3.48 \(\int (a+a \sec (c+d x))^3 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=169 \[ \frac {3 a^3 \tan ^5(c+d x)}{5 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}-\frac {a^3 \tan (c+d x)}{d}+\frac {19 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {a^3 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac {3 a^3 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {17 a^3 \tan (c+d x) \sec (c+d x)}{16 d}+a^3 x \]

[Out]

a^3*x+19/16*a^3*arctanh(sin(d*x+c))/d-a^3*tan(d*x+c)/d-17/16*a^3*sec(d*x+c)*tan(d*x+c)/d-1/8*a^3*sec(d*x+c)^3*
tan(d*x+c)/d+1/3*a^3*tan(d*x+c)^3/d+3/4*a^3*sec(d*x+c)*tan(d*x+c)^3/d+1/6*a^3*sec(d*x+c)^3*tan(d*x+c)^3/d+3/5*
a^3*tan(d*x+c)^5/d

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Rubi [A]  time = 0.22, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3886, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ \frac {3 a^3 \tan ^5(c+d x)}{5 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}-\frac {a^3 \tan (c+d x)}{d}+\frac {19 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {a^3 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac {3 a^3 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {17 a^3 \tan (c+d x) \sec (c+d x)}{16 d}+a^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

a^3*x + (19*a^3*ArcTanh[Sin[c + d*x]])/(16*d) - (a^3*Tan[c + d*x])/d - (17*a^3*Sec[c + d*x]*Tan[c + d*x])/(16*
d) - (a^3*Sec[c + d*x]^3*Tan[c + d*x])/(8*d) + (a^3*Tan[c + d*x]^3)/(3*d) + (3*a^3*Sec[c + d*x]*Tan[c + d*x]^3
)/(4*d) + (a^3*Sec[c + d*x]^3*Tan[c + d*x]^3)/(6*d) + (3*a^3*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 3768

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 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+a \sec (c+d x))^3 \tan ^4(c+d x) \, dx &=\int \left (a^3 \tan ^4(c+d x)+3 a^3 \sec (c+d x) \tan ^4(c+d x)+3 a^3 \sec ^2(c+d x) \tan ^4(c+d x)+a^3 \sec ^3(c+d x) \tan ^4(c+d x)\right ) \, dx\\ &=a^3 \int \tan ^4(c+d x) \, dx+a^3 \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx+\left (3 a^3\right ) \int \sec (c+d x) \tan ^4(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \tan ^4(c+d x) \, dx\\ &=\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {3 a^3 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {a^3 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}-\frac {1}{2} a^3 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx-a^3 \int \tan ^2(c+d x) \, dx-\frac {1}{4} \left (9 a^3\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int x^4 \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {a^3 \tan (c+d x)}{d}-\frac {9 a^3 \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {3 a^3 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {a^3 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac {3 a^3 \tan ^5(c+d x)}{5 d}+\frac {1}{8} a^3 \int \sec ^3(c+d x) \, dx+a^3 \int 1 \, dx+\frac {1}{8} \left (9 a^3\right ) \int \sec (c+d x) \, dx\\ &=a^3 x+\frac {9 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {a^3 \tan (c+d x)}{d}-\frac {17 a^3 \sec (c+d x) \tan (c+d x)}{16 d}-\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {3 a^3 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {a^3 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac {3 a^3 \tan ^5(c+d x)}{5 d}+\frac {1}{16} a^3 \int \sec (c+d x) \, dx\\ &=a^3 x+\frac {19 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {a^3 \tan (c+d x)}{d}-\frac {17 a^3 \sec (c+d x) \tan (c+d x)}{16 d}-\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {3 a^3 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {a^3 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac {3 a^3 \tan ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 1.29, size = 303, normalized size = 1.79 \[ \frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (\sec (c) (210 \sin (2 c+d x)-1440 \sin (c+2 d x)+1200 \sin (3 c+2 d x)-865 \sin (2 c+3 d x)-865 \sin (4 c+3 d x)-1296 \sin (3 c+4 d x)-240 \sin (5 c+4 d x)-435 \sin (4 c+5 d x)-435 \sin (6 c+5 d x)-176 \sin (5 c+6 d x)+2400 d x \cos (c)+1800 d x \cos (c+2 d x)+1800 d x \cos (3 c+2 d x)+720 d x \cos (3 c+4 d x)+720 d x \cos (5 c+4 d x)+120 d x \cos (5 c+6 d x)+120 d x \cos (7 c+6 d x)+1760 \sin (c)+210 \sin (d x))-9120 \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{61440 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^6*(-9120*Cos[c + d*x]^6*(Log[Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*(2400*d*x*Cos[c] + 1800*d*x*Cos[c + 2*d*x]
+ 1800*d*x*Cos[3*c + 2*d*x] + 720*d*x*Cos[3*c + 4*d*x] + 720*d*x*Cos[5*c + 4*d*x] + 120*d*x*Cos[5*c + 6*d*x] +
120*d*x*Cos[7*c + 6*d*x] + 1760*Sin[c] + 210*Sin[d*x] + 210*Sin[2*c + d*x] - 1440*Sin[c + 2*d*x] + 1200*Sin[3
*c + 2*d*x] - 865*Sin[2*c + 3*d*x] - 865*Sin[4*c + 3*d*x] - 1296*Sin[3*c + 4*d*x] - 240*Sin[5*c + 4*d*x] - 435
*Sin[4*c + 5*d*x] - 435*Sin[6*c + 5*d*x] - 176*Sin[5*c + 6*d*x])))/(61440*d)

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fricas [A]  time = 0.82, size = 152, normalized size = 0.90 \[ \frac {480 \, a^{3} d x \cos \left (d x + c\right )^{6} + 285 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 285 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (176 \, a^{3} \cos \left (d x + c\right )^{5} + 435 \, a^{3} \cos \left (d x + c\right )^{4} + 208 \, a^{3} \cos \left (d x + c\right )^{3} - 110 \, a^{3} \cos \left (d x + c\right )^{2} - 144 \, a^{3} \cos \left (d x + c\right ) - 40 \, a^{3}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^4,x, algorithm="fricas")

[Out]

1/480*(480*a^3*d*x*cos(d*x + c)^6 + 285*a^3*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 285*a^3*cos(d*x + c)^6*log(
-sin(d*x + c) + 1) - 2*(176*a^3*cos(d*x + c)^5 + 435*a^3*cos(d*x + c)^4 + 208*a^3*cos(d*x + c)^3 - 110*a^3*cos
(d*x + c)^2 - 144*a^3*cos(d*x + c) - 40*a^3)*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [A]  time = 2.59, size = 164, normalized size = 0.97 \[ \frac {240 \, {\left (d x + c\right )} a^{3} + 285 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 285 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 95 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 366 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1746 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3135 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 525 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^4,x, algorithm="giac")

[Out]

1/240*(240*(d*x + c)*a^3 + 285*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 285*a^3*log(abs(tan(1/2*d*x + 1/2*c) -
1)) - 2*(45*a^3*tan(1/2*d*x + 1/2*c)^11 - 95*a^3*tan(1/2*d*x + 1/2*c)^9 - 366*a^3*tan(1/2*d*x + 1/2*c)^7 + 17
46*a^3*tan(1/2*d*x + 1/2*c)^5 - 3135*a^3*tan(1/2*d*x + 1/2*c)^3 + 525*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x +
1/2*c)^2 - 1)^6)/d

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maple [A]  time = 0.57, size = 193, normalized size = 1.14 \[ \frac {a^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{3} \tan \left (d x +c \right )}{d}+a^{3} x +\frac {a^{3} c}{d}+\frac {19 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{24 d \cos \left (d x +c \right )^{4}}-\frac {19 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{48 d \cos \left (d x +c \right )^{2}}-\frac {19 a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{48 d}-\frac {19 a^{3} \sin \left (d x +c \right )}{16 d}+\frac {19 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {3 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}+\frac {a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*tan(d*x+c)^4,x)

[Out]

1/3*a^3*tan(d*x+c)^3/d-a^3*tan(d*x+c)/d+a^3*x+1/d*a^3*c+19/24/d*a^3*sin(d*x+c)^5/cos(d*x+c)^4-19/48/d*a^3*sin(
d*x+c)^5/cos(d*x+c)^2-19/48*a^3*sin(d*x+c)^3/d-19/16*a^3*sin(d*x+c)/d+19/16/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/
5/d*a^3*sin(d*x+c)^5/cos(d*x+c)^5+1/6/d*a^3*sin(d*x+c)^5/cos(d*x+c)^6

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maxima [A]  time = 0.46, size = 210, normalized size = 1.24 \[ \frac {288 \, a^{3} \tan \left (d x + c\right )^{5} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - 5 \, a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} + 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 90 \, a^{3} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^4,x, algorithm="maxima")

[Out]

1/480*(288*a^3*tan(d*x + c)^5 + 160*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^3 - 5*a^3*(2*(3*sin(d*x
+ c)^5 + 8*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 3*log
(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + 90*a^3*(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 = 2.44, size = 203, normalized size = 1.20 \[ a^3\,x+\frac {19\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}+\frac {-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {19\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {61\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}-\frac {291\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {209\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8}-\frac {35\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4*(a + a/cos(c + d*x))^3,x)

[Out]

a^3*x + (19*a^3*atanh(tan(c/2 + (d*x)/2)))/(8*d) + ((209*a^3*tan(c/2 + (d*x)/2)^3)/8 - (291*a^3*tan(c/2 + (d*x
)/2)^5)/20 + (61*a^3*tan(c/2 + (d*x)/2)^7)/20 + (19*a^3*tan(c/2 + (d*x)/2)^9)/24 - (3*a^3*tan(c/2 + (d*x)/2)^1
1)/8 - (35*a^3*tan(c/2 + (d*x)/2))/8)/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x
)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int 3 \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**3*tan(d*x+c)**4,x)

[Out]

a**3*(Integral(3*tan(c + d*x)**4*sec(c + d*x), x) + Integral(3*tan(c + d*x)**4*sec(c + d*x)**2, x) + Integral(
tan(c + d*x)**4*sec(c + d*x)**3, x) + Integral(tan(c + d*x)**4, x))

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