3.5.45 \(\int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx\) [445]

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

[Out]

b^4*x+2*a*b*(a^2+2*b^2)*arctanh(sin(d*x+c))/d+1/3*a^2*(2*a^2+17*b^2)*tan(d*x+c)/d+4/3*a^3*b*sec(d*x+c)*tan(d*x
+c)/d+1/3*a^2*(a+b*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]
time = 0.17, antiderivative size = 115, 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 = {2871, 3110, 3100, 2814, 3855} \begin {gather*} \frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{3 d}+\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 {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^4 x \end {gather*}

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 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 2871

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(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 3100

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 3110

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] &&
NeQ[a^2 - b^2, 0] && 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))^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*}

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Mathematica [A]
time = 0.34, size = 77, normalized size = 0.67 \begin {gather*} \frac {3 b^4 d x+6 a b \left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))+3 a^2 \left (a^2+6 b^2+2 a b \sec (c+d x)\right ) \tan (c+d x)+a^4 \tan ^3(c+d x)}{3 d} \end {gather*}

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 = 0.23, size = 109, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {-a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{2} b^{2} \tan \left (d x +c \right )+4 a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+b^{4} \left (d x +c \right )}{d}\)	\(109\)
default	\(\frac {-a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{2} b^{2} \tan \left (d x +c \right )+4 a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+b^{4} \left (d x +c \right )}{d}\)	\(109\)
risch	\(b^{4} x -\frac {4 i a^{2} \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-9 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-18 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}-a^{2}-9 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(196\)
norman	\(\frac {b^{4} x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+b^{4} x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b^{4} x -b^{4} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 b^{4} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 b^{4} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 b^{4} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 b^{4} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 a^{2} \left (a^{2}-6 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (a^{2}-2 a b +6 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (a^{2}+2 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a^{2} \left (5 a^{2}-12 a b +18 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a^{2} \left (5 a^{2}+12 a b +18 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (13 a^{2}-30 a b -18 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (13 a^{2}+30 a b -18 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {2 a b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(439\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 125, normalized size = 1.09 \begin {gather*} \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 {gather*}

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 = 0.40, size = 138, normalized size = 1.20 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (109) = 218\).
time = 0.44, size = 221, normalized size = 1.92 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.78, size = 185, normalized size = 1.61 \begin {gather*} \frac {2\,b^4\,\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 {2\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {8\,a\,b^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 {4\,a^3\,b\,\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 {2\,a^3\,b\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {6\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*a^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (a^4*sin(c +
d*x))/(3*d*cos(c + d*x)^3) + (8*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*a^3*b*atanh(sin(c/
2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*a^3*b*sin(c + d*x))/(d*cos(c + d*x)^2) + (6*a^2*b^2*sin(c + d*x))/(d*
cos(c + d*x))

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