3.1367 \(\int \frac{\csc (c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=233 \[ \frac{b^6 \log (a+b \sin (c+d x))}{a d \left (a^2-b^2\right )^3}-\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac{5 a+7 b}{16 d (a+b)^2 (1-\sin (c+d x))}+\frac{5 a-7 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac{1}{16 d (a+b) (1-\sin (c+d x))^2}+\frac{1}{16 d (a-b) (\sin (c+d x)+1)^2}+\frac{\log (\sin (c+d x))}{a d} \]

[Out]

-((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + Log[Sin[c + d*x]]/(a*d) - ((8*a^2 - 21*a
*b + 15*b^2)*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*d) + (b^6*Log[a + b*Sin[c + d*x]])/(a*(a^2 - b^2)^3*d) + 1/(
16*(a + b)*d*(1 - Sin[c + d*x])^2) + (5*a + 7*b)/(16*(a + b)^2*d*(1 - Sin[c + d*x])) + 1/(16*(a - b)*d*(1 + Si
n[c + d*x])^2) + (5*a - 7*b)/(16*(a - b)^2*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.365385, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2837, 12, 894} \[ \frac{b^6 \log (a+b \sin (c+d x))}{a d \left (a^2-b^2\right )^3}-\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac{5 a+7 b}{16 d (a+b)^2 (1-\sin (c+d x))}+\frac{5 a-7 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac{1}{16 d (a+b) (1-\sin (c+d x))^2}+\frac{1}{16 d (a-b) (\sin (c+d x)+1)^2}+\frac{\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + Log[Sin[c + d*x]]/(a*d) - ((8*a^2 - 21*a
*b + 15*b^2)*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*d) + (b^6*Log[a + b*Sin[c + d*x]])/(a*(a^2 - b^2)^3*d) + 1/(
16*(a + b)*d*(1 - Sin[c + d*x])^2) + (5*a + 7*b)/(16*(a + b)^2*d*(1 - Sin[c + d*x])) + 1/(16*(a - b)*d*(1 + Si
n[c + d*x])^2) + (5*a - 7*b)/(16*(a - b)^2*d*(1 + Sin[c + d*x]))

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\csc (c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{b}{x (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^6 \operatorname{Subst}\left (\int \frac{1}{x (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^6 \operatorname{Subst}\left (\int \left (\frac{1}{8 b^4 (a+b) (b-x)^3}+\frac{5 a+7 b}{16 b^5 (a+b)^2 (b-x)^2}+\frac{8 a^2+21 a b+15 b^2}{16 b^6 (a+b)^3 (b-x)}+\frac{1}{a b^6 x}+\frac{1}{a (a-b)^3 (a+b)^3 (a+x)}+\frac{1}{8 b^4 (-a+b) (b+x)^3}+\frac{-5 a+7 b}{16 (a-b)^2 b^5 (b+x)^2}+\frac{8 a^2-21 a b+15 b^2}{16 b^6 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac{\log (\sin (c+d x))}{a d}-\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac{b^6 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )^3 d}+\frac{1}{16 (a+b) d (1-\sin (c+d x))^2}+\frac{5 a+7 b}{16 (a+b)^2 d (1-\sin (c+d x))}+\frac{1}{16 (a-b) d (1+\sin (c+d x))^2}+\frac{5 a-7 b}{16 (a-b)^2 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.81411, size = 220, normalized size = 0.94 \[ \frac{b^6 \left (-\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (c+d x))}{b^6 (a+b)^3}-\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{b^6 (a-b)^3}+\frac{-5 a-7 b}{b^6 (a+b)^2 (\sin (c+d x)-1)}+\frac{5 a-7 b}{b^6 (a-b)^2 (\sin (c+d x)+1)}+\frac{1}{b^6 (a+b) (\sin (c+d x)-1)^2}+\frac{1}{b^6 (a-b) (\sin (c+d x)+1)^2}+\frac{16 \log (\sin (c+d x))}{a b^6}+\frac{16 \log (a+b \sin (c+d x))}{a (a-b)^3 (a+b)^3}\right )}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^6*(-(((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(b^6*(a + b)^3)) + (16*Log[Sin[c + d*x]])/(a*b^6) -
((8*a^2 - 21*a*b + 15*b^2)*Log[1 + Sin[c + d*x]])/((a - b)^3*b^6) + (16*Log[a + b*Sin[c + d*x]])/(a*(a - b)^3*
(a + b)^3) + 1/(b^6*(a + b)*(-1 + Sin[c + d*x])^2) + (-5*a - 7*b)/(b^6*(a + b)^2*(-1 + Sin[c + d*x])) + 1/((a
- b)*b^6*(1 + Sin[c + d*x])^2) + (5*a - 7*b)/((a - b)^2*b^6*(1 + Sin[c + d*x]))))/(16*d)

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Maple [A]  time = 0.098, size = 321, normalized size = 1.4 \begin{align*}{\frac{{b}^{6}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}a}}+{\frac{1}{2\,d \left ( 8\,a+8\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{5\,a}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{7\,b}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ){a}^{2}}{2\,d \left ( a+b \right ) ^{3}}}-{\frac{21\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) ab}{16\,d \left ( a+b \right ) ^{3}}}-{\frac{15\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){b}^{2}}{16\,d \left ( a+b \right ) ^{3}}}+{\frac{1}{2\,d \left ( 8\,a-8\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5\,a}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{7\,b}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ){a}^{2}}{2\,d \left ( a-b \right ) ^{3}}}+{\frac{21\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) ab}{16\,d \left ( a-b \right ) ^{3}}}-{\frac{15\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){b}^{2}}{16\,d \left ( a-b \right ) ^{3}}}+{\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)*sec(d*x+c)^5/(a+b*sin(d*x+c)),x)

[Out]

1/d*b^6/(a+b)^3/(a-b)^3/a*ln(a+b*sin(d*x+c))+1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2-5/16/d/(a+b)^2/(sin(d*x+c)-1)*a-
7/16/d/(a+b)^2/(sin(d*x+c)-1)*b-1/2/d/(a+b)^3*ln(sin(d*x+c)-1)*a^2-21/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-15/16/
d/(a+b)^3*ln(sin(d*x+c)-1)*b^2+1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2+5/16/d/(a-b)^2/(1+sin(d*x+c))*a-7/16/d/(a-b)^2
/(1+sin(d*x+c))*b-1/2/d/(a-b)^3*ln(1+sin(d*x+c))*a^2+21/16/d/(a-b)^3*ln(1+sin(d*x+c))*a*b-15/16/d/(a-b)^3*ln(1
+sin(d*x+c))*b^2+ln(sin(d*x+c))/a/d

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Maxima [A]  time = 1.02392, size = 404, normalized size = 1.73 \begin{align*} \frac{\frac{16 \, b^{6} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac{{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{2 \,{\left ({\left (3 \, a^{2} b - 7 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{3} - 10 \, a b^{2} - 4 \,{\left (a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} -{\left (5 \, a^{2} b - 9 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} + \frac{16 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*b^6*log(b*sin(d*x + c) + a)/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - (8*a^2 - 21*a*b + 15*b^2)*log(sin
(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 21*a*b + 15*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*
b + 3*a*b^2 + b^3) + 2*((3*a^2*b - 7*b^3)*sin(d*x + c)^3 + 6*a^3 - 10*a*b^2 - 4*(a^3 - 2*a*b^2)*sin(d*x + c)^2
- (5*a^2*b - 9*b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 -
2*a^2*b^2 + b^4)*sin(d*x + c)^2) + 16*log(sin(d*x + c))/a)/d

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Fricas [A]  time = 9.77066, size = 790, normalized size = 3.39 \begin{align*} \frac{16 \, b^{6} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{6} - 8 \, a^{4} b^{2} + 4 \, a^{2} b^{4} + 16 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{4} \log \left (-\frac{1}{2} \, \sin \left (d x + c\right )\right ) -{\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (2 \, a^{5} b - 4 \, a^{3} b^{3} + 2 \, a b^{5} +{\left (3 \, a^{5} b - 10 \, a^{3} b^{3} + 7 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \,{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*b^6*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^6 - 8*a^4*b^2 + 4*a^2*b^4 + 16*(a^6 - 3*a^4*b^2 + 3*
a^2*b^4 - b^6)*cos(d*x + c)^4*log(-1/2*sin(d*x + c)) - (8*a^6 + 3*a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4
+ 15*a*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (8*a^6 - 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*a^2*b^4 -
15*a*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 8*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c)^2 - 2*(2*a^5*b
- 4*a^3*b^3 + 2*a*b^5 + (3*a^5*b - 10*a^3*b^3 + 7*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 3*a^5*b^2 + 3*a
^3*b^4 - a*b^6)*d*cos(d*x + c)^4)

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

[Out]

Timed out

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Giac [A]  time = 1.28314, size = 528, normalized size = 2.27 \begin{align*} \frac{\frac{16 \, b^{7} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}} - \frac{{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{16 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} + \frac{2 \,{\left (6 \, a^{5} \sin \left (d x + c\right )^{4} - 18 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} + 18 \, a b^{4} \sin \left (d x + c\right )^{4} + 3 \, a^{4} b \sin \left (d x + c\right )^{3} - 10 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} + 7 \, b^{5} \sin \left (d x + c\right )^{3} - 16 \, a^{5} \sin \left (d x + c\right )^{2} + 48 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 44 \, a b^{4} \sin \left (d x + c\right )^{2} - 5 \, a^{4} b \sin \left (d x + c\right ) + 14 \, a^{2} b^{3} \sin \left (d x + c\right ) - 9 \, b^{5} \sin \left (d x + c\right ) + 12 \, a^{5} - 34 \, a^{3} b^{2} + 28 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*b^7*log(abs(b*sin(d*x + c) + a))/(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7) - (8*a^2 - 21*a*b + 15*b^2)*
log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 21*a*b + 15*b^2)*log(abs(sin(d*x + c) -
1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 16*log(abs(sin(d*x + c)))/a + 2*(6*a^5*sin(d*x + c)^4 - 18*a^3*b^2*sin(d
*x + c)^4 + 18*a*b^4*sin(d*x + c)^4 + 3*a^4*b*sin(d*x + c)^3 - 10*a^2*b^3*sin(d*x + c)^3 + 7*b^5*sin(d*x + c)^
3 - 16*a^5*sin(d*x + c)^2 + 48*a^3*b^2*sin(d*x + c)^2 - 44*a*b^4*sin(d*x + c)^2 - 5*a^4*b*sin(d*x + c) + 14*a^
2*b^3*sin(d*x + c) - 9*b^5*sin(d*x + c) + 12*a^5 - 34*a^3*b^2 + 28*a*b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)
*(sin(d*x + c)^2 - 1)^2))/d