\(\int \sqrt {d \sec (a+b x)} (c \sin (a+b x))^m \, dx\) [485]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.06 (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 = {2666, 2657} \[ \int \sqrt {d \sec (a+b x)} (c \sin (a+b x))^m \, dx=\frac {\cos ^2(a+b x)^{3/4} (d \sec (a+b x))^{3/2} (c \sin (a+b x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(a+b x)\right )}{b c d (m+1)} \]

[In]

Int[Sqrt[d*Sec[a + b*x]]*(c*Sin[a + b*x])^m,x]

[Out]

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

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]

Rule 2666

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(1/b^2)*(b*Co
s[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] && LtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((d \cos (a+b x))^{3/2} (d \sec (a+b x))^{3/2}\right ) \int \frac {(c \sin (a+b x))^m}{\sqrt {d \cos (a+b x)}} \, dx}{d^2} \\ & = \frac {\cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(a+b x)\right ) (d \sec (a+b x))^{3/2} (c \sin (a+b x))^{1+m}}{b c d (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.46 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.38 \[ \int \sqrt {d \sec (a+b x)} (c \sin (a+b x))^m \, dx=-\frac {\csc ^2(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (1-2 m),\frac {1-m}{2},\frac {1}{4} (5-2 m),\sec ^2(a+b x)\right ) \sqrt {d \sec (a+b x)} (c \sin (a+b x))^m \sin (2 (a+b x)) \left (-\tan ^2(a+b x)\right )^{\frac {1-m}{2}}}{b (-1+2 m)} \]

[In]

Integrate[Sqrt[d*Sec[a + b*x]]*(c*Sin[a + b*x])^m,x]

[Out]

-((Csc[a + b*x]^2*Hypergeometric2F1[(1 - 2*m)/4, (1 - m)/2, (5 - 2*m)/4, Sec[a + b*x]^2]*Sqrt[d*Sec[a + b*x]]*
(c*Sin[a + b*x])^m*Sin[2*(a + b*x)]*(-Tan[a + b*x]^2)^((1 - m)/2))/(b*(-1 + 2*m)))

Maple [F]

\[\int \sqrt {d \sec \left (b x +a \right )}\, \left (c \sin \left (b x +a \right )\right )^{m}d x\]

[In]

int((d*sec(b*x+a))^(1/2)*(c*sin(b*x+a))^m,x)

[Out]

int((d*sec(b*x+a))^(1/2)*(c*sin(b*x+a))^m,x)

Fricas [F]

\[ \int \sqrt {d \sec (a+b x)} (c \sin (a+b x))^m \, dx=\int { \sqrt {d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m} \,d x } \]

[In]

integrate((d*sec(b*x+a))^(1/2)*(c*sin(b*x+a))^m,x, algorithm="fricas")

[Out]

integral(sqrt(d*sec(b*x + a))*(c*sin(b*x + a))^m, x)

Sympy [F]

\[ \int \sqrt {d \sec (a+b x)} (c \sin (a+b x))^m \, dx=\int \left (c \sin {\left (a + b x \right )}\right )^{m} \sqrt {d \sec {\left (a + b x \right )}}\, dx \]

[In]

integrate((d*sec(b*x+a))**(1/2)*(c*sin(b*x+a))**m,x)

[Out]

Integral((c*sin(a + b*x))**m*sqrt(d*sec(a + b*x)), x)

Maxima [F]

\[ \int \sqrt {d \sec (a+b x)} (c \sin (a+b x))^m \, dx=\int { \sqrt {d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m} \,d x } \]

[In]

integrate((d*sec(b*x+a))^(1/2)*(c*sin(b*x+a))^m,x, algorithm="maxima")

[Out]

integrate(sqrt(d*sec(b*x + a))*(c*sin(b*x + a))^m, x)

Giac [F]

\[ \int \sqrt {d \sec (a+b x)} (c \sin (a+b x))^m \, dx=\int { \sqrt {d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m} \,d x } \]

[In]

integrate((d*sec(b*x+a))^(1/2)*(c*sin(b*x+a))^m,x, algorithm="giac")

[Out]

integrate(sqrt(d*sec(b*x + a))*(c*sin(b*x + a))^m, x)

Mupad [F(-1)]

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

[In]

int((c*sin(a + b*x))^m*(d/cos(a + b*x))^(1/2),x)

[Out]

int((c*sin(a + b*x))^m*(d/cos(a + b*x))^(1/2), x)