Integrand size = 21, antiderivative size = 140 \[ \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx=-\frac {40 d^3 \csc (a+b x)}{21 b \sqrt {d \tan (a+b x)}}-\frac {20 d^3 \csc ^3(a+b x)}{21 b \sqrt {d \tan (a+b x)}}+\frac {40 d^2 \csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{21 b}+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b} \] Output:
-40/21*d^3*csc(b*x+a)/b/(d*tan(b*x+a))^(1/2)-20/21*d^3*csc(b*x+a)^3/b/(d*t an(b*x+a))^(1/2)+40/21*d^2*csc(b*x+a)*InverseJacobiAM(a-1/4*Pi+b*x,2^(1/2) )*sin(2*b*x+2*a)^(1/2)*(d*tan(b*x+a))^(1/2)/b+2/3*d*csc(b*x+a)^5*(d*tan(b* x+a))^(3/2)/b
Result contains complex when optimal does not.
Time = 1.52 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.93 \[ \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx=-\frac {d^2 \csc (a+b x) \left ((1+10 \cos (2 (a+b x))-5 \cos (4 (a+b x))) \csc ^3(a+b x) \sec (a+b x) \sqrt {\sec ^2(a+b x)}+80 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sqrt {\tan (a+b x)}\right ) \sqrt {d \tan (a+b x)}}{21 b \sqrt {\sec ^2(a+b x)}} \] Input:
Integrate[Csc[a + b*x]^7*(d*Tan[a + b*x])^(5/2),x]
Output:
-1/21*(d^2*Csc[a + b*x]*((1 + 10*Cos[2*(a + b*x)] - 5*Cos[4*(a + b*x)])*Cs c[a + b*x]^3*Sec[a + b*x]*Sqrt[Sec[a + b*x]^2] + 80*(-1)^(1/4)*EllipticF[I *ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Sqrt[Tan[a + b*x]])*Sqrt[d*Ta n[a + b*x]])/(b*Sqrt[Sec[a + b*x]^2])
Time = 0.69 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3073, 3042, 3079, 3042, 3079, 3042, 3081, 3042, 3053, 3042, 3120}
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.
\(\displaystyle \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \tan (a+b x))^{5/2}}{\sin (a+b x)^7}dx\) |
\(\Big \downarrow \) 3073 |
\(\displaystyle \frac {10}{3} d^2 \int \csc ^5(a+b x) \sqrt {d \tan (a+b x)}dx+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {10}{3} d^2 \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)^5}dx+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3079 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \int \csc ^3(a+b x) \sqrt {d \tan (a+b x)}dx-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)^3}dx-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3079 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \left (\frac {2}{3} \int \csc (a+b x) \sqrt {d \tan (a+b x)}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \left (\frac {2}{3} \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \left (\frac {2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\sin (a+b x)}}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \left (\frac {2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\sin (a+b x)}}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \left (\frac {2}{3} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \left (\frac {2}{3} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {10}{3} d^2 \left (\frac {6}{7} \left (\frac {2 \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{3 b}-\frac {2 d \csc (a+b x)}{3 b \sqrt {d \tan (a+b x)}}\right )-\frac {2 d \csc ^3(a+b x)}{7 b \sqrt {d \tan (a+b x)}}\right )+\frac {2 d \csc ^5(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\) |
Input:
Int[Csc[a + b*x]^7*(d*Tan[a + b*x])^(5/2),x]
Output:
(2*d*Csc[a + b*x]^5*(d*Tan[a + b*x])^(3/2))/(3*b) + (10*d^2*((-2*d*Csc[a + b*x]^3)/(7*b*Sqrt[d*Tan[a + b*x]]) + (6*((-2*d*Csc[a + b*x])/(3*b*Sqrt[d* Tan[a + b*x]]) + (2*Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(3*b)))/7))/3
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f }, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1)/ (a^2*f*(n - 1))), x] - Simp[b^2*((m + 2)/(a^2*(n - 1))) Int[(a*Sin[e + f* x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && G tQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2])) && IntegersQ[2*m, 2 *n]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1) /(a^2*f*(m + n + 1))), x] + Simp[(m + 2)/(a^2*(m + n + 1)) Int[(a*Sin[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && L tQ[m, -1] && NeQ[m + n + 1, 0] && IntegersQ[2*m, 2*n]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Time = 231.62 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {\left (-\frac {2 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right ) \left (-20 \cos \left (b x +a \right )-20\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{21}+\frac {40 \cos \left (b x +a \right )^{4}}{21}-\frac {20 \cos \left (b x +a \right )^{2}}{7}+\frac {2}{3}\right ) \sec \left (b x +a \right ) \csc \left (b x +a \right )^{4} \sqrt {d \tan \left (b x +a \right )}\, d^{2}}{b}\) | \(159\) |
Input:
int(csc(b*x+a)^7*(d*tan(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/b*(-2/21*sin(b*x+a)^3*cos(b*x+a)*(-20*cos(b*x+a)-20)*(csc(b*x+a)-cot(b*x +a)+1)^(1/2)*(-2*csc(b*x+a)+2*cot(b*x+a)+2)^(1/2)*(-csc(b*x+a)+cot(b*x+a)) ^(1/2)*EllipticF((csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))+40/21*cos(b* x+a)^4-20/7*cos(b*x+a)^2+2/3)*sec(b*x+a)*csc(b*x+a)^4*(d*tan(b*x+a))^(1/2) *d^2
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.49 \[ \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx=-\frac {2 \, {\left (20 \, {\left (d^{2} \cos \left (b x + a\right )^{5} - 2 \, d^{2} \cos \left (b x + a\right )^{3} + d^{2} \cos \left (b x + a\right )\right )} \sqrt {i \, d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 20 \, {\left (d^{2} \cos \left (b x + a\right )^{5} - 2 \, d^{2} \cos \left (b x + a\right )^{3} + d^{2} \cos \left (b x + a\right )\right )} \sqrt {-i \, d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - {\left (20 \, d^{2} \cos \left (b x + a\right )^{4} - 30 \, d^{2} \cos \left (b x + a\right )^{2} + 7 \, d^{2}\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{21 \, {\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \] Input:
integrate(csc(b*x+a)^7*(d*tan(b*x+a))^(5/2),x, algorithm="fricas")
Output:
-2/21*(20*(d^2*cos(b*x + a)^5 - 2*d^2*cos(b*x + a)^3 + d^2*cos(b*x + a))*s qrt(I*d)*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) + 20*(d^2*c os(b*x + a)^5 - 2*d^2*cos(b*x + a)^3 + d^2*cos(b*x + a))*sqrt(-I*d)*ellipt ic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - (20*d^2*cos(b*x + a)^4 - 30*d^2*cos(b*x + a)^2 + 7*d^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(b*cos( b*x + a)^5 - 2*b*cos(b*x + a)^3 + b*cos(b*x + a))
Timed out. \[ \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)**7*(d*tan(b*x+a))**(5/2),x)
Output:
Timed out
\[ \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right )^{7} \,d x } \] Input:
integrate(csc(b*x+a)^7*(d*tan(b*x+a))^(5/2),x, algorithm="maxima")
Output:
integrate((d*tan(b*x + a))^(5/2)*csc(b*x + a)^7, x)
\[ \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right )^{7} \,d x } \] Input:
integrate(csc(b*x+a)^7*(d*tan(b*x+a))^(5/2),x, algorithm="giac")
Output:
integrate((d*tan(b*x + a))^(5/2)*csc(b*x + a)^7, x)
Timed out. \[ \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}}{{\sin \left (a+b\,x\right )}^7} \,d x \] Input:
int((d*tan(a + b*x))^(5/2)/sin(a + b*x)^7,x)
Output:
int((d*tan(a + b*x))^(5/2)/sin(a + b*x)^7, x)
\[ \int \csc ^7(a+b x) (d \tan (a+b x))^{5/2} \, dx=\sqrt {d}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \csc \left (b x +a \right )^{7} \tan \left (b x +a \right )^{2}d x \right ) d^{2} \] Input:
int(csc(b*x+a)^7*(d*tan(b*x+a))^(5/2),x)
Output:
sqrt(d)*int(sqrt(tan(a + b*x))*csc(a + b*x)**7*tan(a + b*x)**2,x)*d**2