3.5.87 \(\int \frac {(c \sin (a+b x))^m}{(d \sec (a+b x))^{3/2}} \, dx\) [487]

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

[Out]

hypergeom([-1/4, 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)/(cos(b*x+a)^2)^(1/4)/(d
*sec(b*x+a))^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 = {2666, 2657} \begin {gather*} \frac {(c \sin (a+b x))^{m+1} \, _2F_1\left (-\frac {1}{4},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(a+b x)\right )}{b c d (m+1) \sqrt [4]{\cos ^2(a+b x)} \sqrt {d \sec (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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*} \int \frac {(c \sin (a+b x))^m}{(d \sec (a+b x))^{3/2}} \, dx &=\frac {\int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^m \, dx}{d^2 \sqrt {d \cos (a+b x)} \sqrt {d \sec (a+b x)}}\\ &=\frac {\, _2F_1\left (-\frac {1}{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) \sqrt [4]{\cos ^2(a+b x)} \sqrt {d \sec (a+b x)}}\\ \end {align*}

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Mathematica [A]
time = 17.30, size = 116, normalized size = 1.51 \begin {gather*} \frac {2 c \cos (2 (a+b x)) \, _2F_1\left (\frac {1}{4} (-3-2 m),\frac {1-m}{2};\frac {1}{4} (1-2 m);\sec ^2(a+b x)\right ) (c \sin (a+b x))^{-1+m} \left (-\tan ^2(a+b x)\right )^{\frac {1-m}{2}}}{b d (3+2 m) \sqrt {d \sec (a+b x)} \left (-2+\sec ^2(a+b x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (c \sin \left (b x +a \right )\right )^{m}}{\left (d \sec \left (b x +a \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c \sin {\left (a + b x \right )}\right )^{m}}{\left (d \sec {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^m}{{\left (\frac {d}{\cos \left (a+b\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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