*x + c)^6 - 4*cos(d
*x + c)^4 + (cos(d*x + c)^6 - 4*cos(d*x + c)^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 14*(45*cos(d*x + c)^4 -
30*cos(d*x + c)^2 - 16)*sin(d*x + c) + 96)/(3*a^3*d*cos(d*x + c)^6 - 4*a^3*d*cos(d*x + c)^4 + (a^3*d*cos(d*x
+ c)^6 - 4*a^3*d*cos(d*x + c)^4)*sin(d*x + c))

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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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.19368, size = 184, normalized size = 1.08 \begin{align*} \frac{\frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac{10 \,{\left (63 \, \sin \left (d x + c\right )^{2} - 150 \, \sin \left (d x + c\right ) + 91\right )}}{a^{3}{\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac{959 \, \sin \left (d x + c\right )^{5} + 5395 \, \sin \left (d x + c\right )^{4} + 12390 \, \sin \left (d x + c\right )^{3} + 14710 \, \sin \left (d x + c\right )^{2} + 9275 \, \sin \left (d x + c\right ) + 2647}{a^{3}{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{5120 \, 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))^3,x, algorithm="giac")

[Out]

1/5120*(420*log(abs(sin(d*x + c) + 1))/a^3 - 420*log(abs(sin(d*x + c) - 1))/a^3 + 10*(63*sin(d*x + c)^2 - 150*
sin(d*x + c) + 91)/(a^3*(sin(d*x + c) - 1)^2) - (959*sin(d*x + c)^5 + 5395*sin(d*x + c)^4 + 12390*sin(d*x + c)
^3 + 14710*sin(d*x + c)^2 + 9275*sin(d*x + c) + 2647)/(a^3*(sin(d*x + c) + 1)^5))/d