3.1350 \(\int \frac{\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=171 \[ \frac{b^5 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )^2}-\frac{b \log (\sin (c+d x))}{a^2 d}+\frac{1}{4 d (a+b) (1-\sin (c+d x))}-\frac{1}{4 d (a-b) (\sin (c+d x)+1)}-\frac{(3 a+4 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac{(3 a-4 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2}-\frac{\csc (c+d x)}{a d} \]

[Out]

-(Csc[c + d*x]/(a*d)) - ((3*a + 4*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^2*d) - (b*Log[Sin[c + d*x]])/(a^2*d) +
((3*a - 4*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^2*d) + (b^5*Log[a + b*Sin[c + d*x]])/(a^2*(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 = 0.269614, antiderivative size = 171, 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} \[ \frac{b^5 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )^2}-\frac{b \log (\sin (c+d x))}{a^2 d}+\frac{1}{4 d (a+b) (1-\sin (c+d x))}-\frac{1}{4 d (a-b) (\sin (c+d x)+1)}-\frac{(3 a+4 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac{(3 a-4 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2}-\frac{\csc (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

-(Csc[c + d*x]/(a*d)) - ((3*a + 4*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^2*d) - (b*Log[Sin[c + d*x]])/(a^2*d) +
((3*a - 4*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^2*d) + (b^5*Log[a + b*Sin[c + d*x]])/(a^2*(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 ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{b^2}{x^2 (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^5 \operatorname{Subst}\left (\int \frac{1}{x^2 (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^5 \operatorname{Subst}\left (\int \left (\frac{1}{4 b^4 (a+b) (b-x)^2}+\frac{3 a+4 b}{4 b^5 (a+b)^2 (b-x)}+\frac{1}{a b^4 x^2}-\frac{1}{a^2 b^4 x}+\frac{1}{a^2 (a-b)^2 (a+b)^2 (a+x)}+\frac{1}{4 (a-b) b^4 (b+x)^2}+\frac{3 a-4 b}{4 (a-b)^2 b^5 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\csc (c+d x)}{a d}-\frac{(3 a+4 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}-\frac{b \log (\sin (c+d x))}{a^2 d}+\frac{(3 a-4 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac{b^5 \log (a+b \sin (c+d x))}{a^2 \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 = 0.747822, size = 174, normalized size = 1.02 \[ -\frac{\csc (c+d x) (a+b \sin (c+d x)) \left (-\frac{4 b^5 \log (a+b \sin (c+d x))}{a^2 (a-b)^2 (a+b)^2}+\frac{4 b \log (\sin (c+d x))}{a^2}+\frac{1}{(a+b) (\sin (c+d x)-1)}+\frac{1}{(a-b) (\sin (c+d x)+1)}+\frac{(3 a+4 b) \log (1-\sin (c+d x))}{(a+b)^2}-\frac{(3 a-4 b) \log (\sin (c+d x)+1)}{(a-b)^2}+\frac{4 \csc (c+d x)}{a}\right )}{4 d (a \csc (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.099, size = 199, normalized size = 1.2 \begin{align*}{\frac{{b}^{5}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}{a}^{2}}}-{\frac{1}{d \left ( 4\,a+4\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) a}{4\,d \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{d \left ( a+b \right ) ^{2}}}-{\frac{1}{d \left ( 4\,a-4\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{3\,a\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{4\, \left ( a-b \right ) ^{2}d}}-{\frac{b\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{ \left ( a-b \right ) ^{2}d}}-{\frac{1}{da\sin \left ( dx+c \right ) }}-{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.0573, size = 270, normalized size = 1.58 \begin{align*} \frac{\frac{4 \, b^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}} + \frac{{\left (3 \, a - 4 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{{\left (3 \, a + 4 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{2 \,{\left (a b \sin \left (d x + c\right ) -{\left (3 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} + 2 \, a^{2} - 2 \, b^{2}\right )}}{{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} -{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )} - \frac{4 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 6.0214, size = 672, normalized size = 3.93 \begin{align*} \frac{4 \, b^{5} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} - 4 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) +{\left (3 \, a^{5} + 2 \, a^{4} b - 5 \, a^{3} b^{2} - 4 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) -{\left (3 \, a^{5} - 2 \, a^{4} b - 5 \, a^{3} b^{2} + 4 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 2 \,{\left (3 \, a^{5} - 5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.29501, size = 377, normalized size = 2.2 \begin{align*} \frac{\frac{12 \, b^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}} + \frac{3 \,{\left (3 \, a - 4 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{3 \,{\left (3 \, a + 4 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{12 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac{2 \,{\left (2 \, b^{5} \sin \left (d x + c\right )^{3} - 9 \, a^{5} \sin \left (d x + c\right )^{2} + 15 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 6 \, a b^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} b \sin \left (d x + c\right ) - 3 \, a^{2} b^{3} \sin \left (d x + c\right ) - 2 \, b^{5} \sin \left (d x + c\right ) + 6 \, a^{5} - 12 \, a^{3} b^{2} + 6 \, a b^{4}\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/12*(12*b^6*log(abs(b*sin(d*x + c) + a))/(a^6*b - 2*a^4*b^3 + a^2*b^5) + 3*(3*a - 4*b)*log(abs(sin(d*x + c) +
1))/(a^2 - 2*a*b + b^2) - 3*(3*a + 4*b)*log(abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) - 12*b*log(abs(sin(d*x
+ c)))/a^2 + 2*(2*b^5*sin(d*x + c)^3 - 9*a^5*sin(d*x + c)^2 + 15*a^3*b^2*sin(d*x + c)^2 - 6*a*b^4*sin(d*x + c
)^2 + 3*a^4*b*sin(d*x + c) - 3*a^2*b^3*sin(d*x + c) - 2*b^5*sin(d*x + c) + 6*a^5 - 12*a^3*b^2 + 6*a*b^4)/((a^6
- 2*a^4*b^2 + a^2*b^4)*(sin(d*x + c)^3 - sin(d*x + c))))/d