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

Optimal. Leaf size=116 \[ \frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {a^2 \tan (c+d x)}{d}+a^2 x+\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \tan ^3(c+d x) \sec (c+d x)}{2 d}-\frac {3 a b \tan (c+d x) \sec (c+d x)}{4 d}+\frac {b^2 \tan ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3886, 3473, 8, 2611, 3770, 2607, 30} \[ \frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {a^2 \tan (c+d x)}{d}+a^2 x+\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \tan ^3(c+d x) \sec (c+d x)}{2 d}-\frac {3 a b \tan (c+d x) \sec (c+d x)}{4 d}+\frac {b^2 \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [B]  time = 0.97, size = 355, normalized size = 3.06 \[ \frac {\sec ^5(c+d x) \left (-80 a^2 \sin (c+d x)-160 a^2 \sin (3 (c+d x))-80 a^2 \sin (5 (c+d x))+60 a^2 c \cos (5 (c+d x))+60 a^2 d x \cos (5 (c+d x))-60 a b \sin (2 (c+d x))-150 a b \sin (4 (c+d x))-45 a b \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+45 a b \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+150 a \cos (c+d x) \left (4 a (c+d x)-3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+75 a \cos (3 (c+d x)) \left (4 a (c+d x)-3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+120 b^2 \sin (c+d x)-60 b^2 \sin (3 (c+d x))+12 b^2 \sin (5 (c+d x))\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^5*(60*a^2*c*Cos[5*(c + d*x)] + 60*a^2*d*x*Cos[5*(c + d*x)] - 45*a*b*Cos[5*(c + d*x)]*Log[Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2]] + 45*a*b*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 150*a*Cos[c
+ d*x]*(4*a*(c + d*x) - 3*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*
x)/2]]) + 75*a*Cos[3*(c + d*x)]*(4*a*(c + d*x) - 3*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*b*Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]]) - 80*a^2*Sin[c + d*x] + 120*b^2*Sin[c + d*x] - 60*a*b*Sin[2*(c + d*x)] - 160*a
^2*Sin[3*(c + d*x)] - 60*b^2*Sin[3*(c + d*x)] - 150*a*b*Sin[4*(c + d*x)] - 80*a^2*Sin[5*(c + d*x)] + 12*b^2*Si
n[5*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.55, size = 151, normalized size = 1.30 \[ \frac {120 \, a^{2} d x \cos \left (d x + c\right )^{5} + 45 \, a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (75 \, a b \cos \left (d x + c\right )^{3} + 4 \, {\left (20 \, a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 30 \, a b \cos \left (d x + c\right ) - 4 \, {\left (5 \, a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 12 \, b^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(120*a^2*d*x*cos(d*x + c)^5 + 45*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*a*b*cos(d*x + c)^5*log(-s
in(d*x + c) + 1) - 2*(75*a*b*cos(d*x + c)^3 + 4*(20*a^2 - 3*b^2)*cos(d*x + c)^4 - 30*a*b*cos(d*x + c) - 4*(5*a
^2 - 6*b^2)*cos(d*x + c)^2 - 12*b^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [B]  time = 2.39, size = 220, normalized size = 1.90 \[ \frac {60 \, {\left (d x + c\right )} a^{2} + 45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 45 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 210 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 520 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 192 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 210 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 45 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*(d*x + c)*a^2 + 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 - 45*a*b*tan(1/2*d*x + 1/2*c)^9 - 320*a^2*tan(1/2*d*x + 1/2*c)^7 + 210*a*b
*tan(1/2*d*x + 1/2*c)^7 + 520*a^2*tan(1/2*d*x + 1/2*c)^5 - 192*b^2*tan(1/2*d*x + 1/2*c)^5 - 320*a^2*tan(1/2*d*
x + 1/2*c)^3 - 210*a*b*tan(1/2*d*x + 1/2*c)^3 + 60*a^2*tan(1/2*d*x + 1/2*c) + 45*a*b*tan(1/2*d*x + 1/2*c))/(ta
n(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 0.39, size = 164, normalized size = 1.41 \[ \frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \tan \left (d x +c \right )}{d}+a^{2} x +\frac {a^{2} c}{d}+\frac {a b \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}-\frac {a b \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{2}}-\frac {a b \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 a b \sin \left (d x +c \right )}{4 d}+\frac {3 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^2*tan(d*x+c)^4,x)

[Out]

1/3*a^2*tan(d*x+c)^3/d-a^2*tan(d*x+c)/d+a^2*x+1/d*a^2*c+1/2/d*a*b*sin(d*x+c)^5/cos(d*x+c)^4-1/4/d*a*b*sin(d*x+
c)^5/cos(d*x+c)^2-1/4*a*b*sin(d*x+c)^3/d-3/4*a*b*sin(d*x+c)/d+3/4/d*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/5/d*b^2*si
n(d*x+c)^5/cos(d*x+c)^5

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maxima [A]  time = 0.60, size = 118, normalized size = 1.02 \[ \frac {24 \, b^{2} \tan \left (d x + c\right )^{5} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} + 15 \, a b {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*b^2*tan(d*x + c)^5 + 40*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2 + 15*a*b*(2*(5*sin(d*x +
c)^3 - 3*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

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mupad [B]  time = 2.56, size = 332, normalized size = 2.86 \[ \frac {\left (2\,a^2-\frac {3\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (7\,a\,b-\frac {32\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {52\,a^2}{3}-\frac {32\,b^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {32\,a^2}{3}-7\,b\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+\frac {3\,b\,a}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {2\,a^2\,\mathrm {atan}\left (\frac {64\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^6+36\,a^4\,b^2}+\frac {36\,a^4\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^6+36\,a^4\,b^2}\right )}{d}+\frac {3\,a\,b\,\mathrm {atanh}\left (\frac {48\,a^5\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{48\,a^5\,b+27\,a^3\,b^3}+\frac {27\,a^3\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{48\,a^5\,b+27\,a^3\,b^3}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^5*((52*a^2)/3 - (32*b^2)/5) - tan(c/2 + (d*x)/2)^9*((3*a*b)/2 - 2*a^2) - tan(c/2 + (d*x)/2
)^3*(7*a*b + (32*a^2)/3) + tan(c/2 + (d*x)/2)^7*(7*a*b - (32*a^2)/3) + tan(c/2 + (d*x)/2)*((3*a*b)/2 + 2*a^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)) + (2*a^2*atan((64*a^6*tan(c/2 + (d*x)/2))/(64*a^6 + 36*a^4*b^2) + (36*a^4*b^2*tan(c/2
+ (d*x)/2))/(64*a^6 + 36*a^4*b^2)))/d + (3*a*b*atanh((48*a^5*b*tan(c/2 + (d*x)/2))/(48*a^5*b + 27*a^3*b^3) +
(27*a^3*b^3*tan(c/2 + (d*x)/2))/(48*a^5*b + 27*a^3*b^3)))/(2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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