3.484 \(\int (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx\)

Optimal. Leaf size=75 \[ \frac{d \sqrt [4]{\cos ^2(a+b x)} \sqrt{d \sec (a+b x)} (c \sin (a+b x))^{m+1} \, _2F_1\left (\frac{5}{4},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c (m+1)} \]

[Out]

(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))

________________________________________________________________________________________

Rubi [A]  time = 0.107566, antiderivative size = 75, 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 = {2587, 2577} \[ \frac{d \sqrt [4]{\cos ^2(a+b x)} \sqrt{d \sec (a+b x)} (c \sin (a+b x))^{m+1} \, _2F_1\left (\frac{5}{4},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[a + b*x])^(3/2)*(c*Sin[a + b*x])^m,x]

[Out]

(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))

Rule 2587

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[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]

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 (d \sec (a+b x))^{3/2} (c \sin (a+b x))^m \, dx &=\left (d^2 \sqrt{d \cos (a+b x)} \sqrt{d \sec (a+b x)}\right ) \int \frac{(c \sin (a+b x))^m}{(d \cos (a+b x))^{3/2}} \, dx\\ &=\frac{d \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\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)}\\ \end{align*}

Mathematica [A]  time = 1.3574, size = 96, normalized size = 1.28 \[ -\frac{2 \cot (a+b x) (d \sec (a+b x))^{3/2} \left (-\tan ^2(a+b x)\right )^{\frac{1-m}{2}} (c \sin (a+b x))^m \, _2F_1\left (\frac{1}{4} (3-2 m),\frac{1-m}{2};\frac{1}{4} (7-2 m);\sec ^2(a+b x)\right )}{b (2 m-3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[a + b*x])^(3/2)*(c*Sin[a + b*x])^m,x]

[Out]

(-2*Cot[a + b*x]*Hypergeometric2F1[(3 - 2*m)/4, (1 - m)/2, (7 - 2*m)/4, Sec[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))

________________________________________________________________________________________

Maple [F]  time = 0.092, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( bx+a \right ) \right ) ^{{\frac{3}{2}}} \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))^(3/2)*(c*sin(b*x+a))^m,x)

[Out]

int((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x, algorithm="maxima")

[Out]

integrate((d*sec(b*x + a))^(3/2)*(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} d \sec \left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))**(3/2)*(c*sin(b*x+a))**m,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(3/2)*(c*sin(b*x+a))^m,x, algorithm="giac")

[Out]

integrate((d*sec(b*x + a))^(3/2)*(c*sin(b*x + a))^m, x)