3.5.32 \(\int (a+b \cos (c+d x))^3 \sec (c+d x) \, dx\) [432]

Optimal. Leaf size=73 \[ \frac {1}{2} b \left (6 a^2+b^2\right ) x+\frac {a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a b^2 \sin (c+d x)}{2 d}+\frac {b^2 (a+b \cos (c+d x)) \sin (c+d x)}{2 d} \]

[Out]

1/2*b*(6*a^2+b^2)*x+a^3*arctanh(sin(d*x+c))/d+5/2*a*b^2*sin(d*x+c)/d+1/2*b^2*(a+b*cos(d*x+c))*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2872, 3102, 2814, 3855} \begin {gather*} \frac {a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} b x \left (6 a^2+b^2\right )+\frac {5 a b^2 \sin (c+d x)}{2 d}+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(6*a^2 + b^2)*x)/2 + (a^3*ArcTanh[Sin[c + d*x]])/d + (5*a*b^2*Sin[c + d*x])/(2*d) + (b^2*(a + b*Cos[c + d*x
])*Sin[c + d*x])/(2*d)

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2872

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/
(d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a
*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n
- 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m]
|| (EqQ[a, 0] && NeQ[c, 0])))

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^3 \sec (c+d x) \, dx &=\frac {b^2 (a+b \cos (c+d x)) \sin (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^3+b \left (6 a^2+b^2\right ) \cos (c+d x)+5 a b^2 \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {5 a b^2 \sin (c+d x)}{2 d}+\frac {b^2 (a+b \cos (c+d x)) \sin (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^3+b \left (6 a^2+b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} b \left (6 a^2+b^2\right ) x+\frac {5 a b^2 \sin (c+d x)}{2 d}+\frac {b^2 (a+b \cos (c+d x)) \sin (c+d x)}{2 d}+a^3 \int \sec (c+d x) \, dx\\ &=\frac {1}{2} b \left (6 a^2+b^2\right ) x+\frac {a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a b^2 \sin (c+d x)}{2 d}+\frac {b^2 (a+b \cos (c+d x)) \sin (c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.15, size = 105, normalized size = 1.44 \begin {gather*} \frac {2 b \left (6 a^2+b^2\right ) (c+d x)-4 a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a b^2 \sin (c+d x)+b^3 \sin (2 (c+d x))}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(6*a^2 + b^2)*(c + d*x) - 4*a^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*a^3*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]] + 12*a*b^2*Sin[c + d*x] + b^3*Sin[2*(c + d*x)])/(4*d)

________________________________________________________________________________________

Maple [A]
time = 0.13, size = 73, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b \left (d x +c \right )+3 b^{2} a \sin \left (d x +c \right )+b^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(73\)
default	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b \left (d x +c \right )+3 b^{2} a \sin \left (d x +c \right )+b^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(73\)
risch	\(3 a^{2} b x +\frac {b^{3} x}{2}-\frac {3 i b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 i b^{2} a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (2 d x +2 c \right ) b^{3}}{4 d}\)	\(111\)
norman	\(\frac {\left (3 a^{2} b +\frac {1}{2} b^{3}\right ) x +\left (3 a^{2} b +\frac {1}{2} b^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (9 a^{2} b +\frac {3}{2} b^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (9 a^{2} b +\frac {3}{2} b^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b^{2} \left (6 a -b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b^{2} \left (6 a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {12 b^{2} a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 69, normalized size = 0.95 \begin {gather*} \frac {12 \, {\left (d x + c\right )} a^{2} b + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3} + 4 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, a b^{2} \sin \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.45, size = 72, normalized size = 0.99 \begin {gather*} \frac {a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, a^{2} b + b^{3}\right )} d x + {\left (b^{3} \cos \left (d x + c\right ) + 6 \, a b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (67) = 134\).
time = 0.45, size = 137, normalized size = 1.88 \begin {gather*} \frac {2 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (6 \, a^{2} b + b^{3}\right )} {\left (d x + 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((a+b*cos(d*x+c))^3*sec(d*x+c),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.72, size = 123, normalized size = 1.68 \begin {gather*} \frac {2\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {6\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d +
(b^3*sin(2*c + 2*d*x))/(4*d) + (3*a*b^2*sin(c + d*x))/d + (6*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d

________________________________________________________________________________________