3.2.88 \(\int \csc (c+d x) (a+b \sec (c+d x))^3 \, dx\) [188]

Optimal. Leaf size=102 \[ \frac {(a+b)^3 \log (1-\cos (c+d x))}{2 d}-\frac {b \left (3 a^2+b^2\right ) \log (\cos (c+d x))}{d}-\frac {(a-b)^3 \log (1+\cos (c+d x))}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \]

[Out]

1/2*(a+b)^3*ln(1-cos(d*x+c))/d-b*(3*a^2+b^2)*ln(cos(d*x+c))/d-1/2*(a-b)^3*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 = 0.16, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3957, 2916, 12, 1816} \begin {gather*} -\frac {b \left (3 a^2+b^2\right ) \log (\cos (c+d x))}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {(a-b)^3 \log (\cos (c+d x)+1)}{2 d}+\frac {(a+b)^3 \log (1-\cos (c+d x))}{2 d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)^3*Log[1 - Cos[c + d*x]])/(2*d) - (b*(3*a^2 + b^2)*Log[Cos[c + d*x]])/d - ((a - b)^3*Log[1 + Cos[c + d
*x]])/(2*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 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 (c+d x) (a+b \sec (c+d x))^3 \, dx &=-\int (-b-a \cos (c+d x))^3 \csc (c+d x) \sec ^3(c+d x) \, dx\\ &=\frac {a \text {Subst}\left (\int \frac {a^3 (-b+x)^3}{x^3 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^4 \text {Subst}\left (\int \frac {(-b+x)^3}{x^3 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^4 \text {Subst}\left (\int \left (\frac {(a-b)^3}{2 a^4 (a-x)}-\frac {b^3}{a^2 x^3}+\frac {3 b^2}{a^2 x^2}+\frac {b \left (-3 a^2-b^2\right )}{a^4 x}+\frac {(a+b)^3}{2 a^4 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {(a+b)^3 \log (1-\cos (c+d x))}{2 d}-\frac {b \left (3 a^2+b^2\right ) \log (\cos (c+d x))}{d}-\frac {(a-b)^3 \log (1+\cos (c+d x))}{2 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 = 0.20, size = 89, normalized size = 0.87 \begin {gather*} \frac {-2 (a-b)^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 b \left (3 a^2+b^2\right ) \log (\cos (c+d x))+2 (a+b)^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 a b^2 \sec (c+d x)+b^3 \sec ^2(c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - b)^3*Log[Cos[(c + d*x)/2]] - 2*b*(3*a^2 + b^2)*Log[Cos[c + d*x]] + 2*(a + b)^3*Log[Sin[(c + d*x)/2]]
+ 6*a*b^2*Sec[c + d*x] + b^3*Sec[c + d*x]^2)/(2*d)

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Maple [A]
time = 0.10, size = 92, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {b^{3} \left (\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 b^{2} a \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 b \,a^{2} \ln \left (\tan \left (d x +c \right )\right )+a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\)	\(92\)
default	\(\frac {b^{3} \left (\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 b^{2} a \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 b \,a^{2} \ln \left (\tan \left (d x +c \right )\right )+a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\)	\(92\)
norman	\(\frac {\frac {6 b^{2} a}{d}-\frac {2 \left (3 b^{2} a -b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(143\)
risch	\(\frac {2 b^{2} \left (3 a \,{\mathrm e}^{3 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )} a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b \,a^{2}}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2} a}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{3}}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b \,a^{2}}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2} a}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{3}}{d}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 112, normalized size = 1.10 \begin {gather*} -\frac {{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 2 \, {\left (3 \, a^{2} b + b^{3}\right )} \log \left (\cos \left (d x + c\right )\right ) - \frac {6 \, a b^{2} \cos \left (d x + c\right ) + b^{3}}{\cos \left (d x + c\right )^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 7.58, size = 139, normalized size = 1.36 \begin {gather*} \frac {6 \, a b^{2} \cos \left (d x + c\right ) - 2 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + b^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*a*b^2*cos(d*x + c) - 2*(3*a^2*b + b^3)*cos(d*x + c)^2*log(-cos(d*x + c)) - (a^3 - 3*a^2*b + 3*a*b^2 - b
^3)*cos(d*x + c)^2*log(1/2*cos(d*x + c) + 1/2) + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c)^2*log(-1/2*cos(d
*x + c) + 1/2) + b^3)/(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 {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (96) = 192\).
time = 0.52, size = 250, normalized size = 2.45 \begin {gather*} \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + 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 ) - 2 \, {\left (3 \, a^{2} b + 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 {9 \, a^{2} b + 12 \, a b^{2} + 3 \, 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 {2 \, 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 {3 \, b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.13, size = 85, normalized size = 0.83 \begin {gather*} \frac {\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^3}{2}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^3}{2}+\frac {\frac {b^3}{2}+3\,a\,\cos \left (c+d\,x\right )\,b^2}{{\cos \left (c+d\,x\right )}^2}-\ln \left (\cos \left (c+d\,x\right )\right )\,\left (3\,a^2\,b+b^3\right )}{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),x)

[Out]

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

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