Integrand size = 23, antiderivative size = 127 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {15 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d}-\frac {15 a^2}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {5 a \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d} \]
1/4*sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2)/d+15/64*a^(3/2)*arctanh(1/2*(a+a*s in(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-15/32*a^2/d/(a+a*sin(d*x+c))^( 1/2)+5/16*a*sec(d*x+c)^2*(a+a*sin(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.35 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{4 d \sqrt {a+a \sin (c+d x)}} \]
-1/4*(a^2*Hypergeometric2F1[-1/2, 3, 1/2, (1 + Sin[c + d*x])/2])/(d*Sqrt[a + a*Sin[c + d*x]])
Time = 0.51 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3154, 3042, 3154, 3042, 3146, 61, 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.
\(\displaystyle \int \sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{3/2}}{\cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3154 |
\(\displaystyle \frac {5}{8} a \int \sec ^3(c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{8} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\cos (c+d x)^3}dx+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3154 |
\(\displaystyle \frac {5}{8} a \left (\frac {3}{4} a \int \frac {\sec (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{8} a \left (\frac {3}{4} a \int \frac {1}{\cos (c+d x) \sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {5}{8} a \left (\frac {3 a^2 \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{2 a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\) |
(Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2))/(4*d) + (5*a*((Sec[c + d*x]^2* Sqrt[a + a*Sin[c + d*x]])/(2*d) + (3*a^2*(ArcTanh[(Sqrt[a]*Sin[c + d*x])/S qrt[2]]/(Sqrt[2]*a^(3/2)) - 1/(a*Sqrt[a + a*Sin[c + d*x]])))/(4*d)))/8
3.2.25.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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[(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 = 0.54 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80
method | result | size |
default | \(-\frac {2 a^{5} \left (\frac {1}{8 a^{3} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a \left (7 \sin \left (d x +c \right )-11\right )}{8 \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}}{8 a^{3}}\right )}{d}\) | \(101\) |
-2*a^5*(1/8/a^3/(a+a*sin(d*x+c))^(1/2)+1/8/a^3*(1/8*(a+a*sin(d*x+c))^(1/2) *a*(7*sin(d*x+c)-11)/(a*sin(d*x+c)-a)^2-15/16*2^(1/2)/a^(1/2)*arctanh(1/2* (a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))))/d
Time = 0.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.22 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {15 \, {\left (\sqrt {2} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - \sqrt {2} a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (15 \, a \cos \left (d x + c\right )^{2} + 20 \, a \sin \left (d x + c\right ) - 12 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{128 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]
1/128*(15*(sqrt(2)*a*cos(d*x + c)^2*sin(d*x + c) - sqrt(2)*a*cos(d*x + c)^ 2)*sqrt(a)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt( a) + 3*a)/(sin(d*x + c) - 1)) - 4*(15*a*cos(d*x + c)^2 + 20*a*sin(d*x + c) - 12*a)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos( d*x + c)^2)
Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.19 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {15 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{3} - 50 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{4} + 32 \, a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{2}}}{128 \, a d} \]
-1/128*(15*sqrt(2)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a ))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a))) + 4*(15*(a*sin(d*x + c) + a)^2*a^3 - 50*(a*sin(d*x + c) + a)*a^4 + 32*a^5)/((a*sin(d*x + c) + a)^(5 /2) - 4*(a*sin(d*x + c) + a)^(3/2)*a + 4*sqrt(a*sin(d*x + c) + a)*a^2))/(a *d)
Time = 0.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.01 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} a^{\frac {3}{2}} {\left (\frac {2 \, {\left (7 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {16}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - 15 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 15 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{128 \, d} \]
-1/128*sqrt(2)*a^(3/2)*(2*(7*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 9*cos(-1/4 *pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2 + 16/cos( -1/4*pi + 1/2*d*x + 1/2*c) - 15*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 15*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. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]