Integrand size = 23, antiderivative size = 75 \[ \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx=\frac {d \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(a+b x)\right ) \sqrt {d \sec (a+b x)} (c \sin (a+b x))^{1+m}}{b c (1+m)} \] Output:
d*(cos(b*x+a)^2)^(1/4)*hypergeom([5/4, 1/2+1/2*m],[3/2+1/2*m],sin(b*x+a)^2 )*(d*sec(b*x+a))^(1/2)*(c*sin(b*x+a))^(1+m)/b/c/(1+m)
Time = 8.86 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.28 \[ \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx=-\frac {2 \cot (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (3-2 m),\frac {1-m}{2},\frac {1}{4} (7-2 m),\sec ^2(a+b x)\right ) (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \left (-\tan ^2(a+b x)\right )^{\frac {1-m}{2}}}{b (-3+2 m)} \] Input:
Integrate[(d*Sec[a + b*x])^(3/2)*(c*Sin[a + b*x])^m,x]
Output:
(-2*Cot[a + b*x]*Hypergeometric2F1[(3 - 2*m)/4, (1 - m)/2, (7 - 2*m)/4, Se c[a + b*x]^2]*(d*Sec[a + b*x])^(3/2)*(c*Sin[a + b*x])^m*(-Tan[a + b*x]^2)^ ((1 - m)/2))/(b*(-3 + 2*m))
Time = 0.57 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3067, 3042, 3057}
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.
\(\displaystyle \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^mdx\) |
\(\Big \downarrow \) 3067 |
\(\displaystyle d^2 \sqrt {d \cos (a+b x)} \sqrt {d \sec (a+b x)} \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle d^2 \sqrt {d \cos (a+b x)} \sqrt {d \sec (a+b x)} \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{3/2}}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {d \sqrt [4]{\cos ^2(a+b x)} \sqrt {d \sec (a+b x)} (c \sin (a+b x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(a+b x)\right )}{b c (m+1)}\) |
Input:
Int[(d*Sec[a + b*x])^(3/2)*(c*Sin[a + b*x])^m,x]
Output:
(d*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[5/4, (1 + m)/2, (3 + m)/2, Sin [a + b*x]^2]*Sqrt[d*Sec[a + b*x]]*(c*Sin[a + b*x])^(1 + m))/(b*c*(1 + m))
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^2*(b*Cos[e + f*x])^(n - 1)*(b*Sec[e + f*x])^(n - 1) Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n]
\[\int \left (d \sec \left (b x +a \right )\right )^{\frac {3}{2}} \left (c \sin \left (b x +a \right )\right )^{m}d x\]
Input:
int((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x)
Output:
int((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x)
\[ \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx=\int { \left (d \sec \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sin \left (b x + a\right )\right )^{m} \,d x } \] Input:
integrate((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x, algorithm="fricas")
Output:
integral(sqrt(d*sec(b*x + a))*(c*sin(b*x + a))^m*d*sec(b*x + a), x)
Timed out. \[ \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx=\text {Timed out} \] Input:
integrate((d*sec(b*x+a))**(3/2)*(c*sin(b*x+a))**m,x)
Output:
Timed out
\[ \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx=\int { \left (d \sec \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sin \left (b x + a\right )\right )^{m} \,d x } \] Input:
integrate((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x, algorithm="maxima")
Output:
integrate((d*sec(b*x + a))^(3/2)*(c*sin(b*x + a))^m, x)
\[ \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx=\int { \left (d \sec \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sin \left (b x + a\right )\right )^{m} \,d x } \] Input:
integrate((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x, algorithm="giac")
Output:
integrate((d*sec(b*x + a))^(3/2)*(c*sin(b*x + a))^m, x)
Timed out. \[ \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx=\int {\left (c\,\sin \left (a+b\,x\right )\right )}^m\,{\left (\frac {d}{\cos \left (a+b\,x\right )}\right )}^{3/2} \,d x \] Input:
int((c*sin(a + b*x))^m*(d/cos(a + b*x))^(3/2),x)
Output:
int((c*sin(a + b*x))^m*(d/cos(a + b*x))^(3/2), x)
\[ \int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx=\sqrt {d}\, c^{m} \left (\int \sin \left (b x +a \right )^{m} \sqrt {\sec \left (b x +a \right )}\, \sec \left (b x +a \right )d x \right ) d \] Input:
int((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x)
Output:
sqrt(d)*c**m*int(sin(a + b*x)**m*sqrt(sec(a + b*x))*sec(a + b*x),x)*d