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

Optimal. Leaf size=115 \[ \frac{a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{3 d}+\frac{2 a b \left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a^3 b \tan (c+d x) \sec (c+d x)}{3 d}+\frac{a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^4 x \]

[Out]

b^4*x + (2*a*b*(a^2 + 2*b^2)*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*a^2 + 17*b^2)*Tan[c + d*x])/(3*d) + (4*a^3*b*S
ec[c + d*x]*Tan[c + d*x])/(3*d) + (a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.251554, antiderivative size = 115, 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, 3031, 3021, 2735, 3770} \[ \frac{a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{3 d}+\frac{2 a b \left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a^3 b \tan (c+d x) \sec (c+d x)}{3 d}+\frac{a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^4 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^4*x + (2*a*b*(a^2 + 2*b^2)*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*a^2 + 17*b^2)*Tan[c + d*x])/(3*d) + (4*a^3*b*S
ec[c + d*x]*Tan[c + d*x])/(3*d) + (a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 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 3021

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

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

Mathematica [A]  time = 0.387917, size = 77, normalized size = 0.67 \[ \frac{6 a b \left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))+3 a^2 \tan (c+d x) \left (a^2+2 a b \sec (c+d x)+6 b^2\right )+a^4 \tan ^3(c+d x)+3 b^4 d x}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^4*d*x + 6*a*b*(a^2 + 2*b^2)*ArcTanh[Sin[c + d*x]] + 3*a^2*(a^2 + 6*b^2 + 2*a*b*Sec[c + d*x])*Tan[c + d*x]
+ a^4*Tan[c + d*x]^3)/(3*d)

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Maple [A]  time = 0.069, size = 135, normalized size = 1.2 \begin{align*}{\frac{2\,{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+2\,{\frac{{a}^{3}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{a{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{b}^{4}x+{\frac{{b}^{4}c}{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)^4,x)

[Out]

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

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Maxima [A]  time = 0.976064, size = 169, normalized size = 1.47 \begin{align*} \frac{{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} + 3 \,{\left (d x + c\right )} b^{4} - 3 \, a^{3} b{\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 )} + 6 \, a b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, a^{2} b^{2} \tan \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)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.0155, size = 339, normalized size = 2.95 \begin{align*} \frac{3 \, b^{4} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, a^{3} b \cos \left (d x + c\right ) + a^{4} + 2 \,{\left (a^{4} + 9 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \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)^4,x, algorithm="fricas")

[Out]

1/3*(3*b^4*d*x*cos(d*x + c)^3 + 3*(a^3*b + 2*a*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(a^3*b + 2*a*b^3)
*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (6*a^3*b*cos(d*x + c) + a^4 + 2*(a^4 + 9*a^2*b^2)*cos(d*x + c)^2)*sin
(d*x + c))/(d*cos(d*x + c)^3)

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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)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.34265, size = 298, normalized size = 2.59 \begin{align*} \frac{3 \,{\left (d x + c\right )} b^{4} + 6 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 6 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 2 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, a^{2} b^{2} \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)^4,x, algorithm="giac")

[Out]

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