\(\int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx\) [354]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{1+m}}{b c d (1+m) (d \cos (a+b x))^{3/2}} \] Output:

(cos(b*x+a)^2)^(3/4)*hypergeom([7/4, 1/2+1/2*m],[3/2+1/2*m],sin(b*x+a)^2)* 
(c*sin(b*x+a))^(1+m)/b/c/d/(1+m)/(d*cos(b*x+a))^(3/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(a+b x)\right ) (c \sin (a+b x))^m \tan (a+b x)}{b d^2 (1+m) \sqrt {d \cos (a+b x)}} \] Input:

Integrate[(c*Sin[a + b*x])^m/(d*Cos[a + b*x])^(5/2),x]
 

Output:

((Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[7/4, (1 + m)/2, (3 + m)/2, Sin[a 
 + b*x]^2]*(c*Sin[a + b*x])^m*Tan[a + b*x])/(b*d^2*(1 + m)*Sqrt[d*Cos[a + 
b*x]])
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {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.

\(\displaystyle \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}}dx\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {\cos ^2(a+b x)^{3/4} (c \sin (a+b x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(a+b x)\right )}{b c d (m+1) (d \cos (a+b x))^{3/2}}\)

Input:

Int[(c*Sin[a + b*x])^m/(d*Cos[a + b*x])^(5/2),x]
 

Output:

((Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[7/4, (1 + m)/2, (3 + m)/2, Sin[a 
 + b*x]^2]*(c*Sin[a + b*x])^(1 + m))/(b*c*d*(1 + m)*(d*Cos[a + b*x])^(3/2) 
)
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3057
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]
 
Maple [F]

\[\int \frac {\left (c \sin \left (b x +a \right )\right )^{m}}{\left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}d x\]

Input:

int((c*sin(b*x+a))^m/(d*cos(b*x+a))^(5/2),x)
 

Output:

int((c*sin(b*x+a))^m/(d*cos(b*x+a))^(5/2),x)
 

Fricas [F]

\[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{m}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^m/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

integral(sqrt(d*cos(b*x + a))*(c*sin(b*x + a))^m/(d^3*cos(b*x + a)^3), x)
 

Sympy [F]

\[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {\left (c \sin {\left (a + b x \right )}\right )^{m}}{\left (d \cos {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((c*sin(b*x+a))**m/(d*cos(b*x+a))**(5/2),x)
 

Output:

Integral((c*sin(a + b*x))**m/(d*cos(a + b*x))**(5/2), x)
 

Maxima [F]

\[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{m}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^m/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")
 

Output:

integrate((c*sin(b*x + a))^m/(d*cos(b*x + a))^(5/2), x)
 

Giac [F]

\[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{m}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^m/(d*cos(b*x+a))^(5/2),x, algorithm="giac")
 

Output:

integrate((c*sin(b*x + a))^m/(d*cos(b*x + a))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^m}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \] Input:

int((c*sin(a + b*x))^m/(d*cos(a + b*x))^(5/2),x)
 

Output:

int((c*sin(a + b*x))^m/(d*cos(a + b*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\sqrt {d}\, c^{m} \left (\int \frac {\sin \left (b x +a \right )^{m} \sqrt {\cos \left (b x +a \right )}}{\cos \left (b x +a \right )^{3}}d x \right )}{d^{3}} \] Input:

int((c*sin(b*x+a))^m/(d*cos(b*x+a))^(5/2),x)
 

Output:

(sqrt(d)*c**m*int((sin(a + b*x)**m*sqrt(cos(a + b*x)))/cos(a + b*x)**3,x)) 
/d**3