3.1.62 \(\int \frac {1}{(c \cos ^m(a+b x))^{3/2}} \, dx\) [62]

Optimal. Leaf size=89 \[ -\frac {2 \cos ^{1-m}(a+b x) \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2-3 m),\frac {3 (2-m)}{4},\cos ^2(a+b x)\right ) \sin (a+b x)}{b c (2-3 m) \sqrt {c \cos ^m(a+b x)} \sqrt {\sin ^2(a+b x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {3287, 2722} \begin {gather*} -\frac {2 \sin (a+b x) \cos ^{1-m}(a+b x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2-3 m);\frac {3 (2-m)}{4};\cos ^2(a+b x)\right )}{b c (2-3 m) \sqrt {\sin ^2(a+b x)} \sqrt {c \cos ^m(a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Cos[a + b*x]^m)^(-3/2),x]

[Out]

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

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3287

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

________________________________________________________________________________________

Mathematica [A]
time = 0.07, size = 72, normalized size = 0.81 \begin {gather*} -\frac {\cot (a+b x) \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2-3 m),-\frac {3}{4} (-2+m),\cos ^2(a+b x)\right ) \sqrt {\sin ^2(a+b x)}}{\left (b-\frac {3 b m}{2}\right ) \left (c \cos ^m(a+b x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (c \left (\cos ^{m}\left (b x +a \right )\right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cos(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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (c \cos ^{m}{\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*cos(a + b*x)^m)^(3/2),x)

[Out]

int(1/(c*cos(a + b*x)^m)^(3/2), x)

________________________________________________________________________________________