3.302 \(\int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=201 \[ -\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2} b}+\frac {\log \left (\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{2 \sqrt {2} b}-\frac {\log \left (\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{2 \sqrt {2} b} \]

[Out]

-2/3*cos(b*x+a)^(3/2)/b/sin(b*x+a)^(3/2)+1/2*arctan(-1+2^(1/2)*cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2))/b*2^(1/2)+1/
2*arctan(1+2^(1/2)*cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2))/b*2^(1/2)+1/4*ln(1+cot(b*x+a)-2^(1/2)*cos(b*x+a)^(1/2)/s
in(b*x+a)^(1/2))/b*2^(1/2)-1/4*ln(1+cot(b*x+a)+2^(1/2)*cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2))/b*2^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2567, 2575, 297, 1162, 617, 204, 1165, 628} \[ -\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2} b}+\frac {\log \left (\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{2 \sqrt {2} b}-\frac {\log \left (\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{2 \sqrt {2} b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^(5/2)/Sin[a + b*x]^(5/2),x]

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/(Sqrt[2]*b)) + ArcTan[1 + (Sqrt[2]*Sqrt[Cos[a +
b*x]])/Sqrt[Sin[a + b*x]]]/(Sqrt[2]*b) + Log[1 + Cot[a + b*x] - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]
]]/(2*Sqrt[2]*b) - Log[1 + Cot[a + b*x] + (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/(2*Sqrt[2]*b) - (2*
Cos[a + b*x]^(3/2))/(3*b*Sin[a + b*x]^(3/2))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2567

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[e +
f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{k = Denomina
tor[m]}, -Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Cos[e + f*x])^(1/k)/(b*Si
n[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx &=-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}-\int \frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}} \, dx\\ &=-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}+\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{b}\\ &=-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}-\frac {\operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{b}+\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{b}\\ &=-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 \sqrt {2} b}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 \sqrt {2} b}\\ &=\frac {\log \left (1+\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 \sqrt {2} b}-\frac {\log \left (1+\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 \sqrt {2} b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}\\ &=-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}+\frac {\log \left (1+\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 \sqrt {2} b}-\frac {\log \left (1+\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 \sqrt {2} b}-\frac {2 \cos ^{\frac {3}{2}}(a+b x)}{3 b \sin ^{\frac {3}{2}}(a+b x)}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 57, normalized size = 0.28 \[ -\frac {2 \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (-\frac {3}{4},-\frac {3}{4};\frac {1}{4};\sin ^2(a+b x)\right )}{3 b \sin ^{\frac {3}{2}}(a+b x) \sqrt {\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^(5/2)/Sin[a + b*x]^(5/2),x]

[Out]

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

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fricas [B]  time = 25.85, size = 1321, normalized size = 6.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

1/24*(6*(sqrt(2)*b*cos(b*x + a)^2 - sqrt(2)*b)*(b^(-4))^(1/4)*arctan(1/2*((sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x
+ a) + sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b
^3*(b^(-4))^(3/4)*cos(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a))
+ 1)*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + (sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/
4)*cos(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 2*cos(b*x + a)*sin(b*x + a) - 4*(b^2*cos(b*x + a)^4 -
b^2*cos(b*x + a)^2)*sqrt(b^(-4)))/((2*cos(b*x + a)^3 - cos(b*x + a))*sin(b*x + a))) + 6*(sqrt(2)*b*cos(b*x +
a)^2 - sqrt(2)*b)*(b^(-4))^(1/4)*arctan(1/2*((sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/
4)*cos(b*x + a))*sqrt(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*cos(b*x + a
) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 1)*sqrt(cos(b*x + a))*sqrt(
sin(b*x + a)) + (sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x
+ a))*sqrt(sin(b*x + a)) - 2*cos(b*x + a)*sin(b*x + a) + 4*(b^2*cos(b*x + a)^4 - b^2*cos(b*x + a)^2)*sqrt(b^(
-4)))/((2*cos(b*x + a)^3 - cos(b*x + a))*sin(b*x + a))) - 6*(sqrt(2)*b*cos(b*x + a)^2 - sqrt(2)*b)*(b^(-4))^(1
/4)*arctan(1/2*((sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) - sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x
+ a))*sqrt(sin(b*x + a)) + sqrt(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*
cos(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 1)*((sqrt(2)*b^3
*(b^(-4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) +
2*cos(b*x + a)*sin(b*x + a)))/(cos(b*x + a)*sin(b*x + a))) - 6*(sqrt(2)*b*cos(b*x + a)^2 - sqrt(2)*b)*(b^(-4))
^(1/4)*arctan(1/2*((sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) - sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(
b*x + a))*sqrt(sin(b*x + a)) + sqrt(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*(b^(-4))^(3/
4)*cos(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 1)*((sqrt(2)*
b^3*(b^(-4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a))
- 2*cos(b*x + a)*sin(b*x + a)))/(cos(b*x + a)*sin(b*x + a))) - 3*(sqrt(2)*b*cos(b*x + a)^2 - sqrt(2)*b)*(b^(-
4))^(1/4)*log(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*cos(b*x + a) + sqrt
(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 1) + 3*(sqrt(2)*b*cos(b*x + a)^2 -
sqrt(2)*b)*(b^(-4))^(1/4)*log(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*cos
(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 1) + 16*cos(b*x + a
)^(3/2)*sqrt(sin(b*x + a)))/(b*cos(b*x + a)^2 - b)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.12, size = 1261, normalized size = 6.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x)

[Out]

-4/3/b*cos(b*x+a)^(5/2)*(-1+cos(b*x+a))^3*(3*I*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+si
n(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a
))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(b*x+a)*cos(b*x+a)-3*I*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+co
s(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi(((1-cos(b*x+a)+sin(b*x+a)
)/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(b*x+a)*cos(b*x+a)+3*I*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1
/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi(((1-cos(b*x+a)
+sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(b*x+a)-3*I*sin(b*x+a)*((1-cos(b*x+a)+sin(b*x+a))/sin
(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi(((1
-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))+3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2
)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi(((1-cos(b*x+a)+s
in(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(b*x+a)*cos(b*x+a)+3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x
+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi(((1-cos
(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(b*x+a)*cos(b*x+a)-6*((1-cos(b*x+a)+sin(b*x+a)
)/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF
(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*sin(b*x+a)*cos(b*x+a)+3*sin(b*x+a)*((1-cos(b*x+a)+s
in(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*
EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*sin(b*x+a)*((1-cos(b*x+a)+sin
(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*El
lipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-6*sin(b*x+a)*((1-cos(b*x+a)+sin(b
*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*Elli
pticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+2*cos(b*x+a)^2*2^(1/2))/sin(b*x+a)^(3/2)/(-1+c
os(b*x+a)+sin(b*x+a))^3/(-1+cos(b*x+a)-sin(b*x+a))^3*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{\frac {5}{2}}}{\sin \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^(5/2)/sin(b*x + a)^(5/2), x)

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mupad [B]  time = 1.90, size = 44, normalized size = 0.22 \[ -\frac {2\,{\cos \left (a+b\,x\right )}^{7/2}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {7}{4},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (a+b\,x\right )}^2\right )}{7\,b\,{\sin \left (a+b\,x\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^(5/2)/sin(a + b*x)^(5/2),x)

[Out]

-(2*cos(a + b*x)^(7/2)*(sin(a + b*x)^2)^(3/4)*hypergeom([7/4, 7/4], 11/4, cos(a + b*x)^2))/(7*b*sin(a + b*x)^(
3/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**(5/2)/sin(b*x+a)**(5/2),x)

[Out]

Timed out

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