3.889 \(\int \frac{\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=165 \[ -\frac{a^3}{64 d (a \sin (c+d x)+a)^4}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}-\frac{a^2}{24 d (a \sin (c+d x)+a)^3}+\frac{5 a}{128 d (a-a \sin (c+d x))^2}-\frac{5 a}{64 d (a \sin (c+d x)+a)^2}+\frac{15}{128 d (a-a \sin (c+d x))}-\frac{5}{32 d (a \sin (c+d x)+a)}+\frac{35 \tanh ^{-1}(\sin (c+d x))}{128 a d} \]

[Out]

(35*ArcTanh[Sin[c + d*x]])/(128*a*d) + a^2/(96*d*(a - a*Sin[c + d*x])^3) + (5*a)/(128*d*(a - a*Sin[c + d*x])^2
) + 15/(128*d*(a - a*Sin[c + d*x])) - a^3/(64*d*(a + a*Sin[c + d*x])^4) - a^2/(24*d*(a + a*Sin[c + d*x])^3) -
(5*a)/(64*d*(a + a*Sin[c + d*x])^2) - 5/(32*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.125898, antiderivative size = 165, 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} \[ -\frac{a^3}{64 d (a \sin (c+d x)+a)^4}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}-\frac{a^2}{24 d (a \sin (c+d x)+a)^3}+\frac{5 a}{128 d (a-a \sin (c+d x))^2}-\frac{5 a}{64 d (a \sin (c+d x)+a)^2}+\frac{15}{128 d (a-a \sin (c+d x))}-\frac{5}{32 d (a \sin (c+d x)+a)}+\frac{35 \tanh ^{-1}(\sin (c+d x))}{128 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + a*Sin[c + d*x]),x]

[Out]

(35*ArcTanh[Sin[c + d*x]])/(128*a*d) + a^2/(96*d*(a - a*Sin[c + d*x])^3) + (5*a)/(128*d*(a - a*Sin[c + d*x])^2
) + 15/(128*d*(a - a*Sin[c + d*x])) - a^3/(64*d*(a + a*Sin[c + d*x])^4) - a^2/(24*d*(a + a*Sin[c + d*x])^3) -
(5*a)/(64*d*(a + a*Sin[c + d*x])^2) - 5/(32*d*(a + a*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 ^7(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{a^7 \operatorname{Subst}\left (\int \frac{1}{(a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^7 \operatorname{Subst}\left (\int \left (\frac{1}{32 a^5 (a-x)^4}+\frac{5}{64 a^6 (a-x)^3}+\frac{15}{128 a^7 (a-x)^2}+\frac{1}{16 a^4 (a+x)^5}+\frac{1}{8 a^5 (a+x)^4}+\frac{5}{32 a^6 (a+x)^3}+\frac{5}{32 a^7 (a+x)^2}+\frac{35}{128 a^7 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^2}{96 d (a-a \sin (c+d x))^3}+\frac{5 a}{128 d (a-a \sin (c+d x))^2}+\frac{15}{128 d (a-a \sin (c+d x))}-\frac{a^3}{64 d (a+a \sin (c+d x))^4}-\frac{a^2}{24 d (a+a \sin (c+d x))^3}-\frac{5 a}{64 d (a+a \sin (c+d x))^2}-\frac{5}{32 d (a+a \sin (c+d x))}+\frac{35 \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=\frac{35 \tanh ^{-1}(\sin (c+d x))}{128 a d}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}+\frac{5 a}{128 d (a-a \sin (c+d x))^2}+\frac{15}{128 d (a-a \sin (c+d x))}-\frac{a^3}{64 d (a+a \sin (c+d x))^4}-\frac{a^2}{24 d (a+a \sin (c+d x))^3}-\frac{5 a}{64 d (a+a \sin (c+d x))^2}-\frac{5}{32 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.527043, size = 145, normalized size = 0.88 \[ -\frac{\sec ^6(c+d x) \left (-105 \sin ^6(c+d x)-105 \sin ^5(c+d x)+280 \sin ^4(c+d x)+280 \sin ^3(c+d x)-231 \sin ^2(c+d x)-231 \sin (c+d x)-105 \tanh ^{-1}(\sin (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^6 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8+48\right )}{384 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + a*Sin[c + d*x]),x]

[Out]

-(Sec[c + d*x]^6*(48 - 105*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^6*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^8 - 231*Sin[c + d*x] - 231*Sin[c + d*x]^2 + 280*Sin[c + d*x]^3 + 280*Sin[c + d*x]^4 - 105*Sin
[c + d*x]^5 - 105*Sin[c + d*x]^6))/(384*a*d*(1 + Sin[c + d*x]))

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Maple [A]  time = 0.073, size = 162, normalized size = 1. \begin{align*} -{\frac{1}{96\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}+{\frac{5}{128\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{15}{128\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{35\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{256\,da}}-{\frac{1}{64\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{24\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5}{64\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5}{32\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{35\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{256\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96/d/a/(sin(d*x+c)-1)^3+5/128/d/a/(sin(d*x+c)-1)^2-15/128/a/d/(sin(d*x+c)-1)-35/256/a/d*ln(sin(d*x+c)-1)-1/
64/d/a/(1+sin(d*x+c))^4-1/24/d/a/(1+sin(d*x+c))^3-5/64/a/d/(1+sin(d*x+c))^2-5/32/a/d/(1+sin(d*x+c))+35/256*ln(
1+sin(d*x+c))/a/d

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Maxima [A]  time = 1.06108, size = 236, normalized size = 1.43 \begin{align*} -\frac{\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{6} + 105 \, \sin \left (d x + c\right )^{5} - 280 \, \sin \left (d x + c\right )^{4} - 280 \, \sin \left (d x + c\right )^{3} + 231 \, \sin \left (d x + c\right )^{2} + 231 \, \sin \left (d x + c\right ) - 48\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{105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac{105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(2*(105*sin(d*x + c)^6 + 105*sin(d*x + c)^5 - 280*sin(d*x + c)^4 - 280*sin(d*x + c)^3 + 231*sin(d*x + c
)^2 + 231*sin(d*x + c) - 48)/(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) - 105*log(sin(d*x + c) + 1)/a + 105*log(sin(d*x
+ c) - 1)/a)/d

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Fricas [A]  time = 1.51134, size = 462, normalized size = 2.8 \begin{align*} -\frac{210 \, \cos \left (d x + c\right )^{6} - 70 \, \cos \left (d x + c\right )^{4} - 28 \, \cos \left (d x + c\right )^{2} - 105 \,{\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 ) + 105 \,{\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 ) - 14 \,{\left (15 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{2} + 8\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(210*cos(d*x + c)^6 - 70*cos(d*x + c)^4 - 28*cos(d*x + c)^2 - 105*(cos(d*x + c)^6*sin(d*x + c) + cos(d*
x + c)^6)*log(sin(d*x + c) + 1) + 105*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) -
14*(15*cos(d*x + c)^4 + 10*cos(d*x + c)^2 + 8)*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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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)**7/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.32448, size = 184, normalized size = 1.12 \begin{align*} \frac{\frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{2 \,{\left (385 \, \sin \left (d x + c\right )^{3} - 1335 \, \sin \left (d x + c\right )^{2} + 1575 \, \sin \left (d x + c\right ) - 641\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac{875 \, \sin \left (d x + c\right )^{4} + 3980 \, \sin \left (d x + c\right )^{3} + 6930 \, \sin \left (d x + c\right )^{2} + 5548 \, \sin \left (d x + c\right ) + 1771}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3072*(420*log(abs(sin(d*x + c) + 1))/a - 420*log(abs(sin(d*x + c) - 1))/a + 2*(385*sin(d*x + c)^3 - 1335*sin
(d*x + c)^2 + 1575*sin(d*x + c) - 641)/(a*(sin(d*x + c) - 1)^3) - (875*sin(d*x + c)^4 + 3980*sin(d*x + c)^3 +
6930*sin(d*x + c)^2 + 5548*sin(d*x + c) + 1771)/(a*(sin(d*x + c) + 1)^4))/d