3.2.89 \(\int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx\) [189]

Optimal. Leaf size=162 \[ -\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}+\frac {(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac {b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac {(a-4 b) (a-b)^2 \log (1+\cos (c+d x))}{4 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 162, 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, 1819, 1816} \begin {gather*} -\frac {b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac {a^2 \csc ^2(c+d x) \left (a \left (\frac {3 b^2}{a^2}+1\right ) \cos (c+d x)+b \left (\frac {b^2}{a^2}+3\right )\right )}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac {(a-4 b) (a-b)^2 \log (\cos (c+d x)+1)}{4 d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a^2*(b*(3 + b^2/a^2) + a*(1 + (3*b^2)/a^2)*Cos[c + d*x])*Csc[c + d*x]^2)/d + ((a + b)^2*(a + 4*b)*Log[1
- Cos[c + d*x]])/(4*d) - (b*(3*a^2 + 2*b^2)*Log[Cos[c + d*x]])/d - ((a - 4*b)*(a - b)^2*Log[1 + Cos[c + d*x]])
/(4*d) + (3*a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]^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 1816

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

Rule 1819

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

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 \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx &=-\int (-b-a \cos (c+d x))^3 \csc ^3(c+d x) \sec ^3(c+d x) \, dx\\ &=\frac {a^3 \text {Subst}\left (\int \frac {a^3 (-b+x)^3}{x^3 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^6 \text {Subst}\left (\int \frac {(-b+x)^3}{x^3 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}-\frac {a^4 \text {Subst}\left (\int \frac {2 b^3-6 b^2 x+2 b \left (3+\frac {b^2}{a^2}\right ) x^2-\left (1+\frac {3 b^2}{a^2}\right ) x^3}{x^3 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{2 d}\\ &=-\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}-\frac {a^4 \text {Subst}\left (\int \left (-\frac {(a-4 b) (a-b)^2}{2 a^4 (a-x)}+\frac {2 b^3}{a^2 x^3}-\frac {6 b^2}{a^2 x^2}+\frac {2 \left (3 a^2 b+2 b^3\right )}{a^4 x}-\frac {(a+b)^2 (a+4 b)}{2 a^4 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{2 d}\\ &=-\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}+\frac {(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac {b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac {(a-4 b) (a-b)^2 \log (1+\cos (c+d x))}{4 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec ^2(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 4.69, size = 260, normalized size = 1.60 \begin {gather*} \frac {24 a b^2-(a+b)^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )-4 (a-4 b) (a-b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))+4 (a+b)^2 (a+4 b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+(a-b)^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {2 b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {24 a b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {24 a b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a*b^2 - (a + b)^3*Csc[(c + d*x)/2]^2 - 4*(a - 4*b)*(a - b)^2*Log[Cos[(c + d*x)/2]] - 8*b*(3*a^2 + 2*b^2)*L
og[Cos[c + d*x]] + 4*(a + b)^2*(a + 4*b)*Log[Sin[(c + d*x)/2]] + (a - b)^3*Sec[(c + d*x)/2]^2 + (2*b^3)/(Cos[(
c + d*x)/2] - Sin[(c + d*x)/2])^2 + (24*a*b^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*b^3
)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (24*a*b^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
/(8*d)

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Maple [A]
time = 0.14, size = 162, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+3 b^{2} a \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 b \,a^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(162\)
default	\(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+3 b^{2} a \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 b \,a^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(162\)
norman	\(\frac {-\frac {a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}}{8 d}+\frac {\left (a^{3}+15 b^{2} a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {\left (2 a^{3}-3 b \,a^{2}+30 b^{2} a -9 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (a^{3}+6 b \,a^{2}+9 b^{2} a +4 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(250\)
risch	\(\frac {a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+9 b^{2} a \,{\mathrm e}^{7 i \left (d x +c \right )}+6 b \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+3 b^{2} a \,{\mathrm e}^{5 i \left (d x +c \right )}+12 b \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+3 b^{2} a \,{\mathrm e}^{3 i \left (d x +c \right )}+6 b \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a^{3} {\mathrm e}^{i \left (d x +c \right )}+9 b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b \,a^{2}}{d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2} a}{2 d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{3}}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b \,a^{2}}{d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2} a}{2 d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{3}}{d}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(425\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 171, normalized size = 1.06 \begin {gather*} -\frac {{\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 4 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\cos \left (d x + c\right )\right ) + \frac {2 \, {\left (6 \, a b^{2} \cos \left (d x + c\right ) - {\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )}}{\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 4.67, size = 290, normalized size = 1.79 \begin {gather*} -\frac {12 \, a b^{2} \cos \left (d x + c\right ) - 2 \, {\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, b^{3} - 2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) + {\left ({\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\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 ({\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\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 (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\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))^3,x, algorithm="fricas")

[Out]

-1/4*(12*a*b^2*cos(d*x + c) - 2*(a^3 + 9*a*b^2)*cos(d*x + c)^3 + 2*b^3 - 2*(3*a^2*b + 2*b^3)*cos(d*x + c)^2 +
4*((3*a^2*b + 2*b^3)*cos(d*x + c)^4 - (3*a^2*b + 2*b^3)*cos(d*x + c)^2)*log(-cos(d*x + c)) + ((a^3 - 6*a^2*b +
9*a*b^2 - 4*b^3)*cos(d*x + c)^4 - (a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/
2) - ((a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*cos(d*x + c)^4 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*cos(d*x + c)^2)*log
(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^4 - d*cos(d*x + c)^2)

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 482 vs. \(2 (154) = 308\).
time = 0.54, size = 482, normalized size = 2.98 \begin {gather*} -\frac {\frac {a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {3 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) + 8 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} - \frac {2 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {12 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {18 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {8 \, b^{3} {\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 )}}{\cos \left (d x + c\right ) - 1} - \frac {4 \, {\left (9 \, a^{2} b + 12 \, a b^{2} + 6 \, b^{3} + \frac {18 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {9 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{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))^3,x, algorithm="giac")

[Out]

-1/8*(a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a*b^2*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) - b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*(a^3 + 6*a^2*b + 9*a*b^2 + 4
*b^3)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) + 8*(3*a^2*b + 2*b^3)*log(abs(-(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) - 1)) - (a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 12*a^2*b
*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 18*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 8*b^3*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - 4*(9*a^2*b + 12*a*b^2 + 6*b^3 + 18*a^2*b*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b^3*(cos(d*x + c) - 1)/
(cos(d*x + c) + 1) + 9*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 6*b^3*(cos(d*x + c) - 1)^2/(cos(d*x +
c) + 1)^2)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d

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Mupad [B]
time = 1.01, size = 159, normalized size = 0.98 \begin {gather*} \frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2\,\left (a+4\,b\right )}{4\,d}-\frac {\ln \left (\cos \left (c+d\,x\right )\right )\,\left (3\,a^2\,b+2\,b^3\right )}{d}-\frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {a^3}{2}+\frac {9\,a\,b^2}{2}\right )-\frac {b^3}{2}+{\cos \left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{2}+b^3\right )-3\,a\,b^2\,\cos \left (c+d\,x\right )}{d\,\left ({\cos \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^4\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^2\,\left (a-4\,b\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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