Integrand size = 25, antiderivative size = 166 \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{3/2} \, dx=\frac {24 c d^5 \sqrt {c \sec (a+b x)}}{5 b (d \csc (a+b x))^{3/2}}-\frac {12 c d^3 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}{5 b}-\frac {2 c d (d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}}{5 b}-\frac {24 c^2 d^4 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{5 b \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}} \]
24/5*c*d^5*(c*sec(b*x+a))^(1/2)/b/(d*csc(b*x+a))^(3/2)-2/5*c*d*(d*csc(b*x+ a))^(5/2)*(c*sec(b*x+a))^(1/2)/b-12/5*c*d^3*(d*csc(b*x+a))^(1/2)*(c*sec(b* x+a))^(1/2)/b+24/5*c^2*d^4*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*E llipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b/(d*csc(b*x+a))^(1/2)/(c*sec(b*x+a))^ (1/2)/sin(2*b*x+2*a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 11.95 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.69 \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{3/2} \, dx=-\frac {2 c d^3 \sqrt {d \csc (a+b x)} \left (\cot ^2(a+b x) \left (6 \cos (2 (a+b x))+\csc ^2(a+b x)\right )+12 \cos ^2(a+b x) \sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\csc ^2(a+b x)\right )\right ) \sqrt {c \sec (a+b x)} \tan ^2(a+b x)}{5 b} \]
(-2*c*d^3*Sqrt[d*Csc[a + b*x]]*(Cot[a + b*x]^2*(6*Cos[2*(a + b*x)] + Csc[a + b*x]^2) + 12*Cos[a + b*x]^2*(-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[- 1/2, 1/4, 1/2, Csc[a + b*x]^2])*Sqrt[c*Sec[a + b*x]]*Tan[a + b*x]^2)/(5*b)
Time = 0.86 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3105, 3042, 3105, 3042, 3106, 3042, 3110, 3042, 3052, 3042, 3119}
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.
\(\displaystyle \int (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{7/2}dx\) |
\(\Big \downarrow \) 3105 |
\(\displaystyle \frac {6}{5} d^2 \int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}dx-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{5} d^2 \int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}dx-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3105 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \int \frac {(c \sec (a+b x))^{3/2}}{\sqrt {d \csc (a+b x)}}dx-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \int \frac {(c \sec (a+b x))^{3/2}}{\sqrt {d \csc (a+b x)}}dx-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3106 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \left (\frac {2 c d \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-2 c^2 \int \frac {1}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}dx\right )-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \left (\frac {2 c d \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-2 c^2 \int \frac {1}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}dx\right )-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3110 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \left (\frac {2 c d \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c^2 \int \sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}dx}{\sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)}}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \left (\frac {2 c d \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c^2 \int \sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}dx}{\sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)}}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3052 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \left (\frac {2 c d \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c^2 \int \sqrt {\sin (2 a+2 b x)}dx}{\sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \left (\frac {2 c d \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c^2 \int \sqrt {\sin (2 a+2 b x)}dx}{\sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {6}{5} d^2 \left (2 d^2 \left (\frac {2 c d \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c^2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b}\right )-\frac {2 c d \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}{5 b}\) |
(-2*c*d*(d*Csc[a + b*x])^(5/2)*Sqrt[c*Sec[a + b*x]])/(5*b) + (6*d^2*((-2*c *d*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]])/b + 2*d^2*((2*c*d*Sqrt[c*Sec [a + b*x]])/(b*(d*Csc[a + b*x])^(3/2)) - (2*c^2*EllipticE[a - Pi/4 + b*x, 2])/(b*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin[2*a + 2*b*x]]))) )/5
3.3.39.3.1 Defintions of rubi rules used
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] , x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]) Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Simp[(-a)*b*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(m - 1))), x] + Simp[a^2*((m + n - 2)/(m - 1)) Int[(a*Csc[e + f* x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[ m, 1] && IntegersQ[2*m, 2*n] && !GtQ[n, m]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a*b*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(n - 1))), x] + Simp[b^2*((m + n - 2)/(n - 1)) Int[(a*Csc[e + f*x]) ^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(a*Csc[e + f*x])^m*(b*Sec[e + f*x])^n*(a*Sin[e + f*x] )^m*(b*Cos[e + f*x])^n Int[1/((a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[m - 1/2] && IntegerQ[n - 1/ 2]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Time = 1.15 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.40
method | result | size |
default | \(-\frac {\sqrt {2}\, d^{3} c \sqrt {c \sec \left (b x +a \right )}\, \sqrt {d \csc \left (b x +a \right )}\, \left (\left (-24 \cos \left (b x +a \right )-24\right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+\left (12 \cos \left (b x +a \right )+12\right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \left (12 \cos \left (b x +a \right )-6+\csc \left (b x +a \right )^{2}\right )\right )}{5 b}\) | \(232\) |
-1/5/b*2^(1/2)*d^3*c*(c*sec(b*x+a))^(1/2)*(d*csc(b*x+a))^(1/2)*((-24*cos(b *x+a)-24)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)* (cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticE((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/ 2*2^(1/2))+(12*cos(b*x+a)+12)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)- csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)- cot(b*x+a))^(1/2),1/2*2^(1/2))+2^(1/2)*(12*cos(b*x+a)-6+csc(b*x+a)^2))
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.59 \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{3/2} \, dx=\frac {2 \, {\left (3 \, {\left (c d^{3} \cos \left (b x + a\right )^{2} - c d^{3}\right )} \sqrt {-4 i \, c d} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 3 \, {\left (c d^{3} \cos \left (b x + a\right )^{2} - c d^{3}\right )} \sqrt {4 i \, c d} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - 3 \, {\left (c d^{3} \cos \left (b x + a\right )^{2} - c d^{3}\right )} \sqrt {-4 i \, c d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - 3 \, {\left (c d^{3} \cos \left (b x + a\right )^{2} - c d^{3}\right )} \sqrt {4 i \, c d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - {\left (12 \, c d^{3} \cos \left (b x + a\right )^{4} - 18 \, c d^{3} \cos \left (b x + a\right )^{2} + 5 \, c d^{3}\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}}\right )}}{5 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]
2/5*(3*(c*d^3*cos(b*x + a)^2 - c*d^3)*sqrt(-4*I*c*d)*elliptic_e(arcsin(cos (b*x + a) + I*sin(b*x + a)), -1) + 3*(c*d^3*cos(b*x + a)^2 - c*d^3)*sqrt(4 *I*c*d)*elliptic_e(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - 3*(c*d^3*c os(b*x + a)^2 - c*d^3)*sqrt(-4*I*c*d)*elliptic_f(arcsin(cos(b*x + a) + I*s in(b*x + a)), -1) - 3*(c*d^3*cos(b*x + a)^2 - c*d^3)*sqrt(4*I*c*d)*ellipti c_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - (12*c*d^3*cos(b*x + a)^4 - 18*c*d^3*cos(b*x + a)^2 + 5*c*d^3)*sqrt(c/cos(b*x + a))*sqrt(d/sin(b*x + a)))/(b*cos(b*x + a)^2 - b)
Timed out. \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{3/2} \, dx=\text {Timed out} \]
\[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{3/2} \, dx=\int { \left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac {3}{2}} \,d x } \]
\[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{3/2} \, dx=\int { \left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{3/2} \, dx=\int {\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{3/2}\,{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{7/2} \,d x \]