3.1.49 \(\int (a+a \sec (c+d x))^3 \tan ^2(c+d x) \, dx\) [49]

Optimal. Leaf size=98 \[ -a^3 x-\frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {11 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \tan ^3(c+d x)}{d} \]

[Out]

-a^3*x-13/8*a^3*arctanh(sin(d*x+c))/d+a^3*tan(d*x+c)/d+11/8*a^3*sec(d*x+c)*tan(d*x+c)/d+1/4*a^3*sec(d*x+c)^3*t
an(d*x+c)/d+a^3*tan(d*x+c)^3/d

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Rubi [A]
time = 0.12, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3971, 3554, 8, 2691, 3855, 2687, 30, 3853} \begin {gather*} \frac {a^3 \tan ^3(c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}-\frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {11 a^3 \tan (c+d x) \sec (c+d x)}{8 d}-a^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

-(a^3*x) - (13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*Tan[c + d*x])/d + (11*a^3*Sec[c + d*x]*Tan[c + d*x])/(8
*d) + (a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^3*Tan[c + d*x]^3)/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 2687

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 2691

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 3554

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 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 3971

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 ^2(c+d x) \, dx &=\int \left (a^3 \tan ^2(c+d x)+3 a^3 \sec (c+d x) \tan ^2(c+d x)+3 a^3 \sec ^2(c+d x) \tan ^2(c+d x)+a^3 \sec ^3(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^3 \int \tan ^2(c+d x) \, dx+a^3 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx+\left (3 a^3\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {a^3 \tan (c+d x)}{d}+\frac {3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{4} a^3 \int \sec ^3(c+d x) \, dx-a^3 \int 1 \, dx-\frac {1}{2} \left (3 a^3\right ) \int \sec (c+d x) \, dx+\frac {\left (3 a^3\right ) \text {Subst}\left (\int x^2 \, dx,x,\tan (c+d x)\right )}{d}\\ &=-a^3 x-\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {11 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \tan ^3(c+d x)}{d}-\frac {1}{8} a^3 \int \sec (c+d x) \, dx\\ &=-a^3 x-\frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {11 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 \tan ^3(c+d x)}{d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(230\) vs. \(2(98)=196\).
time = 0.88, size = 230, normalized size = 2.35 \begin {gather*} -\frac {a^3 \sec ^4(c+d x) \left (24 d x-39 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+39 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \cos (2 (c+d x)) \left (8 d x-13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos (4 (c+d x)) \left (8 d x-13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-38 \sin (c+d x)-32 \sin (2 (c+d x))-22 \sin (3 (c+d x))\right )}{64 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

-1/64*(a^3*Sec[c + d*x]^4*(24*d*x - 39*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 39*Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]] + 4*Cos[2*(c + d*x)]*(8*d*x - 13*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 13*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]]) + Cos[4*(c + d*x)]*(8*d*x - 13*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 13*Log[Co
s[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 38*Sin[c + d*x] - 32*Sin[2*(c + d*x)] - 22*Sin[3*(c + d*x)]))/d

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Maple [A]
time = 0.09, size = 157, normalized size = 1.60

method	result	size
risch	\(-a^{3} x -\frac {i a^{3} \left (11 \,{\mathrm e}^{7 i \left (d x +c \right )}+16 \,{\mathrm e}^{6 i \left (d x +c \right )}+19 \,{\mathrm e}^{5 i \left (d x +c \right )}-19 \,{\mathrm e}^{3 i \left (d x +c \right )}-16 \,{\mathrm e}^{2 i \left (d x +c \right )}-11 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(139\)
derivativedivides	\(\frac {a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(157\)
default	\(\frac {a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*tan(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 147, normalized size = 1.50 \begin {gather*} \frac {16 \, a^{3} \tan \left (d x + c\right )^{3} - 16 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} + a^{3} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \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 )} - 12 \, a^{3} {\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 )}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^2,x, algorithm="maxima")

[Out]

1/16*(16*a^3*tan(d*x + c)^3 - 16*(d*x + c - tan(d*x + c))*a^3 + a^3*(2*(sin(d*x + c)^3 + sin(d*x + c))/(sin(d*
x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*a^3*(2*sin(d*x + c)/(si
n(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)))/d

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Fricas [A]
time = 3.12, size = 113, normalized size = 1.15 \begin {gather*} -\frac {16 \, a^{3} d x \cos \left (d x + c\right )^{4} + 13 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 13 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (11 \, a^{3} \cos \left (d x + c\right )^{2} + 8 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^2,x, algorithm="fricas")

[Out]

-1/16*(16*a^3*d*x*cos(d*x + c)^4 + 13*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 13*a^3*cos(d*x + c)^4*log(-si
n(d*x + c) + 1) - 2*(11*a^3*cos(d*x + c)^2 + 8*a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.82, size = 132, normalized size = 1.35 \begin {gather*} -\frac {8 \, {\left (d x + c\right )} a^{3} + 13 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 13 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 13 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, 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 )}^{4}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^2,x, algorithm="giac")

[Out]

-1/8*(8*(d*x + c)*a^3 + 13*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 13*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1))
- 2*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 13*a^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*a^3*tan(
1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 1.91, size = 146, normalized size = 1.49 \begin {gather*} \frac {\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {21\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{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 )}-a^3\,x-\frac {13\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(a + a/cos(c + d*x))^3,x)

[Out]

((3*a^3*tan(c/2 + (d*x)/2)^3)/4 - (13*a^3*tan(c/2 + (d*x)/2)^5)/4 + (5*a^3*tan(c/2 + (d*x)/2)^7)/4 + (21*a^3*t
an(c/2 + (d*x)/2))/4)/(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)) - a^3*x - (13*a^3*atanh(tan(c/2 + (d*x)/2)))/(4*d)

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