3.892 \(\int \frac{\csc ^3(c+d x) \sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=232 \[ \frac{a^3}{64 d (a \sin (c+d x)+a)^4}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}+\frac{5 a^2}{48 d (a \sin (c+d x)+a)^3}+\frac{11 a}{128 d (a-a \sin (c+d x))^2}+\frac{29 a}{64 d (a \sin (c+d x)+a)^2}+\frac{69}{128 d (a-a \sin (c+d x))}+\frac{2}{d (a \sin (c+d x)+a)}-\frac{\csc ^2(c+d x)}{2 a d}+\frac{\csc (c+d x)}{a d}-\frac{325 \log (1-\sin (c+d x))}{256 a d}+\frac{5 \log (\sin (c+d x))}{a d}-\frac{955 \log (\sin (c+d x)+1)}{256 a d} \]

[Out]

Csc[c + d*x]/(a*d) - Csc[c + d*x]^2/(2*a*d) - (325*Log[1 - Sin[c + d*x]])/(256*a*d) + (5*Log[Sin[c + d*x]])/(a
*d) - (955*Log[1 + Sin[c + d*x]])/(256*a*d) + a^2/(96*d*(a - a*Sin[c + d*x])^3) + (11*a)/(128*d*(a - a*Sin[c +
d*x])^2) + 69/(128*d*(a - a*Sin[c + d*x])) + a^3/(64*d*(a + a*Sin[c + d*x])^4) + (5*a^2)/(48*d*(a + a*Sin[c +
d*x])^3) + (29*a)/(64*d*(a + a*Sin[c + d*x])^2) + 2/(d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.246226, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac{a^3}{64 d (a \sin (c+d x)+a)^4}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}+\frac{5 a^2}{48 d (a \sin (c+d x)+a)^3}+\frac{11 a}{128 d (a-a \sin (c+d x))^2}+\frac{29 a}{64 d (a \sin (c+d x)+a)^2}+\frac{69}{128 d (a-a \sin (c+d x))}+\frac{2}{d (a \sin (c+d x)+a)}-\frac{\csc ^2(c+d x)}{2 a d}+\frac{\csc (c+d x)}{a d}-\frac{325 \log (1-\sin (c+d x))}{256 a d}+\frac{5 \log (\sin (c+d x))}{a d}-\frac{955 \log (\sin (c+d x)+1)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^3*Sec[c + d*x]^7)/(a + a*Sin[c + d*x]),x]

[Out]

Csc[c + d*x]/(a*d) - Csc[c + d*x]^2/(2*a*d) - (325*Log[1 - Sin[c + d*x]])/(256*a*d) + (5*Log[Sin[c + d*x]])/(a
*d) - (955*Log[1 + Sin[c + d*x]])/(256*a*d) + a^2/(96*d*(a - a*Sin[c + d*x])^3) + (11*a)/(128*d*(a - a*Sin[c +
d*x])^2) + 69/(128*d*(a - a*Sin[c + d*x])) + a^3/(64*d*(a + a*Sin[c + d*x])^4) + (5*a^2)/(48*d*(a + a*Sin[c +
d*x])^3) + (29*a)/(64*d*(a + a*Sin[c + d*x])^2) + 2/(d*(a + a*Sin[c + d*x]))

Rule 2836

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\csc ^3(c+d x) \sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{a^7 \operatorname{Subst}\left (\int \frac{a^3}{(a-x)^4 x^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^{10} \operatorname{Subst}\left (\int \frac{1}{(a-x)^4 x^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^{10} \operatorname{Subst}\left (\int \left (\frac{1}{32 a^8 (a-x)^4}+\frac{11}{64 a^9 (a-x)^3}+\frac{69}{128 a^{10} (a-x)^2}+\frac{325}{256 a^{11} (a-x)}+\frac{1}{a^9 x^3}-\frac{1}{a^{10} x^2}+\frac{5}{a^{11} x}-\frac{1}{16 a^7 (a+x)^5}-\frac{5}{16 a^8 (a+x)^4}-\frac{29}{32 a^9 (a+x)^3}-\frac{2}{a^{10} (a+x)^2}-\frac{955}{256 a^{11} (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\csc (c+d x)}{a d}-\frac{\csc ^2(c+d x)}{2 a d}-\frac{325 \log (1-\sin (c+d x))}{256 a d}+\frac{5 \log (\sin (c+d x))}{a d}-\frac{955 \log (1+\sin (c+d x))}{256 a d}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}+\frac{11 a}{128 d (a-a \sin (c+d x))^2}+\frac{69}{128 d (a-a \sin (c+d x))}+\frac{a^3}{64 d (a+a \sin (c+d x))^4}+\frac{5 a^2}{48 d (a+a \sin (c+d x))^3}+\frac{29 a}{64 d (a+a \sin (c+d x))^2}+\frac{2}{d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.16423, size = 213, normalized size = 0.92 \[ \frac{a^{10} \left (\frac{69}{128 a^{10} (a-a \sin (c+d x))}+\frac{2}{a^{10} (a \sin (c+d x)+a)}+\frac{11}{128 a^9 (a-a \sin (c+d x))^2}+\frac{29}{64 a^9 (a \sin (c+d x)+a)^2}+\frac{1}{96 a^8 (a-a \sin (c+d x))^3}+\frac{5}{48 a^8 (a \sin (c+d x)+a)^3}+\frac{1}{64 a^7 (a \sin (c+d x)+a)^4}-\frac{\csc ^2(c+d x)}{2 a^{11}}+\frac{\csc (c+d x)}{a^{11}}-\frac{325 \log (1-\sin (c+d x))}{256 a^{11}}+\frac{5 \log (\sin (c+d x))}{a^{11}}-\frac{955 \log (\sin (c+d x)+1)}{256 a^{11}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^3*Sec[c + d*x]^7)/(a + a*Sin[c + d*x]),x]

[Out]

(a^10*(Csc[c + d*x]/a^11 - Csc[c + d*x]^2/(2*a^11) - (325*Log[1 - Sin[c + d*x]])/(256*a^11) + (5*Log[Sin[c + d
*x]])/a^11 - (955*Log[1 + Sin[c + d*x]])/(256*a^11) + 1/(96*a^8*(a - a*Sin[c + d*x])^3) + 11/(128*a^9*(a - a*S
in[c + d*x])^2) + 69/(128*a^10*(a - a*Sin[c + d*x])) + 1/(64*a^7*(a + a*Sin[c + d*x])^4) + 5/(48*a^8*(a + a*Si
n[c + d*x])^3) + 29/(64*a^9*(a + a*Sin[c + d*x])^2) + 2/(a^10*(a + a*Sin[c + d*x]))))/d

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Maple [A]  time = 0.108, size = 208, normalized size = 0.9 \begin{align*} -{\frac{1}{96\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}+{\frac{11}{128\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{69}{128\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{325\,\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{5}{48\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{29}{64\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{1}{da \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{955\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{256\,da}}-{\frac{1}{2\,da \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{1}{da\sin \left ( dx+c \right ) }}+5\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^7/(a+a*sin(d*x+c)),x)

[Out]

-1/96/d/a/(sin(d*x+c)-1)^3+11/128/d/a/(sin(d*x+c)-1)^2-69/128/a/d/(sin(d*x+c)-1)-325/256/a/d*ln(sin(d*x+c)-1)+
1/64/d/a/(1+sin(d*x+c))^4+5/48/d/a/(1+sin(d*x+c))^3+29/64/a/d/(1+sin(d*x+c))^2+2/a/d/(1+sin(d*x+c))-955/256*ln
(1+sin(d*x+c))/a/d-1/2/d/a/sin(d*x+c)^2+1/d/a/sin(d*x+c)+5*ln(sin(d*x+c))/a/d

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Maxima [A]  time = 1.08348, size = 293, normalized size = 1.26 \begin{align*} \frac{\frac{2 \,{\left (945 \, \sin \left (d x + c\right )^{8} - 15 \, \sin \left (d x + c\right )^{7} - 3480 \, \sin \left (d x + c\right )^{6} - 120 \, \sin \left (d x + c\right )^{5} + 4479 \, \sin \left (d x + c\right )^{4} + 319 \, \sin \left (d x + c\right )^{3} - 2192 \, \sin \left (d x + c\right )^{2} - 192 \, \sin \left (d x + c\right ) + 192\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 3 \, a \sin \left (d x + c\right )^{7} - 3 \, 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} - a \sin \left (d x + c\right )^{3} - a \sin \left (d x + c\right )^{2}} - \frac{2865 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac{975 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a} + \frac{3840 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{768 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/768*(2*(945*sin(d*x + c)^8 - 15*sin(d*x + c)^7 - 3480*sin(d*x + c)^6 - 120*sin(d*x + c)^5 + 4479*sin(d*x + c
)^4 + 319*sin(d*x + c)^3 - 2192*sin(d*x + c)^2 - 192*sin(d*x + c) + 192)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^8
- 3*a*sin(d*x + c)^7 - 3*a*sin(d*x + c)^6 + 3*a*sin(d*x + c)^5 + 3*a*sin(d*x + c)^4 - a*sin(d*x + c)^3 - a*sin
(d*x + c)^2) - 2865*log(sin(d*x + c) + 1)/a - 975*log(sin(d*x + c) - 1)/a + 3840*log(sin(d*x + c))/a)/d

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Fricas [A]  time = 1.72846, size = 830, normalized size = 3.58 \begin{align*} \frac{1890 \, \cos \left (d x + c\right )^{8} - 600 \, \cos \left (d x + c\right )^{6} - 582 \, \cos \left (d x + c\right )^{4} - 212 \, \cos \left (d x + c\right )^{2} + 3840 \,{\left (\cos \left (d x + c\right )^{8} - \cos \left (d x + c\right )^{6} +{\left (\cos \left (d x + c\right )^{8} - \cos \left (d x + c\right )^{6}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 2865 \,{\left (\cos \left (d x + c\right )^{8} - \cos \left (d x + c\right )^{6} +{\left (\cos \left (d x + c\right )^{8} - \cos \left (d x + c\right )^{6}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 975 \,{\left (\cos \left (d x + c\right )^{8} - \cos \left (d x + c\right )^{6} +{\left (\cos \left (d x + c\right )^{8} - \cos \left (d x + c\right )^{6}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (15 \, \cos \left (d x + c\right )^{6} - 165 \, \cos \left (d x + c\right )^{4} - 34 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 112}{768 \,{\left (a d \cos \left (d x + c\right )^{8} - a d \cos \left (d x + c\right )^{6} +{\left (a d \cos \left (d x + c\right )^{8} - a d \cos \left (d x + c\right )^{6}\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/768*(1890*cos(d*x + c)^8 - 600*cos(d*x + c)^6 - 582*cos(d*x + c)^4 - 212*cos(d*x + c)^2 + 3840*(cos(d*x + c)
^8 - cos(d*x + c)^6 + (cos(d*x + c)^8 - cos(d*x + c)^6)*sin(d*x + c))*log(1/2*sin(d*x + c)) - 2865*(cos(d*x +
c)^8 - cos(d*x + c)^6 + (cos(d*x + c)^8 - cos(d*x + c)^6)*sin(d*x + c))*log(sin(d*x + c) + 1) - 975*(cos(d*x +
c)^8 - cos(d*x + c)^6 + (cos(d*x + c)^8 - cos(d*x + c)^6)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 2*(15*cos(d*
x + c)^6 - 165*cos(d*x + c)^4 - 34*cos(d*x + c)^2 - 8)*sin(d*x + c) - 112)/(a*d*cos(d*x + c)^8 - a*d*cos(d*x +
c)^6 + (a*d*cos(d*x + c)^8 - a*d*cos(d*x + c)^6)*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(csc(d*x+c)**3*sec(d*x+c)**7/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.39307, size = 246, normalized size = 1.06 \begin{align*} -\frac{\frac{11460 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac{3900 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac{15360 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} + \frac{1536 \,{\left (15 \, \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )}}{a \sin \left (d x + c\right )^{2}} - \frac{2 \,{\left (3575 \, \sin \left (d x + c\right )^{3} - 11553 \, \sin \left (d x + c\right )^{2} + 12513 \, \sin \left (d x + c\right ) - 4551\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac{23875 \, \sin \left (d x + c\right )^{4} + 101644 \, \sin \left (d x + c\right )^{3} + 163074 \, \sin \left (d x + c\right )^{2} + 117036 \, \sin \left (d x + c\right ) + 31779}{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(csc(d*x+c)^3*sec(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/3072*(11460*log(abs(sin(d*x + c) + 1))/a + 3900*log(abs(sin(d*x + c) - 1))/a - 15360*log(abs(sin(d*x + c)))
/a + 1536*(15*sin(d*x + c)^2 - 2*sin(d*x + c) + 1)/(a*sin(d*x + c)^2) - 2*(3575*sin(d*x + c)^3 - 11553*sin(d*x
+ c)^2 + 12513*sin(d*x + c) - 4551)/(a*(sin(d*x + c) - 1)^3) - (23875*sin(d*x + c)^4 + 101644*sin(d*x + c)^3
+ 163074*sin(d*x + c)^2 + 117036*sin(d*x + c) + 31779)/(a*(sin(d*x + c) + 1)^4))/d