Integrand size = 21, antiderivative size = 73 \[ \int (a+a \sec (c+d x))^{3/2} \tan (c+d x) \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d} \] Output:
-2*a^(3/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d+2*a*(a+a*sec(d*x+c))^ (1/2)/d+2/3*(a+a*sec(d*x+c))^(3/2)/d
Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int (a+a \sec (c+d x))^{3/2} \tan (c+d x) \, dx=\frac {2 (a (1+\sec (c+d x)))^{3/2} \left (-3 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} (4+\sec (c+d x))\right )}{3 d (1+\sec (c+d x))^{3/2}} \] Input:
Integrate[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x],x]
Output:
(2*(a*(1 + Sec[c + d*x]))^(3/2)*(-3*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqrt [1 + Sec[c + d*x]]*(4 + Sec[c + d*x])))/(3*d*(1 + Sec[c + d*x])^(3/2))
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 25, 4368, 60, 60, 73, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \tan (c+d x) (a \sec (c+d x)+a)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{3/2}dx\) |
\(\Big \downarrow \) 4368 |
\(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a)^{3/2}d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}d\sec (c+d x)+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \left (a \int \frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+2 \sqrt {a \sec (c+d x)+a}\right )+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {a \left (2 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}+2 \sqrt {a \sec (c+d x)+a}\right )+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}}{d}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {a \left (2 \sqrt {a \sec (c+d x)+a}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )\right )+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}}{d}\) |
Input:
Int[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x],x]
Output:
((2*(a + a*Sec[c + d*x])^(3/2))/3 + a*(-2*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]] + 2*Sqrt[a + a*Sec[c + d*x]]))/d
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(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]
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {a +a \sec \left (d x +c \right )}\, a -2 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\) | \(57\) |
default | \(\frac {\frac {2 \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {a +a \sec \left (d x +c \right )}\, a -2 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\) | \(57\) |
Input:
int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/d*(2/3*(a+a*sec(d*x+c))^(3/2)+2*(a+a*sec(d*x+c))^(1/2)*a-2*a^(3/2)*arcta nh((a+a*sec(d*x+c))^(1/2)/a^(1/2)))
Time = 0.13 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.26 \[ \int (a+a \sec (c+d x))^{3/2} \tan (c+d x) \, dx=\left [\frac {3 \, a^{\frac {3}{2}} \cos \left (d x + c\right ) \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (4 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{6 \, d \cos \left (d x + c\right )}, \frac {3 \, \sqrt {-a} a \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right ) + 2 \, {\left (4 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{3 \, d \cos \left (d x + c\right )}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c),x, algorithm="fricas")
Output:
[1/6*(3*a^(3/2)*cos(d*x + c)*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*(4*a*cos(d*x + c) + a)*sqrt((a*cos(d*x + c) + a)/cos(d* x + c)))/(d*cos(d*x + c)), 1/3*(3*sqrt(-a)*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*(4*a*cos(d*x + c) + a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/( d*cos(d*x + c))]
\[ \int (a+a \sec (c+d x))^{3/2} \tan (c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \tan {\left (c + d x \right )}\, dx \] Input:
integrate((a+a*sec(d*x+c))**(3/2)*tan(d*x+c),x)
Output:
Integral((a*(sec(c + d*x) + 1))**(3/2)*tan(c + d*x), x)
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.18 \[ \int (a+a \sec (c+d x))^{3/2} \tan (c+d x) \, dx=\frac {3 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 2 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} + 6 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a}{3 \, d} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c),x, algorithm="maxima")
Output:
1/3*(3*a^(3/2)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d* x + c)) + sqrt(a))) + 2*(a + a/cos(d*x + c))^(3/2) + 6*sqrt(a + a/cos(d*x + c))*a)/d
Time = 0.47 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.67 \[ \int (a+a \sec (c+d x))^{3/2} \tan (c+d x) \, dx=\frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a - 2 \, a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{3 \, d} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c),x, algorithm="giac")
Output:
1/3*sqrt(2)*(3*sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) + 2*(3*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a - 2*a^2)/ ((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))*a*sg n(cos(d*x + c))/d
Time = 12.78 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.92 \[ \int (a+a \sec (c+d x))^{3/2} \tan (c+d x) \, dx=\frac {2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{3\,d}-\frac {2\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{\sqrt {a}}\right )}{d}+\frac {2\,a\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{d} \] Input:
int(tan(c + d*x)*(a + a/cos(c + d*x))^(3/2),x)
Output:
(2*(a + a/cos(c + d*x))^(3/2))/(3*d) - (2*a^(3/2)*atanh((a + a/cos(c + d*x ))^(1/2)/a^(1/2)))/d + (2*a*(a + a/cos(c + d*x))^(1/2))/d
\[ \int (a+a \sec (c+d x))^{3/2} \tan (c+d x) \, dx=\frac {\sqrt {a}\, a \left (2 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )+8 \sqrt {\sec \left (d x +c \right )+1}+3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) d \right )}{3 d} \] Input:
int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c),x)
Output:
(sqrt(a)*a*(2*sqrt(sec(c + d*x) + 1)*sec(c + d*x) + 8*sqrt(sec(c + d*x) + 1) + 3*int((sqrt(sec(c + d*x) + 1)*tan(c + d*x))/(sec(c + d*x) + 1),x)*d)) /(3*d)