Integrand size = 25, antiderivative size = 106 \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx=-\frac {64 c^3 d^3 \sqrt {d \csc (a+b x)}}{15 b \sqrt {c \sec (a+b x)}}+\frac {16 c d^3 \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}{15 b}-\frac {2 c d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{3/2}}{5 b} \] Output:
-64/15*c^3*d^3*(d*csc(b*x+a))^(1/2)/b/(c*sec(b*x+a))^(1/2)+16/15*c*d^3*(d* csc(b*x+a))^(1/2)*(c*sec(b*x+a))^(3/2)/b-2/5*c*d*(d*csc(b*x+a))^(5/2)*(c*s ec(b*x+a))^(3/2)/b
Time = 0.49 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.54 \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx=-\frac {2 c d^3 \left (-5+32 \cos ^2(a+b x)+3 \cot ^2(a+b x)\right ) \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}{15 b} \] Input:
Integrate[(d*Csc[a + b*x])^(7/2)*(c*Sec[a + b*x])^(5/2),x]
Output:
(-2*c*d^3*(-5 + 32*Cos[a + b*x]^2 + 3*Cot[a + b*x]^2)*Sqrt[d*Csc[a + b*x]] *(c*Sec[a + b*x])^(3/2))/(15*b)
Time = 0.47 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3105, 3042, 3106, 3042, 3099}
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.
\(\displaystyle \int (c \sec (a+b x))^{5/2} (d \csc (a+b x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c \sec (a+b x))^{5/2} (d \csc (a+b x))^{7/2}dx\) |
\(\Big \downarrow \) 3105 |
\(\displaystyle \frac {8}{5} d^2 \int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}dx-\frac {2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{5} d^2 \int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}dx-\frac {2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3106 |
\(\displaystyle \frac {8}{5} d^2 \left (\frac {4}{3} c^2 \int (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}dx+\frac {2 c d (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}{3 b}\right )-\frac {2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{5} d^2 \left (\frac {4}{3} c^2 \int (d \csc (a+b x))^{3/2} \sqrt {c \sec (a+b x)}dx+\frac {2 c d (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}{3 b}\right )-\frac {2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{5/2}}{5 b}\) |
\(\Big \downarrow \) 3099 |
\(\displaystyle \frac {8}{5} d^2 \left (\frac {2 c d (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}{3 b}-\frac {8 c^3 d \sqrt {d \csc (a+b x)}}{3 b \sqrt {c \sec (a+b x)}}\right )-\frac {2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{5/2}}{5 b}\) |
Input:
Int[(d*Csc[a + b*x])^(7/2)*(c*Sec[a + b*x])^(5/2),x]
Output:
(-2*c*d*(d*Csc[a + b*x])^(5/2)*(c*Sec[a + b*x])^(3/2))/(5*b) + (8*d^2*((-8 *c^3*d*Sqrt[d*Csc[a + b*x]])/(3*b*Sqrt[c*Sec[a + b*x]]) + (2*c*d*Sqrt[d*Cs c[a + b*x]]*(c*Sec[a + b*x])^(3/2))/(3*b)))/5
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] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 2, 0] && NeQ[n, 1]
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]
Time = 0.76 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64
\[\frac {2 \left (32 \cos \left (b x +a \right )^{4}-40 \cos \left (b x +a \right )^{2}+5\right ) \sqrt {c \sec \left (b x +a \right )}\, c^{2} d^{3} \sqrt {d \csc \left (b x +a \right )}\, \sec \left (b x +a \right ) \csc \left (b x +a \right )^{2}}{15 b}\]
Input:
int((d*csc(b*x+a))^(7/2)*(c*sec(b*x+a))^(5/2),x)
Output:
2/15/b*(32*cos(b*x+a)^4-40*cos(b*x+a)^2+5)*(c*sec(b*x+a))^(1/2)*c^2*d^3*(d *csc(b*x+a))^(1/2)*sec(b*x+a)*csc(b*x+a)^2
Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.84 \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx=-\frac {2 \, {\left (32 \, c^{2} d^{3} \cos \left (b x + a\right )^{4} - 40 \, c^{2} d^{3} \cos \left (b x + a\right )^{2} + 5 \, c^{2} d^{3}\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}}}{15 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \] Input:
integrate((d*csc(b*x+a))^(7/2)*(c*sec(b*x+a))^(5/2),x, algorithm="fricas")
Output:
-2/15*(32*c^2*d^3*cos(b*x + a)^4 - 40*c^2*d^3*cos(b*x + a)^2 + 5*c^2*d^3)* sqrt(c/cos(b*x + a))*sqrt(d/sin(b*x + a))/(b*cos(b*x + a)^3 - b*cos(b*x + a))
Timed out. \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate((d*csc(b*x+a))**(7/2)*(c*sec(b*x+a))**(5/2),x)
Output:
Timed out
\[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx=\int { \left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((d*csc(b*x+a))^(7/2)*(c*sec(b*x+a))^(5/2),x, algorithm="maxima")
Output:
integrate((d*csc(b*x + a))^(7/2)*(c*sec(b*x + a))^(5/2), x)
\[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx=\int { \left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((d*csc(b*x+a))^(7/2)*(c*sec(b*x+a))^(5/2),x, algorithm="giac")
Output:
integrate((d*csc(b*x + a))^(7/2)*(c*sec(b*x + a))^(5/2), x)
Time = 11.45 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06 \[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx=\frac {16\,c^2\,d^3\,\sqrt {\frac {c}{\cos \left (a+b\,x\right )}}\,\sqrt {\frac {d}{\sin \left (a+b\,x\right )}}\,\left (5\,\cos \left (a+b\,x\right )-3\,\cos \left (3\,a+3\,b\,x\right )-4\,\cos \left (5\,a+5\,b\,x\right )+2\,\cos \left (7\,a+7\,b\,x\right )\right )}{15\,b\,\left (\cos \left (2\,a+2\,b\,x\right )+2\,\cos \left (4\,a+4\,b\,x\right )-\cos \left (6\,a+6\,b\,x\right )-2\right )} \] Input:
int((c/cos(a + b*x))^(5/2)*(d/sin(a + b*x))^(7/2),x)
Output:
(16*c^2*d^3*(c/cos(a + b*x))^(1/2)*(d/sin(a + b*x))^(1/2)*(5*cos(a + b*x) - 3*cos(3*a + 3*b*x) - 4*cos(5*a + 5*b*x) + 2*cos(7*a + 7*b*x)))/(15*b*(co s(2*a + 2*b*x) + 2*cos(4*a + 4*b*x) - cos(6*a + 6*b*x) - 2))
\[ \int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx=\sqrt {d}\, \sqrt {c}\, \left (\int \sqrt {\sec \left (b x +a \right )}\, \sqrt {\csc \left (b x +a \right )}\, \csc \left (b x +a \right )^{3} \sec \left (b x +a \right )^{2}d x \right ) c^{2} d^{3} \] Input:
int((d*csc(b*x+a))^(7/2)*(c*sec(b*x+a))^(5/2),x)
Output:
sqrt(d)*sqrt(c)*int(sqrt(sec(a + b*x))*sqrt(csc(a + b*x))*csc(a + b*x)**3* sec(a + b*x)**2,x)*c**2*d**3