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

Optimal. Leaf size=77 \[ \frac{\cos ^2(a+b x)^{3/4} (d \sec (a+b x))^{3/2} (c \sin (a+b x))^{m+1} \, _2F_1\left (\frac{3}{4},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c d (m+1)} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.0952061, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2586, 2577} \[ \frac{\cos ^2(a+b x)^{3/4} (d \sec (a+b x))^{3/2} (c \sin (a+b x))^{m+1} \, _2F_1\left (\frac{3}{4},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c d (m+1)} \]

Antiderivative was successfully verified.

[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 2586

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(1*(b*Cos[e +
f*x])^(n + 1)*(b*Sec[e + f*x])^(n + 1))/b^2, 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]

Rule 2577

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)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

\begin{align*} \int \sqrt{d \sec (a+b x)} (c \sin (a+b x))^m \, dx &=\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} \, _2F_1\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]  time = 1.37376, size = 106, normalized size = 1.38 \[ -\frac{\sin (2 (a+b x)) \csc ^2(a+b x) \sqrt{d \sec (a+b x)} \left (-\tan ^2(a+b x)\right )^{\frac{1-m}{2}} (c \sin (a+b x))^m \, _2F_1\left (\frac{1}{4} (1-2 m),\frac{1-m}{2};\frac{1}{4} (5-2 m);\sec ^2(a+b x)\right )}{b (2 m-1)} \]

Antiderivative was successfully verified.

[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]  time = 0.102, size = 0, normalized size = 0. \begin{align*} \int \sqrt{d\sec \left ( bx+a \right ) } \left ( c\sin \left ( bx+a \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sin{\left (a + b x \right )}\right )^{m} \sqrt{d \sec{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)