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

Optimal. Leaf size=107 \[ \frac{b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{3 d}+2 a b x \left (2 a^2+b^2\right )+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a b^3 \sin (c+d x) \cos (c+d x)}{3 d}+\frac{b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

[Out]

2*a*b*(2*a^2 + b^2)*x + (a^4*ArcTanh[Sin[c + d*x]])/d + (b^2*(17*a^2 + 2*b^2)*Sin[c + d*x])/(3*d) + (4*a*b^3*C
os[c + d*x]*Sin[c + d*x])/(3*d) + (b^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.22764, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2793, 3033, 3023, 2735, 3770} \[ \frac{b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{3 d}+2 a b x \left (2 a^2+b^2\right )+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a b^3 \sin (c+d x) \cos (c+d x)}{3 d}+\frac{b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*b*(2*a^2 + b^2)*x + (a^4*ArcTanh[Sin[c + d*x]])/d + (b^2*(17*a^2 + 2*b^2)*Sin[c + d*x])/(3*d) + (4*a*b^3*C
os[c + d*x]*Sin[c + d*x])/(3*d) + (b^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(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 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

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 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 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 (c+d x) \, dx &=\frac{b^2 (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x)) \left (3 a^3+b \left (9 a^2+2 b^2\right ) \cos (c+d x)+8 a b^2 \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{4 a b^3 \cos (c+d x) \sin (c+d x)}{3 d}+\frac{b^2 (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{6} \int \left (6 a^4+12 a b \left (2 a^2+b^2\right ) \cos (c+d x)+2 b^2 \left (17 a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{3 d}+\frac{4 a b^3 \cos (c+d x) \sin (c+d x)}{3 d}+\frac{b^2 (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{6} \int \left (6 a^4+12 a b \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=2 a b \left (2 a^2+b^2\right ) x+\frac{b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{3 d}+\frac{4 a b^3 \cos (c+d x) \sin (c+d x)}{3 d}+\frac{b^2 (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+a^4 \int \sec (c+d x) \, dx\\ &=2 a b \left (2 a^2+b^2\right ) x+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{3 d}+\frac{4 a b^3 \cos (c+d x) \sin (c+d x)}{3 d}+\frac{b^2 (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.159085, size = 128, normalized size = 1.2 \[ \frac{24 a b \left (2 a^2+b^2\right ) (c+d x)+9 b^2 \left (8 a^2+b^2\right ) \sin (c+d x)-12 a^4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a^4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 a b^3 \sin (2 (c+d x))+b^4 \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a*b*(2*a^2 + b^2)*(c + d*x) - 12*a^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*a^4*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] + 9*b^2*(8*a^2 + b^2)*Sin[c + d*x] + 12*a*b^3*Sin[2*(c + d*x)] + b^4*Sin[3*(c + d*x)])/(
12*d)

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Maple [A]  time = 0.058, size = 131, normalized size = 1.2 \begin{align*}{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{a}^{3}bx+4\,{\frac{{a}^{3}bc}{d}}+6\,{\frac{{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{a{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+2\,a{b}^{3}x+2\,{\frac{a{b}^{3}c}{d}}+{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){b}^{4}}{3\,d}}+{\frac{2\,{b}^{4}\sin \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.961537, size = 128, normalized size = 1.2 \begin{align*} \frac{12 \,{\left (d x + c\right )} a^{3} b + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{3} -{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{4} + 3 \, a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 18 \, a^{2} b^{2} \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(12*(d*x + c)*a^3*b + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b^3 - (sin(d*x + c)^3 - 3*sin(d*x + c))*b^4 + 3
*a^4*log(sec(d*x + c) + tan(d*x + c)) + 18*a^2*b^2*sin(d*x + c))/d

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Fricas [A]  time = 2.05108, size = 239, normalized size = 2.23 \begin{align*} \frac{3 \, a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \,{\left (2 \, a^{3} b + a b^{3}\right )} d x + 2 \,{\left (b^{4} \cos \left (d x + c\right )^{2} + 6 \, a b^{3} \cos \left (d x + c\right ) + 18 \, a^{2} b^{2} + 2 \, b^{4}\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.36577, size = 286, normalized size = 2.67 \begin{align*} \frac{3 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 6 \,{\left (2 \, a^{3} b + a b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, b^{4} \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 )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(2*a^3*b + a*b^3)
*(d*x + c) + 2*(18*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*b^4*tan(1/2*d*x + 1/2*c
)^5 + 36*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*b^4*tan(1/2*d*x + 1/2*c)^3 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c) + 6*a
*b^3*tan(1/2*d*x + 1/2*c) + 3*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d