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

Optimal. Leaf size=168 \[ \frac {a b^2}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac {2 a b \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac {\csc ^2(c+d x) \left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^2}+\frac {(a-b) \log (1-\cos (c+d x))}{4 d (a+b)^3}-\frac {(a+b) \log (\cos (c+d x)+1)}{4 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 168, 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 = {3872, 2837, 12, 1647, 1629} \[ \frac {a b^2}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac {2 a b \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac {\csc ^2(c+d x) \left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^2}+\frac {(a-b) \log (1-\cos (c+d x))}{4 d (a+b)^3}-\frac {(a+b) \log (\cos (c+d x)+1)}{4 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^2)/((a^2 - b^2)^2*d*(b + a*Cos[c + d*x])) + ((2*a*b - (a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 -
b^2)^2*d) + ((a - b)*Log[1 - Cos[c + d*x]])/(4*(a + b)^3*d) - ((a + b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^3*d)
+ (2*a*b*(a^2 + b^2)*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^3*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 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 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]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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))^2} \, dx &=\int \frac {\cot ^2(c+d x) \csc (c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\frac {a^3 \operatorname {Subst}\left (\int \frac {x^2}{a^2 (-b+x)^2 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a \operatorname {Subst}\left (\int \frac {x^2}{(-b+x)^2 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {\left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^2 d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {a^2 b^2 \left (a^2+b^2\right )}{\left (a^2-b^2\right )^2}+\frac {2 a^2 b x}{a^2-b^2}+\frac {a^2 \left (a^2+b^2\right ) x^2}{\left (a^2-b^2\right )^2}}{(-b+x)^2 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{2 a d}\\ &=\frac {\left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^2 d}+\frac {\operatorname {Subst}\left (\int \left (\frac {a (a+b)}{2 (a-b)^3 (a-x)}+\frac {2 a^2 b^2}{(a-b)^2 (a+b)^2 (b-x)^2}-\frac {4 a^2 b \left (a^2+b^2\right )}{(a-b)^3 (a+b)^3 (b-x)}+\frac {a (a-b)}{2 (a+b)^3 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{2 a d}\\ &=\frac {a b^2}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^2 d}+\frac {(a-b) \log (1-\cos (c+d x))}{4 (a+b)^3 d}-\frac {(a+b) \log (1+\cos (c+d x))}{4 (a-b)^3 d}+\frac {2 a b \left (a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.31, size = 224, normalized size = 1.33 \[ \frac {\sec ^2(c+d x) (a \cos (c+d x)+b) \left (\frac {16 a b \left (a^2+b^2\right ) (a \cos (c+d x)+b) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^3}+\frac {8 a b^2}{(a-b)^2 (a+b)^2}+\frac {4 (a+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{(b-a)^3}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a+b)^2}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a-b)^2}+\frac {4 (a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{(a+b)^3}\right )}{8 d (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*((8*a*b^2)/((a - b)^2*(a + b)^2) - ((b + a*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/(a + b)^2 +
(4*(a + b)*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2]])/(-a + b)^3 + (16*a*b*(a^2 + b^2)*(b + a*Cos[c + d*x])*
Log[b + a*Cos[c + d*x]])/(a^2 - b^2)^3 + (4*(a - b)*(b + a*Cos[c + d*x])*Log[Sin[(c + d*x)/2]])/(a + b)^3 + ((
b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a - b)^2)*Sec[c + d*x]^2)/(8*d*(a + b*Sec[c + d*x])^2)

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fricas [B]  time = 0.76, size = 630, normalized size = 3.75 \[ -\frac {8 \, a^{3} b^{2} - 8 \, a b^{4} - 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 8 \, {\left (a^{3} b^{2} + a b^{4} - {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5} - {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5} - {\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{3} + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) - {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*a^3*b^2 - 8*a*b^4 - 2*(a^5 + 2*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*cos(d*x
+ c) + 8*(a^3*b^2 + a*b^4 - (a^4*b + a^2*b^3)*cos(d*x + c)^3 - (a^3*b^2 + a*b^4)*cos(d*x + c)^2 + (a^4*b + a^
2*b^3)*cos(d*x + c))*log(a*cos(d*x + c) + b) - (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5 - (a^5 + 4*a^4*b
+ 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*cos(d*x + c)^3 - (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*cos(d*x + c
)^2 + (a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^4*b - 4*a
^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5 - (a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*cos(d*x + c)^3 - (a^4*b -
4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*cos(d*x + c)^2 + (a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*cos(d
*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^3 + (a^6*b - 3*a^
4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^2 - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c) - (a^6*b - 3*
a^4*b^3 + 3*a^2*b^5 - b^7)*d)

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giac [B]  time = 0.32, size = 456, normalized size = 2.71 \[ \frac {\frac {2 \, {\left (a - b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {16 \, {\left (a^{3} b + a b^{3}\right )} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {a^{3} - a^{2} b - a b^{2} + b^{3} - \frac {8 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {3 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {3 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}} - \frac {\cos \left (d x + c\right ) - 1}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(a - b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 16*(a^3*b +
a*b^3)*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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (a^3 - a^2*b - a*b^2 + b^3 - 8*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
) + 8*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 3*a^2*b*(c
os(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 3*a*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - b^3*(cos(d*x + c
) - 1)^2/(cos(d*x + c) + 1)^2)/((a^4 - 2*a^2*b^2 + b^4)*(a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x
+ c) - 1)/(cos(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - b*(cos(d*x + c) - 1)^2/(cos(d*x +
c) + 1)^2)) - (cos(d*x + c) - 1)/((a^2 - 2*a*b + b^2)*(cos(d*x + c) + 1)))/d

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maple [A]  time = 0.65, size = 224, normalized size = 1.33 \[ \frac {b^{2} a}{d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 a^{3} b \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {2 a \,b^{3} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{4 d \left (a +b \right )^{2} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right ) a}{4 d \left (a +b \right )^{3}}-\frac {\ln \left (-1+\cos \left (d x +c \right )\right ) b}{4 d \left (a +b \right )^{3}}+\frac {1}{4 d \left (a -b \right )^{2} \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right ) a}{4 d \left (a -b \right )^{3}}-\frac {\ln \left (1+\cos \left (d x +c \right )\right ) b}{4 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.80, size = 274, normalized size = 1.63 \[ \frac {\frac {8 \, {\left (a^{3} b + a b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (a + b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (a - b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a b^{2} - {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )}}{a^{4} b - 2 \, a^{2} b^{3} + b^{5} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*(a^3*b + a*b^3)*log(a*cos(d*x + c) + b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (a + b)*log(cos(d*x + c)
+ 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (a - b)*log(cos(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(4*a*
b^2 - (a^3 + 3*a*b^2)*cos(d*x + c)^2 + (a^2*b - b^3)*cos(d*x + c))/(a^4*b - 2*a^2*b^3 + b^5 - (a^5 - 2*a^3*b^2
+ a*b^4)*cos(d*x + c)^3 - (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)^2 + (a^5 - 2*a^3*b^2 + a*b^4)*cos(d*x + c)))
/d

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mupad [B]  time = 1.47, size = 228, normalized size = 1.36 \[ \frac {\frac {2\,a\,b^2}{{\left (a^2-b^2\right )}^2}+\frac {b\,\cos \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}-\frac {{\cos \left (c+d\,x\right )}^2\,\left (a^3+3\,a\,b^2\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left (-a\,{\cos \left (c+d\,x\right )}^3-b\,{\cos \left (c+d\,x\right )}^2+a\,\cos \left (c+d\,x\right )+b\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,\left (\frac {b}{2\,{\left (a+b\right )}^3}-\frac {1}{4\,{\left (a+b\right )}^2}\right )}{d}+\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (2\,a^3\,b+2\,a\,b^3\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,\left (a+b\right )}{4\,d\,{\left (a-b\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^3*(a + b/cos(c + d*x))^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)**3/(a + b*sec(c + d*x))**2, x)

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