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

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

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 132, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.53, size = 132, normalized size = 1.00 \[ \frac {b^4 \left (\frac {\csc (c+d x)}{a^2 b^3}+\frac {\log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )}+\frac {\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 b^4}-\frac {\csc ^2(c+d x)}{2 a b^4}-\frac {\log (1-\sin (c+d x))}{2 b^4 (a+b)}-\frac {\log (\sin (c+d x)+1)}{2 b^4 (a-b)}\right )}{d} \]

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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fricas [A]  time = 1.24, size = 224, normalized size = 1.70 \[ \frac {a^{4} - a^{2} b^{2} + 2 \, {\left (b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (a^{4} - b^{4} - {\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (a^{4} + a^{3} b - {\left (a^{4} + a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} - a^{3} b - {\left (a^{4} - a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{5} - a^{3} b^{2}\right )} d\right )}} \]

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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giac [A]  time = 0.23, size = 148, normalized size = 1.12 \[ \frac {\frac {2 \, b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b - a^{3} b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2} + 3 \, b^{2} \sin \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) + a^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

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

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maple [A]  time = 0.48, size = 144, normalized size = 1.09 \[ -\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{d \left (2 a +2 b \right )}+\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{3} \left (a +b \right ) \left (a -b \right )}-\frac {1}{2 d a \sin \left (d x +c \right )^{2}}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a d}+\frac {b^{2} \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {b}{d \,a^{2} \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{d \left (2 a -2 b \right )} \]

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

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maxima [A]  time = 0.31, size = 114, normalized size = 0.86 \[ \frac {\frac {2 \, b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5} - a^{3} b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} + \frac {2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

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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mupad [B]  time = 11.85, size = 125, normalized size = 0.95 \[ \frac {\ln \left (\sin \left (c+d\,x\right )\right )\,\left (a^2+b^2\right )}{a^3\,d}-\frac {\frac {1}{2\,a}-\frac {b\,\sin \left (c+d\,x\right )}{a^2}}{d\,{\sin \left (c+d\,x\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,\left (a+b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,\left (a-b\right )}+\frac {b^4\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^5-a^3\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*sin(c + d*x)^3*(a + b*sin(c + d*x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

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]

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

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