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]
-(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))
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Rubi [A] time = 0.115262, antiderivative size = 201, normalized size of antiderivative = 1.,
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 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]
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 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 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 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 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 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]
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*}
Mathematica [C] time = 0.034081, 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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Maple [C] time = 0.095, size = 1281, normalized size = 6.4 \begin{align*} \text{result too large to display} \end{align*}
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*2^(1/2)*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)+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)*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)*Ellipti
cPi((-(-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*(-(-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)*Ell
ipticPi((-(-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*(-(-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))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*E
llipticPi((-(-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*(-(-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*(-(-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))*si
n(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+c
os(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)-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
)+2*cos(b*x+a)^2*2^(1/2))/sin(b*x+a)^(3/2)/(-1+cos(b*x+a)+sin(b*x+a))^3/(-1+cos(b*x+a)-sin(b*x+a))^3
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{\frac{5}{2}}}{\sin \left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}
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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Fricas [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(cos(b*x+a)^(5/2)/sin(b*x+a)^(5/2),x, algorithm="fricas")
[Out]
Timed out
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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(cos(b*x+a)**(5/2)/sin(b*x+a)**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{\frac{5}{2}}}{\sin \left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}
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]
integrate(cos(b*x + a)^(5/2)/sin(b*x + a)^(5/2), x)