3.443 \(\int (a+b \cos (c+d x))^4 \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=114 \[ -\frac{2 a b \left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac{b^2 \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} b^2 x \left (12 a^2+b^2\right )+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \tan (c+d x) (a+b \cos (c+d x))^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.234006, antiderivative size = 114, normalized size of antiderivative = 1., 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, 3033, 3023, 2735, 3770} \[ -\frac{2 a b \left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac{b^2 \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} b^2 x \left (12 a^2+b^2\right )+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \tan (c+d x) (a+b \cos (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^2,x]

[Out]

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

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

Mathematica [A]  time = 0.633494, size = 119, normalized size = 1.04 \[ \frac{2 b \left (b \left (12 a^2+b^2\right ) (c+d x)-8 a^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+8 a^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+4 a^4 \tan (c+d x)+16 a b^3 \sin (c+d x)+b^4 \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^2,x]

[Out]

(2*b*(b*(12*a^2 + b^2)*(c + d*x) - 8*a^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*a^3*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) + 16*a*b^3*Sin[c + d*x] + b^4*Sin[2*(c + d*x)] + 4*a^4*Tan[c + d*x])/(4*d)

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Maple [A]  time = 0.064, size = 109, normalized size = 1. \begin{align*}{\frac{{a}^{4}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{a}^{2}{b}^{2}x+6\,{\frac{{a}^{2}{b}^{2}c}{d}}+4\,{\frac{a{b}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{4}x}{2}}+{\frac{{b}^{4}c}{2\,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)^2,x)

[Out]

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

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Maxima [A]  time = 0.985013, size = 122, normalized size = 1.07 \begin{align*} \frac{24 \,{\left (d x + c\right )} a^{2} b^{2} +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4} + 8 \, a^{3} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, a b^{3} \sin \left (d x + c\right ) + 4 \, a^{4} \tan \left (d x + c\right )}{4 \, 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)^2,x, algorithm="maxima")

[Out]

1/4*(24*(d*x + c)*a^2*b^2 + (2*d*x + 2*c + sin(2*d*x + 2*c))*b^4 + 8*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*
x + c) - 1)) + 16*a*b^3*sin(d*x + c) + 4*a^4*tan(d*x + c))/d

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Fricas [A]  time = 1.96007, size = 294, normalized size = 2.58 \begin{align*} \frac{4 \, a^{3} b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, a^{3} b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (12 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (b^{4} \cos \left (d x + c\right )^{2} + 8 \, a b^{3} \cos \left (d x + c\right ) + 2 \, a^{4}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \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)^2,x, algorithm="fricas")

[Out]

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

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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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.39385, size = 230, normalized size = 2.02 \begin{align*} \frac{8 \, a^{3} b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, a^{3} b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{4 \, a^{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 (12 \, a^{2} b^{2} + b^{4}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (8 \, 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 )^{3} + 8 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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 )}^{2}}}{2 \, 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)^2,x, algorithm="giac")

[Out]

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