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

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 197, normalized size of antiderivative = 1.00, 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 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 ^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*}

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Mathematica [A]  time = 1.44, size = 168, normalized size = 0.85 \[ -\frac {\frac {4 b^6 \log (a+b \sin (c+d x))}{a^3 (a-b)^2 (a+b)^2}-\frac {4 b \csc (c+d x)}{a^2}-\frac {4 \left (2 a^2+b^2\right ) \log (\sin (c+d x))}{a^3}+\frac {1}{(a+b) (\sin (c+d x)-1)}-\frac {1}{(a-b) (\sin (c+d x)+1)}+\frac {(4 a+5 b) \log (1-\sin (c+d x))}{(a+b)^2}+\frac {(4 a-5 b) \log (\sin (c+d x)+1)}{(a-b)^2}+\frac {2 \csc ^2(c+d x)}{a}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((-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])))/d

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fricas [B]  time = 3.05, size = 440, normalized size = 2.23 \[ -\frac {2 \, a^{6} - 2 \, a^{4} b^{2} - 2 \, {\left (2 \, a^{6} - 3 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (b^{6} \cos \left (d x + c\right )^{4} - b^{6} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left ({\left (2 \, a^{6} - 3 \, a^{4} b^{2} + b^{6}\right )} \cos \left (d x + c\right )^{4} - {\left (2 \, a^{6} - 3 \, a^{4} b^{2} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left ({\left (4 \, a^{6} + 3 \, a^{5} b - 6 \, a^{4} b^{2} - 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (4 \, a^{6} + 3 \, a^{5} b - 6 \, a^{4} b^{2} - 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (4 \, a^{6} - 3 \, a^{5} b - 6 \, a^{4} b^{2} + 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (4 \, a^{6} - 3 \, a^{5} b - 6 \, a^{4} b^{2} + 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{5} b - a^{3} b^{3} - {\left (3 \, a^{5} b - 5 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right )^{4} - {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right )^{2}\right )}} \]

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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giac [A]  time = 0.27, size = 275, normalized size = 1.40 \[ -\frac {\frac {4 \, b^{7} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}} + \frac {{\left (4 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (4 \, a + 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (2 \, a^{3} \sin \left (d x + c\right )^{2} - 3 \, a b^{2} \sin \left (d x + c\right )^{2} + a^{2} b \sin \left (d x + c\right ) - b^{3} \sin \left (d x + c\right ) - 3 \, a^{3} + 4 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}} - \frac {4 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} + \frac {2 \, {\left (6 \, a^{2} \sin \left (d x + c\right )^{2} + 3 \, b^{2} \sin \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) + a^{2}\right )}}{a^{3} \sin \left (d x + c\right )^{2}}}{4 \, d} \]

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

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maple [A]  time = 0.54, size = 231, normalized size = 1.17 \[ -\frac {1}{d \left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a}{d \left (a +b \right )^{2}}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right ) b}{4 d \left (a +b \right )^{2}}-\frac {b^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{3} \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{2 d a \sin \left (d x +c \right )^{2}}+\frac {2 \ln \left (\sin \left (d x +c \right )\right )}{a d}+\frac {b^{2} \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {b}{d \,a^{2} \sin \left (d x +c \right )}+\frac {1}{d \left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}-\frac {a \ln \left (1+\sin \left (d x +c \right )\right )}{\left (a -b \right )^{2} d}+\frac {5 b \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2} d} \]

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/(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/a^3*b^6/(a+b
)^2/(a-b)^2*ln(a+b*sin(d*x+c))-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/si
n(d*x+c)+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

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maxima [A]  time = 0.32, size = 244, normalized size = 1.24 \[ -\frac {\frac {4 \, b^{6} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}} + \frac {{\left (4 \, a - 5 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (4 \, a + 5 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left ({\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + a^{3} - a b^{2} - {\left (2 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{2} - 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )^{4} - {\left (a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}} - \frac {4 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{4 \, d} \]

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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mupad [B]  time = 12.48, size = 240, normalized size = 1.22 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b}{4\,{\left (a-b\right )}^2}-\frac {1}{a-b}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{a+b}\right )}{d}-\frac {\frac {1}{2\,a}-\frac {b\,\sin \left (c+d\,x\right )}{a^2}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (2\,a^2-b^2\right )}{2\,a\,\left (a^2-b^2\right )}+\frac {b\,{\sin \left (c+d\,x\right )}^3\,\left (3\,a^2-2\,b^2\right )}{2\,a^2\,\left (a^2-b^2\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^2-{\sin \left (c+d\,x\right )}^4\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )\right )\,\left (2\,a^2+b^2\right )}{a^3\,d}-\frac {b^6\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^7-2\,a^5\,b^2+a^3\,b^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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