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

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

Optimal result

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

[Out]

(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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2657} \[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\cos ^2(a+b x)^{3/4} (c \sin (a+b x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(a+b x)\right )}{b c d (m+1) (d \cos (a+b x))^{3/2}} \]

[In]

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

[Out]

((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))

Rule 2657

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*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(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]

Rubi steps \begin{align*} \text {integral}& = \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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {(c \sin (a+b x))^m}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(a+b x)\right ) (c \sin (a+b x))^m \tan (a+b x)}{b d^2 (1+m) \sqrt {d \cos (a+b x)}} \]

[In]

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

[Out]

((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]])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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