3.5.56 \(\int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx\) [456]

Optimal. Leaf size=135 \[ \frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (5 a^2+4 b^2\right ) \tan (c+d x)}{5 d}+\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {b^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\left (5 a^2+4 b^2\right ) \tan ^3(c+d x)}{15 d} \]

[Out]

3/4*a*b*arctanh(sin(d*x+c))/d+1/5*(5*a^2+4*b^2)*tan(d*x+c)/d+3/4*a*b*sec(d*x+c)*tan(d*x+c)/d+1/2*a*b*sec(d*x+c
)^3*tan(d*x+c)/d+1/5*b^2*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(5*a^2+4*b^2)*tan(d*x+c)^3/d

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3873, 3853, 3855, 4131, 3852} \begin {gather*} \frac {\left (5 a^2+4 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (5 a^2+4 b^2\right ) \tan (c+d x)}{5 d}+\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a b \tan (c+d x) \sec (c+d x)}{4 d}+\frac {b^2 \tan (c+d x) \sec ^4(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*b*ArcTanh[Sin[c + d*x]])/(4*d) + ((5*a^2 + 4*b^2)*Tan[c + d*x])/(5*d) + (3*a*b*Sec[c + d*x]*Tan[c + d*x])
/(4*d) + (a*b*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) + (b^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((5*a^2 + 4*b^2)*
Tan[c + d*x]^3)/(15*d)

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]

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx &=(2 a b) \int \sec ^5(c+d x) \, dx+\int \sec ^4(c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {b^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} (3 a b) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (5 a^2+4 b^2\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {b^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 a b) \int \sec (c+d x) \, dx-\frac {\left (5 a^2+4 b^2\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (5 a^2+4 b^2\right ) \tan (c+d x)}{5 d}+\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {b^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\left (5 a^2+4 b^2\right ) \tan ^3(c+d x)}{15 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.62, size = 118, normalized size = 0.87 \begin {gather*} \frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {3 a b \left (\tanh ^{-1}(\sin (c+d x))+\sec (c+d x) \tan (c+d x)\right )}{4 d}+\frac {a^2 \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d}+\frac {b^2 \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) + (3*a*b*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x]))/(4*d) +
(a^2*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d + (b^2*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/d

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 110, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 b a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(110\)
default	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 b a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(110\)
risch	\(-\frac {i \left (45 a b \,{\mathrm e}^{9 i \left (d x +c \right )}+210 b a \,{\mathrm e}^{7 i \left (d x +c \right )}-120 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-280 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-320 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-210 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-200 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-160 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-45 b a \,{\mathrm e}^{i \left (d x +c \right )}-40 a^{2}-32 b^{2}\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {3 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}\)	\(194\)
norman	\(\frac {-\frac {4 \left (25 a^{2}+29 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {\left (4 a^{2}-5 b a +4 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {\left (4 a^{2}+5 b a +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {\left (16 a^{2}-3 b a +8 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (16 a^{2}+3 b a +8 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {3 b a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {3 b a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-a^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+2*b*a*(-(-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)))-b^2*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 132, normalized size = 0.98 \begin {gather*} \frac {40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + 8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} b^{2} - 15 \, a b {\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 )}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/120*(40*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 + 8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*b
^2 - 15*a*b*(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)))/d

________________________________________________________________________________________

Fricas [A]
time = 3.68, size = 136, normalized size = 1.01 \begin {gather*} \frac {45 \, a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (45 \, a b \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right ) + 4 \, {\left (5 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 12 \, b^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/120*(45*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*a*b*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(45*a*b*
cos(d*x + c)^3 + 8*(5*a^2 + 4*b^2)*cos(d*x + c)^4 + 30*a*b*cos(d*x + c) + 4*(5*a^2 + 4*b^2)*cos(d*x + c)^2 + 1
2*b^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*sec(c + d*x)**4, x)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (123) = 246\).
time = 0.50, size = 272, normalized size = 2.01 \begin {gather*} \frac {45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 200 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 232 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, 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 )}^{5}}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(45*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(60*a^2*tan(1/
2*d*x + 1/2*c)^9 - 75*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*b^2*tan(1/2*d*x + 1/2*c)^9 - 160*a^2*tan(1/2*d*x + 1/2*c
)^7 + 30*a*b*tan(1/2*d*x + 1/2*c)^7 - 80*b^2*tan(1/2*d*x + 1/2*c)^7 + 200*a^2*tan(1/2*d*x + 1/2*c)^5 + 232*b^2
*tan(1/2*d*x + 1/2*c)^5 - 160*a^2*tan(1/2*d*x + 1/2*c)^3 - 30*a*b*tan(1/2*d*x + 1/2*c)^3 - 80*b^2*tan(1/2*d*x
+ 1/2*c)^3 + 60*a^2*tan(1/2*d*x + 1/2*c) + 75*a*b*tan(1/2*d*x + 1/2*c) + 60*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2
*d*x + 1/2*c)^2 - 1)^5)/d

________________________________________________________________________________________

Mupad [B]
time = 3.82, size = 221, normalized size = 1.64 \begin {gather*} \frac {3\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\left (2\,a^2-\frac {5\,a\,b}{2}+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {16\,a^2}{3}+a\,b-\frac {8\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,a^2}{3}+\frac {116\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,a^2}{3}-a\,b-\frac {8\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+\frac {5\,a\,b}{2}+2\,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 )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^2/cos(c + d*x)^4,x)

[Out]

(3*a*b*atanh(tan(c/2 + (d*x)/2)))/(2*d) - (tan(c/2 + (d*x)/2)^5*((20*a^2)/3 + (116*b^2)/15) + tan(c/2 + (d*x)/
2)^9*(2*a^2 - (5*a*b)/2 + 2*b^2) - tan(c/2 + (d*x)/2)^3*(a*b + (16*a^2)/3 + (8*b^2)/3) - tan(c/2 + (d*x)/2)^7*
((16*a^2)/3 - a*b + (8*b^2)/3) + tan(c/2 + (d*x)/2)*((5*a*b)/2 + 2*a^2 + 2*b^2))/(d*(5*tan(c/2 + (d*x)/2)^2 -
10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))

________________________________________________________________________________________