\(\int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx\) [136]

Optimal result

Integrand size = 23, antiderivative size = 99 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3965, 81, 52, 65, 213} \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a^2 d}+\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}-\frac {2 (a \sec (c+d x)+a)^{3/2}}{3 a d}-\frac {2 \sqrt {a \sec (c+d x)+a}}{d} \]

[In]

Int[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^3,x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 3965

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(d*b^(m - 1)
)^(-1), Subst[Int[(-a + b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a,
b, c, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(-a+a x) (a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d} \\ & = \frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac {\text {Subst}\left (\int \frac {(a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{a d} \\ & = -\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac {\text {Subst}\left (\int \frac {\sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {2 \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac {a \text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {2 \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac {2 \text {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d} \\ & = \frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.81 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\frac {2 \sqrt {a (1+\sec (c+d x))} \left (15 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} \left (-17+\sec (c+d x)+3 \sec ^2(c+d x)\right )\right )}{15 d \sqrt {1+\sec (c+d x)}} \]

[In]

Integrate[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^3,x]

[Out]

(2*Sqrt[a*(1 + Sec[c + d*x])]*(15*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqrt[1 + Sec[c + d*x]]*(-17 + Sec[c + d*x]
+ 3*Sec[c + d*x]^2)))/(15*d*Sqrt[1 + Sec[c + d*x]])

Maple [A] (verified)

Time = 4.74 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (15 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+17-\sec \left (d x +c \right )-3 \sec \left (d x +c \right )^{2}\right )}{15 d}\)	\(81\)

[In]

int((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

-2/15/d*(a*(1+sec(d*x+c)))^(1/2)*(15*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)+17-sec(d*x+c)-3*sec(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.62 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\left [\frac {15 \, \sqrt {a} \cos \left (d x + c\right )^{2} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, {\left (17 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{30 \, d \cos \left (d x + c\right )^{2}}, -\frac {15 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{2} + 2 \, {\left (17 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}}\right ] \]

[In]

integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

[1/30*(15*sqrt(a)*cos(d*x + c)^2*log(-8*a*cos(d*x + c)^2 - 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt((a
*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) - 4*(17*cos(d*x + c)^2 - cos(d*x + c) - 3)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2), -1/15*(15*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + a))*cos(d*x + c)^2 + 2*(17*cos(d*x + c)^2 - cos(d*x + c) -
3)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2)]

Sympy [F]

\[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \tan ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate((a+a*sec(d*x+c))**(1/2)*tan(d*x+c)**3,x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*tan(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=-\frac {15 \, \sqrt {a} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 30 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \frac {6 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a^{2}} + \frac {10 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a}}{15 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/15*(15*sqrt(a)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a))) + 30*sqrt(a +
a/cos(d*x + c)) - 6*(a + a/cos(d*x + c))^(5/2)/a^2 + 10*(a + a/cos(d*x + c))^(3/2)/a)/d

Giac [F]

\[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \tan \left (d x + c\right )^{3} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]

[In]

int(tan(c + d*x)^3*(a + a/cos(c + d*x))^(1/2),x)

[Out]

int(tan(c + d*x)^3*(a + a/cos(c + d*x))^(1/2), x)