3.1337 \(\int \frac{\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=132 \[ \frac{b^4 \log (a+b \sin (c+d x))}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac{\csc ^2(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.202267, antiderivative size = 132, 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} \[ \frac{b^4 \log (a+b \sin (c+d x))}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac{\csc ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{b^3}{x^3 (a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{x^3 (a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^4 \operatorname{Subst}\left (\int \left (\frac{1}{2 b^4 (a+b) (b-x)}+\frac{1}{a b^2 x^3}-\frac{1}{a^2 b^2 x^2}+\frac{a^2+b^2}{a^3 b^4 x}+\frac{1}{a^3 (a-b) (a+b) (a+x)}+\frac{1}{2 b^4 (-a+b) (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{\log (1-\sin (c+d x))}{2 (a+b) d}+\frac{\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac{\log (1+\sin (c+d x))}{2 (a-b) d}+\frac{b^4 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 0.534019, size = 132, normalized size = 1. \[ \frac{b^4 \left (\frac{\csc (c+d x)}{a^2 b^3}+\frac{\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 b^4}+\frac{\log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )}-\frac{\csc ^2(c+d x)}{2 a b^4}-\frac{\log (1-\sin (c+d x))}{2 b^4 (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 b^4 (a-b)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.092, size = 144, normalized size = 1.1 \begin{align*}{\frac{{b}^{4}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) \left ( a-b \right ){a}^{3}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{d \left ( 2\,a+2\,b \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d \left ( 2\,a-2\,b \right ) }}-{\frac{1}{2\,da \left ( \sin \left ( dx+c \right ) \right ) ^{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)/(a+b*sin(d*x+c)),x)

[Out]

1/d*b^4/(a+b)/(a-b)/a^3*ln(a+b*sin(d*x+c))-1/d/(2*a+2*b)*ln(sin(d*x+c)-1)-1/d/(2*a-2*b)*ln(1+sin(d*x+c))-1/2/d
/a/sin(d*x+c)^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 = 0.993531, size = 154, normalized size = 1.17 \begin{align*} \frac{\frac{2 \, b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5} - a^{3} b^{2}} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac{2 \,{\left (a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} + \frac{2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.16071, size = 501, normalized size = 3.8 \begin{align*} \frac{a^{4} - a^{2} b^{2} + 2 \,{\left (b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \,{\left (a^{4} - b^{4} -{\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \sin \left (d x + c\right )\right ) +{\left (a^{4} + a^{3} b -{\left (a^{4} + a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{4} - a^{3} b -{\left (a^{4} - a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{5} - a^{3} b^{2}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.23968, size = 200, normalized size = 1.52 \begin{align*} \frac{\frac{2 \, b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b - a^{3} b^{3}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac{\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} + \frac{2 \,{\left (a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac{3 \, 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}}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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