3.2.93 \(\int \csc ^2(c+d x) (a+b \sec (c+d x))^3 \, dx\) [193]

Optimal. Leaf size=133 \[ \frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {a^3 \cot (c+d x)}{d}-\frac {3 a b^2 \cot (c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {3 b^3 \csc (c+d x)}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d}+\frac {3 a b^2 \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3957, 2991, 3852, 8, 2701, 327, 213, 2700, 14, 294} \begin {gather*} -\frac {a^3 \cot (c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {3 a b^2 \tan (c+d x)}{d}-\frac {3 a b^2 \cot (c+d x)}{d}-\frac {3 b^3 \csc (c+d x)}{2 d}+\frac {3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (3*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cot[c + d*x])/d - (3*a*b^2*Cot[
c + d*x])/d - (3*a^2*b*Csc[c + d*x])/d - (3*b^3*Csc[c + d*x])/(2*d) + (b^3*Csc[c + d*x]*Sec[c + d*x]^2)/(2*d)
+ (3*a*b^2*Tan[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2701

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2991

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/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 ^2(c+d x) (a+b \sec (c+d x))^3 \, dx &=-\int (-b-a \cos (c+d x))^3 \csc ^2(c+d x) \sec ^3(c+d x) \, dx\\ &=\int \left (a^3 \csc ^2(c+d x)+3 a^2 b \csc ^2(c+d x) \sec (c+d x)+3 a b^2 \csc ^2(c+d x) \sec ^2(c+d x)+b^3 \csc ^2(c+d x) \sec ^3(c+d x)\right ) \, dx\\ &=a^3 \int \csc ^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \csc ^2(c+d x) \sec (c+d x) \, dx+\left (3 a b^2\right ) \int \csc ^2(c+d x) \sec ^2(c+d x) \, dx+b^3 \int \csc ^2(c+d x) \sec ^3(c+d x) \, dx\\ &=-\frac {a^3 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac {\left (3 a^2 b\right ) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (3 a b^2\right ) \text {Subst}\left (\int \frac {1+x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac {b^3 \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac {a^3 \cot (c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d}-\frac {\left (3 a^2 b\right ) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (3 a b^2\right ) \text {Subst}\left (\int \left (1+\frac {1}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{2 d}\\ &=\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^3 \cot (c+d x)}{d}-\frac {3 a b^2 \cot (c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {3 b^3 \csc (c+d x)}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d}+\frac {3 a b^2 \tan (c+d x)}{d}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{2 d}\\ &=\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {a^3 \cot (c+d x)}{d}-\frac {3 a b^2 \cot (c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {3 b^3 \csc (c+d x)}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d}+\frac {3 a b^2 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(406\) vs. \(2(133)=266\).
time = 0.44, size = 406, normalized size = 3.05 \begin {gather*} -\frac {\csc ^5\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (12 a^2 b+2 b^3+6 a \left (a^2+2 b^2\right ) \cos (c+d x)+6 \left (2 a^2 b+b^3\right ) \cos (2 (c+d x))+2 a^3 \cos (3 (c+d x))+12 a b^2 \cos (3 (c+d x))+6 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+3 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-6 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-3 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+6 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+3 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-6 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-3 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))\right )}{16 d \left (-1+\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(Csc[(c + d*x)/2]^5*Sec[(c + d*x)/2]*(12*a^2*b + 2*b^3 + 6*a*(a^2 + 2*b^2)*Cos[c + d*x] + 6*(2*a^2*b + b
^3)*Cos[2*(c + d*x)] + 2*a^3*Cos[3*(c + d*x)] + 12*a*b^2*Cos[3*(c + d*x)] + 6*a^2*b*Log[Cos[(c + d*x)/2] - Sin
[(c + d*x)/2]]*Sin[c + d*x] + 3*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 6*a^2*b*Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 3*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 6*
a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 3*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2
]]*Sin[3*(c + d*x)] - 6*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 3*b^3*Log[Cos[(c + d
*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)]))/(d*(-1 + Cot[(c + d*x)/2]^2)^2)

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Maple [A]
time = 0.09, size = 129, normalized size = 0.97

method	result	size
derivativedivides	\(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 b^{2} a \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+3 b \,a^{2} \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-a^{3} \cot \left (d x +c \right )}{d}\)	\(129\)
default	\(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 b^{2} a \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+3 b \,a^{2} \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-a^{3} \cot \left (d x +c \right )}{d}\)	\(129\)
norman	\(\frac {-\frac {a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}}{2 d}+\frac {\left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 \left (a^{3}-b \,a^{2}+7 b^{2} a -b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {3 \left (a^{3}+b \,a^{2}+7 b^{2} a +b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 b \left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {3 b \left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(216\)
risch	\(-\frac {i \left (6 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+2 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+12 b \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+4 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+12 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{i \left (d x +c \right )}+2 a^{3}+12 b^{2} a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}+\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}-\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(262\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 139, normalized size = 1.05 \begin {gather*} -\frac {b^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} b {\left (\frac {2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a b^{2} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + \frac {4 \, a^{3}}{\tan \left (d x + c\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.85, size = 151, normalized size = 1.14 \begin {gather*} \frac {3 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 3 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 12 \, a b^{2} \cos \left (d x + c\right ) - 4 \, {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, b^{3} - 6 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}}{4 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.52, size = 225, normalized size = 1.69 \begin {gather*} \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, {\left (2 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.47, size = 181, normalized size = 1.36 \begin {gather*} \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,a^2\,b+3\,b^3\right )}{d}-\frac {3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^3+6\,a^2\,b+18\,a\,b^2+4\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^3+3\,a^2\,b+15\,a\,b^2-b^3\right )}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a-b\right )}^3}{2\,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)^2,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*(6*a^2*b + 3*b^3))/d - (3*a*b^2 - tan(c/2 + (d*x)/2)^2*(18*a*b^2 + 6*a^2*b + 2*a^3
+ 4*b^3) + 3*a^2*b + a^3 + b^3 + tan(c/2 + (d*x)/2)^4*(15*a*b^2 + 3*a^2*b + a^3 - b^3))/(d*(2*tan(c/2 + (d*x)/
2) - 4*tan(c/2 + (d*x)/2)^3 + 2*tan(c/2 + (d*x)/2)^5)) + (tan(c/2 + (d*x)/2)*(a - b)^3)/(2*d)

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