3.3.1 \(\int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx\) [201]

Optimal. Leaf size=116 \[ \frac {(b-a \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {a \log (1-\cos (c+d x))}{4 (a+b)^2 d}-\frac {a \log (1+\cos (c+d x))}{4 (a-b)^2 d}+\frac {a^2 b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^2 d} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3957, 2916, 12, 837, 815} \begin {gather*} \frac {a^2 b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^2}+\frac {\csc ^2(c+d x) (b-a \cos (c+d x))}{2 d \left (a^2-b^2\right )}+\frac {a \log (1-\cos (c+d x))}{4 d (a+b)^2}-\frac {a \log (\cos (c+d x)+1)}{4 d (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b - a*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)*d) + (a*Log[1 - Cos[c + d*x]])/(4*(a + b)^2*d) - (a*Log[1
+ Cos[c + d*x]])/(4*(a - b)^2*d) + (a^2*b*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^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 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 2916

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/b)*x)^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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac {\cot (c+d x) \csc ^2(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac {a^3 \text {Subst}\left (\int \frac {x}{a (-b+x) \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^2 \text {Subst}\left (\int \frac {x}{(-b+x) \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {(b-a \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {\text {Subst}\left (\int \frac {a^2 b+a^2 x}{(-b+x) \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=\frac {(b-a \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {\text {Subst}\left (\int \left (\frac {a (a+b)}{2 (a-b) (a-x)}-\frac {2 a^2 b}{(a-b) (a+b) (b-x)}+\frac {a (a-b)}{2 (a+b) (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=\frac {(b-a \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {a \log (1-\cos (c+d x))}{4 (a+b)^2 d}-\frac {a \log (1+\cos (c+d x))}{4 (a-b)^2 d}+\frac {a^2 b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 123, normalized size = 1.06 \begin {gather*} \frac {-(a-b)^2 (a+b) \csc ^2\left (\frac {1}{2} (c+d x)\right )-4 a \left ((a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 a b \log (b+a \cos (c+d x))-(a-b)^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+(a-b) (a+b)^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 (a-b)^2 (a+b)^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a - b)^2*(a + b)*Csc[(c + d*x)/2]^2) - 4*a*((a + b)^2*Log[Cos[(c + d*x)/2]] - 2*a*b*Log[b + a*Cos[c + d*x]
] - (a - b)^2*Log[Sin[(c + d*x)/2]]) + (a - b)*(a + b)^2*Sec[(c + d*x)/2]^2)/(8*(a - b)^2*(a + b)^2*d)

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Maple [A]
time = 0.15, size = 110, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\frac {1}{\left (4 a -4 b \right ) \left (1+\cos \left (d x +c \right )\right )}-\frac {a \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (d x +c \right )\right )}+\frac {a \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{2}}+\frac {b \,a^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\)	\(110\)
default	\(\frac {\frac {1}{\left (4 a -4 b \right ) \left (1+\cos \left (d x +c \right )\right )}-\frac {a \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (d x +c \right )\right )}+\frac {a \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{2}}+\frac {b \,a^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\)	\(110\)
norman	\(\frac {-\frac {1}{8 d \left (a +b \right )}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \left (a -b \right )}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \,a^{2} \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \left (a^{2}+2 b a +b^{2}\right )}\)	\(137\)
risch	\(-\frac {i a x}{2 \left (a^{2}+2 b a +b^{2}\right )}-\frac {i a c}{2 d \left (a^{2}+2 b a +b^{2}\right )}+\frac {i a x}{2 a^{2}-4 b a +2 b^{2}}+\frac {i a c}{2 d \left (a^{2}-2 b a +b^{2}\right )}-\frac {2 i b \,a^{2} x}{a^{4}-2 b^{2} a^{2}+b^{4}}-\frac {2 i b \,a^{2} c}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {a \,{\mathrm e}^{3 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )} a}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \left (a^{2}+2 b a +b^{2}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \left (a^{2}-2 b a +b^{2}\right )}+\frac {b \,a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}\)	\(311\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 132, normalized size = 1.14 \begin {gather*} \frac {\frac {4 \, a^{2} b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {a \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {a \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a \cos \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.50, size = 216, normalized size = 1.86 \begin {gather*} -\frac {2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) - 4 \, {\left (a^{2} b \cos \left (d x + c\right )^{2} - a^{2} b\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2} - {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.48, size = 202, normalized size = 1.74 \begin {gather*} \frac {\frac {8 \, a^{2} b \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, a \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {{\left (a + b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {\cos \left (d x + c\right ) - 1}{{\left (a - b\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*a^2*b*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
)))/(a^4 - 2*a^2*b^2 + b^4) + 2*a*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^2 + 2*a*b + b^2) + (a +
b - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/((a^2 + 2*a*b + b^2)*(cos(d*x + c) - 1)) -
(cos(d*x + c) - 1)/((a - b)*(cos(d*x + c) + 1)))/d

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Mupad [B]
time = 0.30, size = 133, normalized size = 1.15 \begin {gather*} \frac {a\,\ln \left (\cos \left (c+d\,x\right )-1\right )}{4\,d\,{\left (a+b\right )}^2}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (\frac {a}{4\,{\left (a+b\right )}^2}-\frac {a}{4\,{\left (a-b\right )}^2}\right )}{d}-\frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^2-1\right )}-\frac {a\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{4\,d\,{\left (a-b\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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