3.1351 \(\int \frac{\csc ^3(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=197 \[ -\frac{b^6 \log (a+b \sin (c+d x))}{a^3 d \left (a^2-b^2\right )^2}+\frac{\left (2 a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}+\frac{1}{4 d (a+b) (1-\sin (c+d x))}+\frac{1}{4 d (a-b) (\sin (c+d x)+1)}-\frac{(4 a+5 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}-\frac{(4 a-5 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2}-\frac{\csc ^2(c+d x)}{2 a d} \]

[Out]

(b*Csc[c + d*x])/(a^2*d) - Csc[c + d*x]^2/(2*a*d) - ((4*a + 5*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^2*d) + ((2*
a^2 + b^2)*Log[Sin[c + d*x]])/(a^3*d) - ((4*a - 5*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^2*d) - (b^6*Log[a + b*S
in[c + d*x]])/(a^3*(a^2 - b^2)^2*d) + 1/(4*(a + b)*d*(1 - Sin[c + d*x])) + 1/(4*(a - b)*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.312379, antiderivative size = 197, 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 = {2837, 12, 894} \[ -\frac{b^6 \log (a+b \sin (c+d x))}{a^3 d \left (a^2-b^2\right )^2}+\frac{\left (2 a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}+\frac{1}{4 d (a+b) (1-\sin (c+d x))}+\frac{1}{4 d (a-b) (\sin (c+d x)+1)}-\frac{(4 a+5 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}-\frac{(4 a-5 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2}-\frac{\csc ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Csc[c + d*x])/(a^2*d) - Csc[c + d*x]^2/(2*a*d) - ((4*a + 5*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^2*d) + ((2*
a^2 + b^2)*Log[Sin[c + d*x]])/(a^3*d) - ((4*a - 5*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^2*d) - (b^6*Log[a + b*S
in[c + d*x]])/(a^3*(a^2 - b^2)^2*d) + 1/(4*(a + b)*d*(1 - Sin[c + d*x])) + 1/(4*(a - b)*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 ^3(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{b^3}{x^3 (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^6 \operatorname{Subst}\left (\int \frac{1}{x^3 (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^6 \operatorname{Subst}\left (\int \left (\frac{1}{4 b^5 (a+b) (b-x)^2}+\frac{4 a+5 b}{4 b^6 (a+b)^2 (b-x)}+\frac{1}{a b^4 x^3}-\frac{1}{a^2 b^4 x^2}+\frac{2 a^2+b^2}{a^3 b^6 x}-\frac{1}{a^3 (a-b)^2 (a+b)^2 (a+x)}-\frac{1}{4 (a-b) b^5 (b+x)^2}+\frac{-4 a+5 b}{4 (a-b)^2 b^6 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b \csc (c+d x)}{a^2 d}-\frac{\csc ^2(c+d x)}{2 a d}-\frac{(4 a+5 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac{\left (2 a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac{(4 a-5 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac{b^6 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )^2 d}+\frac{1}{4 (a+b) d (1-\sin (c+d x))}+\frac{1}{4 (a-b) d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.4216, size = 168, normalized size = 0.85 \[ -\frac{\frac{4 b^6 \log (a+b \sin (c+d x))}{a^3 (a-b)^2 (a+b)^2}-\frac{4 \left (2 a^2+b^2\right ) \log (\sin (c+d x))}{a^3}-\frac{4 b \csc (c+d x)}{a^2}+\frac{1}{(a+b) (\sin (c+d x)-1)}-\frac{1}{(a-b) (\sin (c+d x)+1)}+\frac{(4 a+5 b) \log (1-\sin (c+d x))}{(a+b)^2}+\frac{(4 a-5 b) \log (\sin (c+d x)+1)}{(a-b)^2}+\frac{2 \csc ^2(c+d x)}{a}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-4*b*Csc[c + d*x])/a^2 + (2*Csc[c + d*x]^2)/a + ((4*a + 5*b)*Log[1 - Sin[c + d*x]])/(a + b)^2 - (4*(2*a^2 +
b^2)*Log[Sin[c + d*x]])/a^3 + ((4*a - 5*b)*Log[1 + Sin[c + d*x]])/(a - b)^2 + (4*b^6*Log[a + b*Sin[c + d*x]])
/(a^3*(a - b)^2*(a + b)^2) + 1/((a + b)*(-1 + Sin[c + d*x])) - 1/((a - b)*(1 + Sin[c + d*x])))/(4*d)

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Maple [A]  time = 0.111, size = 231, normalized size = 1.2 \begin{align*} -{\frac{{b}^{6}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}{a}^{3}}}-{\frac{1}{d \left ( 4\,a+4\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) a}{d \left ( a+b \right ) ^{2}}}-{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{4\,d \left ( a+b \right ) ^{2}}}+{\frac{1}{d \left ( 4\,a-4\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{a\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{ \left ( a-b \right ) ^{2}d}}+{\frac{5\,b\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{4\, \left ( a-b \right ) ^{2}d}}-{\frac{1}{2\,da \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}}+{\frac{{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{{a}^{3}d}}+{\frac{b}{d{a}^{2}\sin \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*b^6/(a+b)^2/(a-b)^2/a^3*ln(a+b*sin(d*x+c))-1/d/(4*a+4*b)/(sin(d*x+c)-1)-1/d/(a+b)^2*ln(sin(d*x+c)-1)*a-5/
4/d/(a+b)^2*ln(sin(d*x+c)-1)*b+1/d/(4*a-4*b)/(1+sin(d*x+c))-a*ln(1+sin(d*x+c))/(a-b)^2/d+5/4*b*ln(1+sin(d*x+c)
)/(a-b)^2/d-1/2/d/a/sin(d*x+c)^2+2*ln(sin(d*x+c))/a/d+b^2*ln(sin(d*x+c))/a^3/d+1/d/a^2*b/sin(d*x+c)

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Maxima [A]  time = 1.02816, size = 329, normalized size = 1.67 \begin{align*} -\frac{\frac{4 \, b^{6} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}} + \frac{{\left (4 \, a - 5 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{{\left (4 \, a + 5 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{2 \,{\left ({\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + a^{3} - a b^{2} -{\left (2 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{2} - 2 \,{\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )^{4} -{\left (a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}} - \frac{4 \,{\left (2 \, a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*b^6*log(b*sin(d*x + c) + a)/(a^7 - 2*a^5*b^2 + a^3*b^4) + (4*a - 5*b)*log(sin(d*x + c) + 1)/(a^2 - 2*a
*b + b^2) + (4*a + 5*b)*log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2) - 2*((3*a^2*b - 2*b^3)*sin(d*x + c)^3 + a^3
- a*b^2 - (2*a^3 - a*b^2)*sin(d*x + c)^2 - 2*(a^2*b - b^3)*sin(d*x + c))/((a^4 - a^2*b^2)*sin(d*x + c)^4 - (a^
4 - a^2*b^2)*sin(d*x + c)^2) - 4*(2*a^2 + b^2)*log(sin(d*x + c))/a^3)/d

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Fricas [B]  time = 8.90208, size = 973, normalized size = 4.94 \begin{align*} -\frac{2 \, a^{6} - 2 \, a^{4} b^{2} - 2 \,{\left (2 \, a^{6} - 3 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (b^{6} \cos \left (d x + c\right )^{4} - b^{6} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \,{\left ({\left (2 \, a^{6} - 3 \, a^{4} b^{2} + b^{6}\right )} \cos \left (d x + c\right )^{4} -{\left (2 \, a^{6} - 3 \, a^{4} b^{2} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \sin \left (d x + c\right )\right ) +{\left ({\left (4 \, a^{6} + 3 \, a^{5} b - 6 \, a^{4} b^{2} - 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{4} -{\left (4 \, a^{6} + 3 \, a^{5} b - 6 \, a^{4} b^{2} - 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (4 \, a^{6} - 3 \, a^{5} b - 6 \, a^{4} b^{2} + 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{4} -{\left (4 \, a^{6} - 3 \, a^{5} b - 6 \, a^{4} b^{2} + 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{5} b - a^{3} b^{3} -{\left (3 \, a^{5} b - 5 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right )^{4} -{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.29979, size = 371, normalized size = 1.88 \begin{align*} -\frac{\frac{4 \, b^{7} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}} + \frac{{\left (4 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{{\left (4 \, a + 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{2 \,{\left (2 \, a^{3} \sin \left (d x + c\right )^{2} - 3 \, a b^{2} \sin \left (d x + c\right )^{2} + a^{2} b \sin \left (d x + c\right ) - b^{3} \sin \left (d x + c\right ) - 3 \, a^{3} + 4 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}} - \frac{4 \,{\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} + \frac{2 \,{\left (6 \, a^{2} \sin \left (d x + c\right )^{2} + 3 \, b^{2} \sin \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) + a^{2}\right )}}{a^{3} \sin \left (d x + c\right )^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*b^7*log(abs(b*sin(d*x + c) + a))/(a^7*b - 2*a^5*b^3 + a^3*b^5) + (4*a - 5*b)*log(abs(sin(d*x + c) + 1)
)/(a^2 - 2*a*b + b^2) + (4*a + 5*b)*log(abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) - 2*(2*a^3*sin(d*x + c)^2 -
3*a*b^2*sin(d*x + c)^2 + a^2*b*sin(d*x + c) - b^3*sin(d*x + c) - 3*a^3 + 4*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(s
in(d*x + c)^2 - 1)) - 4*(2*a^2 + b^2)*log(abs(sin(d*x + c)))/a^3 + 2*(6*a^2*sin(d*x + c)^2 + 3*b^2*sin(d*x + c
)^2 - 2*a*b*sin(d*x + c) + a^2)/(a^3*sin(d*x + c)^2))/d