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

Optimal. Leaf size=119 \[ \frac{a^2 \tan ^5(c+d x)}{5 d}+\frac{a^2 \tan ^3(c+d x)}{3 d}-\frac{a^2 \tan (c+d x)}{d}+\frac{3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 \tan ^3(c+d x) \sec (c+d x)}{2 d}-\frac{3 a^2 \tan (c+d x) \sec (c+d x)}{4 d}+a^2 x \]

[Out]

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

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Rubi [A]  time = 0.139138, antiderivative size = 119, normalized size of antiderivative = 1., 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} \[ \frac{a^2 \tan ^5(c+d x)}{5 d}+\frac{a^2 \tan ^3(c+d x)}{3 d}-\frac{a^2 \tan (c+d x)}{d}+\frac{3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 \tan ^3(c+d x) \sec (c+d x)}{2 d}-\frac{3 a^2 \tan (c+d x) \sec (c+d x)}{4 d}+a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 8

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

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 3770

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

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 30

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

Rubi steps

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

Mathematica [B]  time = 5.40286, size = 558, normalized size = 4.69 \[ \frac{1}{960} a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\cos \left (\frac{c}{2}\right ) \left (\frac{151}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{36}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{151}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{36}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}\right )}{d}+\frac{149 \sin \left (\frac{c}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{149 \sin \left (\frac{c}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{24 \sin \left (\frac{c}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{24 \sin \left (\frac{c}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{180 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{180 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{\sec (c) \sin \left (\frac{d x}{2}\right ) \left (333 \cos \left (2 c+\frac{3 d x}{2}\right )+287 \cos \left (2 c+\frac{5 d x}{2}\right )+67 \cos \left (4 c+\frac{7 d x}{2}\right )+68 \cos \left (4 c+\frac{9 d x}{2}\right )+293 \cos \left (\frac{d x}{2}\right )\right ) \sec ^5(c+d x)}{2 d}+240 x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(240*x - (180*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (180*
Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d - ((293*Cos[(d*x)/2] + 333*Cos[2*c + (3*d*x)/2] + 287*Cos[2*c + (5
*d*x)/2] + 67*Cos[4*c + (7*d*x)/2] + 68*Cos[4*c + (9*d*x)/2])*Sec[c]*Sec[c + d*x]^5*Sin[(d*x)/2])/(2*d) - (24*
Sin[c/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + (149*Sin[c/2])/(d*(Cos[c/2] - Si
n[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (24*Sin[c/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^4) + (149*Sin[c/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (Cos
[c/2]*(36/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) - 151/((Cos[c/2] - Sin[c/2])*(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2])^2) - 36/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + 151/(
(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)))/d))/960

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Maple [A]  time = 0.046, size = 169, normalized size = 1.4 \begin{align*}{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d}}+{a}^{2}x+{\frac{{a}^{2}c}{d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{a}^{2}\sin \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.76646, size = 161, normalized size = 1.35 \begin{align*} \frac{24 \, a^{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^{2}{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*a^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^2*(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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Fricas [A]  time = 1.02987, size = 360, normalized size = 3.03 \begin{align*} \frac{120 \, a^{2} d x \cos \left (d x + c\right )^{5} + 45 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (68 \, a^{2} \cos \left (d x + c\right )^{4} + 75 \, a^{2} \cos \left (d x + c\right )^{3} + 4 \, a^{2} \cos \left (d x + c\right )^{2} - 30 \, a^{2} \cos \left (d x + c\right ) - 12 \, a^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*a^2*cos(d*x + c)^5*log(-s
in(d*x + c) + 1) - 2*(68*a^2*cos(d*x + c)^4 + 75*a^2*cos(d*x + c)^3 + 4*a^2*cos(d*x + c)^2 - 30*a^2*cos(d*x +
c) - 12*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.71275, size = 200, normalized size = 1.68 \begin{align*} \frac{60 \,{\left (d x + c\right )} a^{2} + 45 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a^{2} \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 )^{9} - 110 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 328 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 530 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, a^{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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*(d*x + c)*a^2 + 45*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1))
+ 2*(15*a^2*tan(1/2*d*x + 1/2*c)^9 - 110*a^2*tan(1/2*d*x + 1/2*c)^7 + 328*a^2*tan(1/2*d*x + 1/2*c)^5 - 530*a^
2*tan(1/2*d*x + 1/2*c)^3 + 105*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d