3.23 \(\int \frac {1}{(c \sin ^m(a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=89 \[ \frac {2 \cos (a+b x) \sin ^{1-m}(a+b x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2-3 m);\frac {3 (2-m)}{4};\sin ^2(a+b x)\right )}{b c (2-3 m) \sqrt {\cos ^2(a+b x)} \sqrt {c \sin ^m(a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3208, 2643} \[ \frac {2 \cos (a+b x) \sin ^{1-m}(a+b x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2-3 m);\frac {3 (2-m)}{4};\sin ^2(a+b x)\right )}{b c (2-3 m) \sqrt {\cos ^2(a+b x)} \sqrt {c \sin ^m(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x]^m)^(-3/2),x]

[Out]

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

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3208

Int[(u_.)*((b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sin[e + f*x
])^n)^FracPart[p])/(c*Sin[e + f*x])^(n*FracPart[p]), Int[ActivateTrig[u]*(c*Sin[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 71, normalized size = 0.80 \[ \frac {\sqrt {\cos ^2(a+b x)} \tan (a+b x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2-3 m);-\frac {3}{4} (m-2);\sin ^2(a+b x)\right )}{\left (b-\frac {3 b m}{2}\right ) \left (c \sin ^m(a+b x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sin(b*x+a)^m)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \sin \left (b x + a\right )^{m}\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sin(b*x+a)^m)^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.55, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \left (\sin ^{m}\left (b x +a \right )\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*sin(b*x+a)^m)^(3/2),x)

[Out]

int(1/(c*sin(b*x+a)^m)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \sin \left (b x + a\right )^{m}\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sin(b*x+a)^m)^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*sin(a + b*x)^m)^(3/2),x)

[Out]

int(1/(c*sin(a + b*x)^m)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \sin ^{m}{\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sin(b*x+a)**m)**(3/2),x)

[Out]

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

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