\(\int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx\) [426]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {\left (3 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {\left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a b \tan ^3(c+d x)}{3 d} \]

[Out]

1/8*(3*a^2+4*b^2)*arctanh(sin(d*x+c))/d+2*a*b*tan(d*x+c)/d+1/8*(3*a^2+4*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/4*a^2*s
ec(d*x+c)^3*tan(d*x+c)/d+2/3*a*b*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2868, 3852, 3091, 3853, 3855} \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {\left (3 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (3 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {2 a b \tan (c+d x)}{d} \]

[In]

Int[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^5,x]

[Out]

((3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a*b*Tan[c + d*x])/d + ((3*a^2 + 4*b^2)*Sec[c + d*x]*Tan[c +
d*x])/(8*d) + (a^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (2*a*b*Tan[c + d*x]^3)/(3*d)

Rule 2868

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[2*c*(d/b)
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3091

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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*} \text {integral}& = (2 a b) \int \sec ^4(c+d x) \, dx+\int \left (a^2+b^2 \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx \\ & = \frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \left (3 a^2+4 b^2\right ) \int \sec ^3(c+d x) \, dx-\frac {(2 a b) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {2 a b \tan (c+d x)}{d}+\frac {\left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {1}{8} \left (3 a^2+4 b^2\right ) \int \sec (c+d x) \, dx \\ & = \frac {\left (3 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {\left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a b \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {3 \left (3 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 \left (3 a^2+4 b^2\right ) \sec (c+d x)+6 a^2 \sec ^3(c+d x)+16 a b \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^5,x]

[Out]

(3*(3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(3*a^2 + 4*b^2)*Sec[c + d*x] + 6*a^2*Sec[c + d*x]^3
+ 16*a*b*(3 + Tan[c + d*x]^2)))/(24*d)

Maple [A] (verified)

Time = 3.76 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(111\)
default	\(\frac {a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(111\)
parts	\(\frac {a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {2 a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(116\)
parallelrisch	\(\frac {-36 \left (a^{2}+\frac {4 b^{2}}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+36 \left (a^{2}+\frac {4 b^{2}}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (18 a^{2}+24 b^{2}\right ) \sin \left (3 d x +3 c \right )+128 a b \sin \left (2 d x +2 c \right )+32 a b \sin \left (4 d x +4 c \right )+\left (66 a^{2}+24 b^{2}\right ) \sin \left (d x +c \right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(186\)
risch	\(-\frac {i \left (9 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+33 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-96 a b \,{\mathrm e}^{4 i \left (d x +c \right )}-33 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-12 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-128 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2} {\mathrm e}^{i \left (d x +c \right )}-12 b^{2} {\mathrm e}^{i \left (d x +c \right )}-32 a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{2 d}\)	\(248\)
norman	\(\frac {\frac {\left (5 a^{2}-16 a b +4 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (5 a^{2}+16 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (21 a^{2}-16 a b -12 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (21 a^{2}+16 a b -12 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (39 a^{2}-16 a b +12 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (39 a^{2}+16 a b +12 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (3 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(274\)

[In]

int((a+cos(d*x+c)*b)^2*sec(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-2*a*b*(-2/3-1/3*sec(d*
x+c)^2)*tan(d*x+c)+b^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.21 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, a b \cos \left (d x + c\right )^{3} + 16 \, a b \cos \left (d x + c\right ) + 3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(3*(3*a^2 + 4*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*a^2 + 4*b^2)*cos(d*x + c)^4*log(-sin(d*x +
c) + 1) + 2*(32*a*b*cos(d*x + c)^3 + 16*a*b*cos(d*x + c) + 3*(3*a^2 + 4*b^2)*cos(d*x + c)^2 + 6*a^2)*sin(d*x
+ c))/(d*cos(d*x + c)^4)

Sympy [F]

\[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \sec ^{5}{\left (c + d x \right )}\, dx \]

[In]

integrate((a+b*cos(d*x+c))**2*sec(d*x+c)**5,x)

[Out]

Integral((a + b*cos(c + d*x))**2*sec(c + d*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.31 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b - 3 \, a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b - 3*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4
- 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*b^2*(2*sin(d*x + c)/(sin(d*x
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (102) = 204\).

Time = 0.32 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.35 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(3*(3*a^2 + 4*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*a^2 + 4*b^2)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) + 2*(15*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*b^2*tan(1/2*d*x + 1/2*c)^7 + 9*a^2
*tan(1/2*d*x + 1/2*c)^5 + 80*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*tan(1/2*d*x +
1/2*c)^3 - 80*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*b^2*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c) + 48*a*
b*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 18.04 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.67 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2}{4}+b^2\right )}{d}+\frac {\left (\frac {5\,a^2}{4}-4\,a\,b+b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a^2}{4}+\frac {20\,a\,b}{3}-b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,a^2}{4}-\frac {20\,a\,b}{3}-b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a^2}{4}+4\,a\,b+b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int((a + b*cos(c + d*x))^2/cos(c + d*x)^5,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((3*a^2)/4 + b^2))/d + (tan(c/2 + (d*x)/2)^5*((20*a*b)/3 + (3*a^2)/4 - b^2) + tan(c
/2 + (d*x)/2)*(4*a*b + (5*a^2)/4 + b^2) + tan(c/2 + (d*x)/2)^7*((5*a^2)/4 - 4*a*b + b^2) - tan(c/2 + (d*x)/2)^
3*((20*a*b)/3 - (3*a^2)/4 + b^2))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6
+ tan(c/2 + (d*x)/2)^8 + 1))