3.444 \(\int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=108 \[ \frac {3 a^3 b \tan (c+d x)}{d}-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac {a^2 \left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+4 a b^3 x \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2792, 3031, 3023, 2735, 3770} \[ -\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac {a^2 \left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {3 a^3 b \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+4 a b^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^3,x]

[Out]

4*a*b^3*x + (a^2*(a^2 + 12*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) - (b^2*(a^2 - 2*b^2)*Sin[c + d*x])/(2*d) + (3*a^3
*b*Tan[c + d*x])/d + (a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 2735

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 2792

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

Rule 3023

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 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3770

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))^4 \sec ^3(c+d x) \, dx &=\frac {a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x)) \left (6 a^2 b+a \left (a^2+6 b^2\right ) \cos (c+d x)-b \left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {3 a^3 b \tan (c+d x)}{d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-a^2 \left (a^2+12 b^2\right )-8 a b^3 \cos (c+d x)+b^2 \left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a^3 b \tan (c+d x)}{d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-a^2 \left (a^2+12 b^2\right )-8 a b^3 \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=4 a b^3 x-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a^3 b \tan (c+d x)}{d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a^2 \left (a^2+12 b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=4 a b^3 x+\frac {a^2 \left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a^3 b \tan (c+d x)}{d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 2.44, size = 174, normalized size = 1.61 \[ \frac {16 a^3 b \tan (c+d x)+a \left (\frac {a^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^3}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-2 a \left (a^2+12 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a \left (a^2+12 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+16 b^3 c+16 b^3 d x\right )+4 b^4 \sin (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^3,x]

[Out]

(a*(16*b^3*c + 16*b^3*d*x - 2*a*(a^2 + 12*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*a*(a^2 + 12*b^2)*L
og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + a^3/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - a^3/(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^2) + 4*b^4*Sin[c + d*x] + 16*a^3*b*Tan[c + d*x])/(4*d)

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fricas [A]  time = 0.75, size = 130, normalized size = 1.20 \[ \frac {16 \, a b^{3} d x \cos \left (d x + c\right )^{2} + {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{3} b \cos \left (d x + c\right ) + a^{4}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*(16*a*b^3*d*x*cos(d*x + c)^2 + (a^4 + 12*a^2*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (a^4 + 12*a^2*b^2
)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*b^4*cos(d*x + c)^2 + 8*a^3*b*cos(d*x + c) + a^4)*sin(d*x + c))/
(d*cos(d*x + c)^2)

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giac [A]  time = 0.86, size = 177, normalized size = 1.64 \[ \frac {8 \, {\left (d x + c\right )} a b^{3} + \frac {4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 114, normalized size = 1.06 \[ \frac {a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {4 a^{3} b \tan \left (d x +c \right )}{d}+\frac {6 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+4 a \,b^{3} x +\frac {4 a \,b^{3} c}{d}+\frac {b^{4} \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*sec(d*x+c)^3,x)

[Out]

1/2*a^4*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+4*a^3*b*tan(d*x+c)/d+6/d*a^2*b^2*ln(sec(d*
x+c)+tan(d*x+c))+4*a*b^3*x+4/d*a*b^3*c+1/d*b^4*sin(d*x+c)

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maxima [A]  time = 0.86, size = 115, normalized size = 1.06 \[ \frac {16 \, {\left (d x + c\right )} a b^{3} - a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, b^{4} \sin \left (d x + c\right ) + 16 \, a^{3} b \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

1/4*(16*(d*x + c)*a*b^3 - a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) + 12*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*b^4*sin(d*x + c) + 16*a^3*b*tan(d*x + c
))/d

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mupad [B]  time = 0.72, size = 152, normalized size = 1.41 \[ \frac {b^4\,\sin \left (c+d\,x\right )}{d}+\frac {a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {12\,a^2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,a\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,a^3\,b\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^4/cos(c + d*x)^3,x)

[Out]

(b^4*sin(c + d*x))/d + (a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (a^4*sin(c + d*x))/(2*d*cos(c +
d*x)^2) + (12*a^2*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*a*b^3*atan(sin(c/2 + (d*x)/2)/cos(c
/2 + (d*x)/2)))/d + (4*a^3*b*sin(c + d*x))/(d*cos(c + d*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**4*sec(d*x+c)**3,x)

[Out]

Timed out

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