Integrand size = 23, antiderivative size = 73 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}+\frac {\sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{2 d} \] Output:
1/4*2^(1/2)*a^(3/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d+ 1/2*sec(d*x+c)^2*(a+a*sin(d*x+c))^(3/2)/d
Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {a \left (\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right )-\frac {2 \sqrt {a (1+\sin (c+d x))}}{-1+\sin (c+d x)}\right )}{4 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2),x]
Output:
(a*(Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])] - (2*Sqrt[a*(1 + Sin[c + d*x])])/(-1 + Sin[c + d*x])))/(4*d)
Time = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3154, 3042, 3146, 73, 219}
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.
\(\displaystyle \int \sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{3/2}}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3154 |
\(\displaystyle \frac {1}{4} a \int \sec (c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\cos (c+d x)}dx+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a^2 \int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {a^2 \int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{2 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{2 \sqrt {2} d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\) |
Input:
Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2),x]
Output:
(a^(3/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[2]])/(2*Sqrt[2]*d) + (Sec[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2))/(2*d)
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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))), x] + Simp[a*((m + p + 1)/(g^2*(p + 1))) Int[(g* Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ [m + 1/2, 2*p]
Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {2 a^{3} \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 a \left (-a +a \sin \left (d x +c \right )\right )}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{d}\) | \(70\) |
Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
2*a^3*(-1/4*(a+a*sin(d*x+c))^(1/2)/a/(-a+a*sin(d*x+c))+1/8/a^(3/2)*2^(1/2) *arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))/d
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.32 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (a \sin \left (d x + c\right ) - a\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 4 \, \sqrt {\frac {1}{2}} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} a}{4 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/4*(sqrt(1/2)*(a*sin(d*x + c) - a)*sqrt(a)*log(-(a*sin(d*x + c) + 4*sqrt( 1/2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) - 2*sqrt( a*sin(d*x + c) + a)*a)/(d*sin(d*x + c) - d)
Timed out. \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))**(3/2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.29 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{3}}{a \sin \left (d x + c\right ) - a}}{8 \, a d} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
Output:
-1/8*(sqrt(2)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(s qrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a))) + 4*sqrt(a*sin(d*x + c) + a)*a ^3/(a*sin(d*x + c) - a))/(a*d)
Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} a^{\frac {3}{2}} {\left (\frac {2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{8 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
Output:
-1/8*sqrt(2)*a^(3/2)*(2*cos(-1/4*pi + 1/2*d*x + 1/2*c)/(cos(-1/4*pi + 1/2* d*x + 1/2*c)^2 - 1) - log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + log(-cos(- 1/4*pi + 1/2*d*x + 1/2*c) + 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d
Timed out. \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:
int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^3,x)
Output:
int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^3, x)
\[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3} \sin \left (d x +c \right )d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) \] Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*sec(c + d*x)**3*sin(c + d*x),x) + in t(sqrt(sin(c + d*x) + 1)*sec(c + d*x)**3,x))