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

Optimal. Leaf size=229 \[ \frac {b^2 \left (3 a^2+b^2\right )}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}+\frac {\csc ^2(c+d x) \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac {b^3}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {b \left (3 a^4+8 a^2 b^2+b^4\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}+\frac {(a-2 b) \log (1-\cos (c+d x))}{4 d (a+b)^4}-\frac {(a+2 b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

-1/2*b^3/(a^2-b^2)^2/d/(b+a*cos(d*x+c))^2+b^2*(3*a^2+b^2)/(a^2-b^2)^3/d/(b+a*cos(d*x+c))+1/2*(b*(3*a^2+b^2)-a*
(a^2+3*b^2)*cos(d*x+c))*csc(d*x+c)^2/(a^2-b^2)^3/d+1/4*(a-2*b)*ln(1-cos(d*x+c))/(a+b)^4/d-1/4*(a+2*b)*ln(1+cos
(d*x+c))/(a-b)^4/d+b*(3*a^4+8*a^2*b^2+b^4)*ln(b+a*cos(d*x+c))/(a^2-b^2)^4/d

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Rubi [A]  time = 0.51, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3872, 2721, 1647, 1629} \[ -\frac {b^3}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {b^2 \left (3 a^2+b^2\right )}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}+\frac {b \left (8 a^2 b^2+3 a^4+b^4\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}+\frac {\csc ^2(c+d x) \left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}+\frac {(a-2 b) \log (1-\cos (c+d x))}{4 d (a+b)^4}-\frac {(a+2 b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-b^3/(2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) + (b^2*(3*a^2 + b^2))/((a^2 - b^2)^3*d*(b + a*Cos[c + d*x])) +
((b*(3*a^2 + b^2) - a*(a^2 + 3*b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)^3*d) + ((a - 2*b)*Log[1 - Co
s[c + d*x]])/(4*(a + b)^4*d) - ((a + 2*b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) + (b*(3*a^4 + 8*a^2*b^2 + b^4
)*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^4*d)

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 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

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))^3} \, dx &=-\int \frac {\cot ^3(c+d x)}{(-b-a \cos (c+d x))^3} \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^3}{(-b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {\left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}+\frac {\operatorname {Subst}\left (\int \frac {\frac {a^4 b^3 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3}-\frac {a^2 b^2 \left (3 a^4+3 a^2 b^2-2 b^4\right ) x}{\left (a^2-b^2\right )^3}+\frac {a^4 b \left (3 a^2-7 b^2\right ) x^2}{\left (a^2-b^2\right )^3}+\frac {a^4 \left (a^2+3 b^2\right ) x^3}{\left (a^2-b^2\right )^3}}{(-b+x)^3 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{2 a^2 d}\\ &=\frac {\left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}+\frac {\operatorname {Subst}\left (\int \left (\frac {a^2 (a+2 b)}{2 (a-b)^4 (a-x)}-\frac {2 a^2 b^3}{\left (a^2-b^2\right )^2 (b-x)^3}+\frac {2 a^2 b^2 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3 (b-x)^2}-\frac {2 a^2 b \left (3 a^4+8 a^2 b^2+b^4\right )}{\left (a^2-b^2\right )^4 (b-x)}+\frac {a^2 (a-2 b)}{2 (a+b)^4 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{2 a^2 d}\\ &=-\frac {b^3}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}+\frac {b^2 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac {\left (b \left (3 a^2+b^2\right )-a \left (a^2+3 b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}+\frac {(a-2 b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}-\frac {(a+2 b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}+\frac {b \left (3 a^4+8 a^2 b^2+b^4\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d}\\ \end {align*}

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Mathematica [C]  time = 6.34, size = 332, normalized size = 1.45 \[ -\frac {b^2 \left (3 a^2+b^2\right )}{d (b-a)^3 (a+b)^3 (a \cos (c+d x)+b)}-\frac {2 i \left (3 a^4 b+8 a^2 b^3+b^5\right ) (c+d x)}{d (a-b)^4 (a+b)^4}+\frac {\left (3 a^4 b+8 a^2 b^3+b^5\right ) \log (a \cos (c+d x)+b)}{d \left (b^2-a^2\right )^4}-\frac {b^3}{2 d (b-a)^2 (a+b)^2 (a \cos (c+d x)+b)^2}-\frac {i (-a-2 b) \tan ^{-1}(\tan (c+d x))}{2 d (b-a)^4}-\frac {i (a-2 b) \tan ^{-1}(\tan (c+d x))}{2 d (a+b)^4}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d (a+b)^3}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d (b-a)^3}+\frac {(a-2 b) \log \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )}{4 d (a+b)^4}+\frac {(-a-2 b) \log \left (\cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{4 d (b-a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*(3*a^4*b + 8*a^2*b^3 + b^5)*(c + d*x))/((a - b)^4*(a + b)^4*d) - ((I/2)*(-a - 2*b)*ArcTan[Tan[c + d*x]
])/((-a + b)^4*d) - ((I/2)*(a - 2*b)*ArcTan[Tan[c + d*x]])/((a + b)^4*d) - b^3/(2*(-a + b)^2*(a + b)^2*d*(b +
a*Cos[c + d*x])^2) - (b^2*(3*a^2 + b^2))/((-a + b)^3*(a + b)^3*d*(b + a*Cos[c + d*x])) - Csc[(c + d*x)/2]^2/(8
*(a + b)^3*d) + ((-a - 2*b)*Log[Cos[(c + d*x)/2]^2])/(4*(-a + b)^4*d) + ((3*a^4*b + 8*a^2*b^3 + b^5)*Log[b + a
*Cos[c + d*x]])/((-a^2 + b^2)^4*d) + ((a - 2*b)*Log[Sin[(c + d*x)/2]^2])/(4*(a + b)^4*d) - Sec[(c + d*x)/2]^2/
(8*(-a + b)^3*d)

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fricas [B]  time = 0.89, size = 1071, normalized size = 4.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(16*a^4*b^3 - 8*a^2*b^5 - 8*b^7 - 2*(a^7 + 8*a^5*b^2 - 7*a^3*b^4 - 2*a*b^6)*cos(d*x + c)^3 + 2*(a^6*b - 1
1*a^4*b^3 + 7*a^2*b^5 + 3*b^7)*cos(d*x + c)^2 + 2*(11*a^5*b^2 - 10*a^3*b^4 - a*b^6)*cos(d*x + c) + 4*(3*a^4*b^
3 + 8*a^2*b^5 + b^7 - (3*a^6*b + 8*a^4*b^3 + a^2*b^5)*cos(d*x + c)^4 - 2*(3*a^5*b^2 + 8*a^3*b^4 + a*b^6)*cos(d
*x + c)^3 + (3*a^6*b + 5*a^4*b^3 - 7*a^2*b^5 - b^7)*cos(d*x + c)^2 + 2*(3*a^5*b^2 + 8*a^3*b^4 + a*b^6)*cos(d*x
+ c))*log(a*cos(d*x + c) + b) - (a^5*b^2 + 6*a^4*b^3 + 14*a^3*b^4 + 16*a^2*b^5 + 9*a*b^6 + 2*b^7 - (a^7 + 6*a
^6*b + 14*a^5*b^2 + 16*a^4*b^3 + 9*a^3*b^4 + 2*a^2*b^5)*cos(d*x + c)^4 - 2*(a^6*b + 6*a^5*b^2 + 14*a^4*b^3 + 1
6*a^3*b^4 + 9*a^2*b^5 + 2*a*b^6)*cos(d*x + c)^3 + (a^7 + 6*a^6*b + 13*a^5*b^2 + 10*a^4*b^3 - 5*a^3*b^4 - 14*a^
2*b^5 - 9*a*b^6 - 2*b^7)*cos(d*x + c)^2 + 2*(a^6*b + 6*a^5*b^2 + 14*a^4*b^3 + 16*a^3*b^4 + 9*a^2*b^5 + 2*a*b^6
)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^5*b^2 - 6*a^4*b^3 + 14*a^3*b^4 - 16*a^2*b^5 + 9*a*b^6 - 2*b^7
- (a^7 - 6*a^6*b + 14*a^5*b^2 - 16*a^4*b^3 + 9*a^3*b^4 - 2*a^2*b^5)*cos(d*x + c)^4 - 2*(a^6*b - 6*a^5*b^2 + 1
4*a^4*b^3 - 16*a^3*b^4 + 9*a^2*b^5 - 2*a*b^6)*cos(d*x + c)^3 + (a^7 - 6*a^6*b + 13*a^5*b^2 - 10*a^4*b^3 - 5*a^
3*b^4 + 14*a^2*b^5 - 9*a*b^6 + 2*b^7)*cos(d*x + c)^2 + 2*(a^6*b - 6*a^5*b^2 + 14*a^4*b^3 - 16*a^3*b^4 + 9*a^2*
b^5 - 2*a*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^
8)*d*cos(d*x + c)^4 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)^3 - (a^10 - 5*a^8*b
^2 + 10*a^6*b^4 - 10*a^4*b^6 + 5*a^2*b^8 - b^10)*d*cos(d*x + c)^2 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b
^7 + a*b^9)*d*cos(d*x + c) - (a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d)

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giac [B]  time = 2.65, size = 800, normalized size = 3.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(a - 2*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)
+ 8*(3*a^4*b + 8*a^2*b^3 + b^5)*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^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) + (a + b - 2*a*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) + 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4
*a*b^3 + b^4)*(cos(d*x + c) - 1)) - (cos(d*x + c) - 1)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(cos(d*x + c) + 1)) -
4*(9*a^6*b + 6*a^5*b^2 + 9*a^4*b^3 + 28*a^3*b^4 + 11*a^2*b^5 - 2*a*b^6 + 3*b^7 + 18*a^6*b*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - 12*a^5*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 26*a^4*b^3*(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) + 4*a^3*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 38*a^2*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
+ 8*a*b^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 6*b^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*a^6*b*(cos(
d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 18*a^5*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 33*a^4*b^3*(cos(
d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 48*a^3*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 27*a^2*b^5*(cos(
d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 6*a*b^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 3*b^7*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(a + b + a*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2))/d

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maple [A]  time = 0.68, size = 322, normalized size = 1.41 \[ -\frac {b^{3}}{2 d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )^{2}}+\frac {3 b \ln \left (b +a \cos \left (d x +c \right )\right ) a^{4}}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {8 b^{3} \ln \left (b +a \cos \left (d x +c \right )\right ) a^{2}}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{5} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {3 b^{2} a^{2}}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}+\frac {b^{4}}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}+\frac {1}{4 d \left (a +b \right )^{3} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right ) a}{4 d \left (a +b \right )^{4}}-\frac {\ln \left (-1+\cos \left (d x +c \right )\right ) b}{2 d \left (a +b \right )^{4}}+\frac {1}{4 d \left (a -b \right )^{3} \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right ) a}{4 d \left (a -b \right )^{4}}-\frac {\ln \left (1+\cos \left (d x +c \right )\right ) b}{2 d \left (a -b \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.79, size = 435, normalized size = 1.90 \[ \frac {\frac {4 \, {\left (3 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac {{\left (a + 2 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {2 \, {\left (8 \, a^{2} b^{3} + 4 \, b^{5} - {\left (a^{5} + 9 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{4} b - 10 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (11 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )}}{a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8} - {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(3*a^4*b + 8*a^2*b^3 + b^5)*log(a*cos(d*x + c) + b)/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) - (
a + 2*b)*log(cos(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) + (a - 2*b)*log(cos(d*x + c) - 1)/(
a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + 2*(8*a^2*b^3 + 4*b^5 - (a^5 + 9*a^3*b^2 + 2*a*b^4)*cos(d*x + c)^3
+ (a^4*b - 10*a^2*b^3 - 3*b^5)*cos(d*x + c)^2 + (11*a^3*b^2 + a*b^4)*cos(d*x + c))/(a^6*b^2 - 3*a^4*b^4 + 3*a
^2*b^6 - b^8 - (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*cos(d*x + c)^4 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b
^7)*cos(d*x + c)^3 + (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3
*a^3*b^5 - a*b^7)*cos(d*x + c)))/d

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mupad [B]  time = 1.70, size = 378, normalized size = 1.65 \[ \frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (\frac {3\,b}{4\,{\left (a+b\right )}^4}-\frac {1}{4\,{\left (a+b\right )}^3}+\frac {3\,b}{4\,{\left (a-b\right )}^4}+\frac {1}{4\,{\left (a-b\right )}^3}\right )}{d}-\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,\left (\frac {3\,b}{4\,{\left (a+b\right )}^4}-\frac {1}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {\frac {{\cos \left (c+d\,x\right )}^3\,\left (a^5+9\,a^3\,b^2+2\,a\,b^4\right )}{2\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {{\cos \left (c+d\,x\right )}^2\,\left (-a^4\,b+10\,a^2\,b^3+3\,b^5\right )}{2\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {2\,b\,\left (2\,a^2\,b^2+b^4\right )}{\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {a\,\cos \left (c+d\,x\right )\,\left (11\,a^2\,b^2+b^4\right )}{2\,\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^2\,\left (a^2-b^2\right )+b^2-a^2\,{\cos \left (c+d\,x\right )}^4+2\,a\,b\,\cos \left (c+d\,x\right )-2\,a\,b\,{\cos \left (c+d\,x\right )}^3\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,\left (a+2\,b\right )}{4\,d\,{\left (a-b\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(b + a*cos(c + d*x))*((3*b)/(4*(a + b)^4) - 1/(4*(a + b)^3) + (3*b)/(4*(a - b)^4) + 1/(4*(a - b)^3)))/d -
(log(cos(c + d*x) - 1)*((3*b)/(4*(a + b)^4) - 1/(4*(a + b)^3)))/d - ((cos(c + d*x)^3*(2*a*b^4 + a^5 + 9*a^3*b^
2))/(2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (cos(c + d*x)^2*(3*b^5 - a^4*b + 10*a^2*b^3))/(2*(a^6 - b^6 + 3*
a^2*b^4 - 3*a^4*b^2)) - (2*b*(b^4 + 2*a^2*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (a*cos(c + d*x)*(b^4 +
11*a^2*b^2))/(2*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)))/(d*(cos(c + d*x)^2*(a^2 - b^2) + b^2 - a^2*cos(c + d*x)
^4 + 2*a*b*cos(c + d*x) - 2*a*b*cos(c + d*x)^3)) - (log(cos(c + d*x) + 1)*(a + 2*b))/(4*d*(a - b)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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