3.2.81 \(\int \frac {\cos ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) [181]
Optimal. Leaf size=28 \[ -\frac {\cos ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]
[Out]
-1/3*cos(b*x+a)^3/b/sin(2*b*x+2*a)^(3/2)
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Rubi [A]
time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps
used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {4376}
\begin {gather*} -\frac {\cos ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(5/2),x]
[Out]
-1/3*Cos[a + b*x]^3/(b*Sin[2*a + 2*b*x]^(3/2))
Rule 4376
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-(e*Cos[a +
b*x])^m)*((g*Sin[c + d*x])^(p + 1)/(b*g*m)), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] &&
EqQ[d/b, 2] && !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
Rubi steps
\begin {align*} \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx &=-\frac {\cos ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 27, normalized size = 0.96 \begin {gather*} -\frac {\csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 (a+b x))}{24 b} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(5/2),x]
[Out]
-1/24*(Csc[a + b*x]^3*Sin[2*(a + b*x)]^(3/2))/b
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 175.78, size = 192, normalized size = 6.86
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| method |
result |
size |
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| default |
\(\frac {\sqrt {-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1}}\, \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \left (4 \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2}\, \sqrt {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{24 \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}\, \sqrt {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, b}\) |
\(192\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x,method=_RETURNVERBOSE)
[Out]
1/24*(-tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)^2-1)/tan(1/2*b*x+1/2*a)*(4*(tan(
1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2
*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)+tan(1/2*b*x+1/2*a)^4-1)/(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2
-1))^(1/2)/(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)/b
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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(cos(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")
[Out]
integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(5/2), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 53 vs.
\(2 (24) = 48\).
time = 2.70, size = 53, normalized size = 1.89 \begin {gather*} \frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) + \cos \left (b x + a\right )^{2} - 1}{12 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")
[Out]
1/12*(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*cos(b*x + a) + cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^2 - b)
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)**3/sin(2*b*x+2*a)**(5/2),x)
[Out]
Timed out
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 15292 vs.
\(2 (24) = 48\).
time = 92.58, size = 15292, normalized size = 546.14 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="giac")
[Out]
1/48*sqrt(2)*sqrt(-tan(1/2*b*x)^4*tan(1/2*a)^3 - tan(1/2*b*x)^3*tan(1/2*a)^4 + tan(1/2*b*x)^4*tan(1/2*a) + 6*t
an(1/2*b*x)^3*tan(1/2*a)^2 + 6*tan(1/2*b*x)^2*tan(1/2*a)^3 + tan(1/2*b*x)*tan(1/2*a)^4 - tan(1/2*b*x)^3 - 6*ta
n(1/2*b*x)^2*tan(1/2*a) - 6*tan(1/2*b*x)*tan(1/2*a)^2 - tan(1/2*a)^3 + tan(1/2*b*x) + tan(1/2*a))*(((((((sqrt(
2)*tan(1/2*a)^57 + 50*sqrt(2)*tan(1/2*a)^55 + 516*sqrt(2)*tan(1/2*a)^53 + 1750*sqrt(2)*tan(1/2*a)^51 - 5215*sq
rt(2)*tan(1/2*a)^49 - 77796*sqrt(2)*tan(1/2*a)^47 - 386576*sqrt(2)*tan(1/2*a)^45 - 1186876*sqrt(2)*tan(1/2*a)^
43 - 2515251*sqrt(2)*tan(1/2*a)^41 - 3759930*sqrt(2)*tan(1/2*a)^39 - 3812844*sqrt(2)*tan(1/2*a)^37 - 2196894*s
qrt(2)*tan(1/2*a)^35 - 36499*sqrt(2)*tan(1/2*a)^33 + 824296*sqrt(2)*tan(1/2*a)^31 - 824296*sqrt(2)*tan(1/2*a)^
27 + 36499*sqrt(2)*tan(1/2*a)^25 + 2196894*sqrt(2)*tan(1/2*a)^23 + 3812844*sqrt(2)*tan(1/2*a)^21 + 3759930*sqr
t(2)*tan(1/2*a)^19 + 2515251*sqrt(2)*tan(1/2*a)^17 + 1186876*sqrt(2)*tan(1/2*a)^15 + 386576*sqrt(2)*tan(1/2*a)
^13 + 77796*sqrt(2)*tan(1/2*a)^11 + 5215*sqrt(2)*tan(1/2*a)^9 - 1750*sqrt(2)*tan(1/2*a)^7 - 516*sqrt(2)*tan(1/
2*a)^5 - 50*sqrt(2)*tan(1/2*a)^3 - sqrt(2)*tan(1/2*a))*tan(1/2*b*x)/(tan(1/2*a)^51 + 23*tan(1/2*a)^49 + 252*ta
n(1/2*a)^47 + 1748*tan(1/2*a)^45 + 8602*tan(1/2*a)^43 + 31878*tan(1/2*a)^41 + 92092*tan(1/2*a)^39 + 211508*tan
(1/2*a)^37 + 389367*tan(1/2*a)^35 + 572033*tan(1/2*a)^33 + 653752*tan(1/2*a)^31 + 534888*tan(1/2*a)^29 + 20801
2*tan(1/2*a)^27 - 208012*tan(1/2*a)^25 - 534888*tan(1/2*a)^23 - 653752*tan(1/2*a)^21 - 572033*tan(1/2*a)^19 -
389367*tan(1/2*a)^17 - 211508*tan(1/2*a)^15 - 92092*tan(1/2*a)^13 - 31878*tan(1/2*a)^11 - 8602*tan(1/2*a)^9 -
1748*tan(1/2*a)^7 - 252*tan(1/2*a)^5 - 23*tan(1/2*a)^3 - tan(1/2*a)) + 24*(sqrt(2)*tan(1/2*a)^56 - 9*sqrt(2)*t
an(1/2*a)^54 - 293*sqrt(2)*tan(1/2*a)^52 - 2499*sqrt(2)*tan(1/2*a)^50 - 11234*sqrt(2)*tan(1/2*a)^48 - 28814*sq
rt(2)*tan(1/2*a)^46 - 30750*sqrt(2)*tan(1/2*a)^44 + 62254*sqrt(2)*tan(1/2*a)^42 + 338619*sqrt(2)*tan(1/2*a)^40
+ 727149*sqrt(2)*tan(1/2*a)^38 + 874209*sqrt(2)*tan(1/2*a)^36 + 376295*sqrt(2)*tan(1/2*a)^34 - 693804*sqrt(2)
*tan(1/2*a)^32 - 1611124*sqrt(2)*tan(1/2*a)^30 - 1611124*sqrt(2)*tan(1/2*a)^28 - 693804*sqrt(2)*tan(1/2*a)^26
+ 376295*sqrt(2)*tan(1/2*a)^24 + 874209*sqrt(2)*tan(1/2*a)^22 + 727149*sqrt(2)*tan(1/2*a)^20 + 338619*sqrt(2)*
tan(1/2*a)^18 + 62254*sqrt(2)*tan(1/2*a)^16 - 30750*sqrt(2)*tan(1/2*a)^14 - 28814*sqrt(2)*tan(1/2*a)^12 - 1123
4*sqrt(2)*tan(1/2*a)^10 - 2499*sqrt(2)*tan(1/2*a)^8 - 293*sqrt(2)*tan(1/2*a)^6 - 9*sqrt(2)*tan(1/2*a)^4 + sqrt
(2)*tan(1/2*a)^2)/(tan(1/2*a)^51 + 23*tan(1/2*a)^49 + 252*tan(1/2*a)^47 + 1748*tan(1/2*a)^45 + 8602*tan(1/2*a)
^43 + 31878*tan(1/2*a)^41 + 92092*tan(1/2*a)^39 + 211508*tan(1/2*a)^37 + 389367*tan(1/2*a)^35 + 572033*tan(1/2
*a)^33 + 653752*tan(1/2*a)^31 + 534888*tan(1/2*a)^29 + 208012*tan(1/2*a)^27 - 208012*tan(1/2*a)^25 - 534888*ta
n(1/2*a)^23 - 653752*tan(1/2*a)^21 - 572033*tan(1/2*a)^19 - 389367*tan(1/2*a)^17 - 211508*tan(1/2*a)^15 - 9209
2*tan(1/2*a)^13 - 31878*tan(1/2*a)^11 - 8602*tan(1/2*a)^9 - 1748*tan(1/2*a)^7 - 252*tan(1/2*a)^5 - 23*tan(1/2*
a)^3 - tan(1/2*a)))*tan(1/2*b*x) - 3*(sqrt(2)*tan(1/2*a)^57 + 178*sqrt(2)*tan(1/2*a)^55 + 2052*sqrt(2)*tan(1/2
*a)^53 + 7254*sqrt(2)*tan(1/2*a)^51 - 20575*sqrt(2)*tan(1/2*a)^49 - 310500*sqrt(2)*tan(1/2*a)^47 - 1498640*sqr
t(2)*tan(1/2*a)^45 - 4335420*sqrt(2)*tan(1/2*a)^43 - 8189235*sqrt(2)*tan(1/2*a)^41 - 9438650*sqrt(2)*tan(1/2*a
)^39 - 3297260*sqrt(2)*tan(1/2*a)^37 + 9834210*sqrt(2)*tan(1/2*a)^35 + 20470125*sqrt(2)*tan(1/2*a)^33 + 171965
20*sqrt(2)*tan(1/2*a)^31 - 17196520*sqrt(2)*tan(1/2*a)^27 - 20470125*sqrt(2)*tan(1/2*a)^25 - 9834210*sqrt(2)*t
an(1/2*a)^23 + 3297260*sqrt(2)*tan(1/2*a)^21 + 9438650*sqrt(2)*tan(1/2*a)^19 + 8189235*sqrt(2)*tan(1/2*a)^17 +
4335420*sqrt(2)*tan(1/2*a)^15 + 1498640*sqrt(2)*tan(1/2*a)^13 + 310500*sqrt(2)*tan(1/2*a)^11 + 20575*sqrt(2)*
tan(1/2*a)^9 - 7254*sqrt(2)*tan(1/2*a)^7 - 2052*sqrt(2)*tan(1/2*a)^5 - 178*sqrt(2)*tan(1/2*a)^3 - sqrt(2)*tan(
1/2*a))/(tan(1/2*a)^51 + 23*tan(1/2*a)^49 + 252*tan(1/2*a)^47 + 1748*tan(1/2*a)^45 + 8602*tan(1/2*a)^43 + 3187
8*tan(1/2*a)^41 + 92092*tan(1/2*a)^39 + 211508*tan(1/2*a)^37 + 389367*tan(1/2*a)^35 + 572033*tan(1/2*a)^33 + 6
53752*tan(1/2*a)^31 + 534888*tan(1/2*a)^29 + 208012*tan(1/2*a)^27 - 208012*tan(1/2*a)^25 - 534888*tan(1/2*a)^2
3 - 653752*tan(1/2*a)^21 - 572033*tan(1/2*a)^19 - 389367*tan(1/2*a)^17 - 211508*tan(1/2*a)^15 - 92092*tan(1/2*
a)^13 - 31878*tan(1/2*a)^11 - 8602*tan(1/2*a)^9 - 1748*tan(1/2*a)^7 - 252*tan(1/2*a)^5 - 23*tan(1/2*a)^3 - tan
(1/2*a)))*tan(1/2*b*x) - 16*(3*sqrt(2)*tan(1/2*a)^56 - 91*sqrt(2)*tan(1/2*a)^54 - 1967*sqrt(2)*tan(1/2*a)^52 -
15945*sqrt(2)*tan(1/2*a)^50 - 72870*sqrt(2)*tan(1/2*a)^48 - 204330*sqrt(2)*tan(1/2*a)^46 - 320730*sqrt(2)*tan
(1/2*a)^44 - 47990*sqrt(2)*tan(1/2*a)^42 + 1098545*sqrt(2)*tan(1/2*a)^40 + 2914695*sqrt(2)*tan(1/2*a)^38 + 384
2275*sqrt(2)*tan(1/2*a)^36 + 1955765*sqrt(2)*tan(1/2*a)^34 - 2577540*sqrt(2)*tan(1/2*a)^32 - 6569820*sqrt(2)*t
an(1/2*a)^30 - 6569820*sqrt(2)*tan(1/2*a)^28 - ...
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Mupad [B]
time = 1.17, size = 94, normalized size = 3.36 \begin {gather*} \frac {\sqrt {\sin \left (2\,a+2\,b\,x\right )}\,\left (\frac {2\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{3}-{\sin \left (\frac {3\,a}{2}+\frac {3\,b\,x}{2}\right )}^2+\frac {{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2}{3}\right )}{b\,\left (30\,{\sin \left (a+b\,x\right )}^2-12\,{\sin \left (2\,a+2\,b\,x\right )}^2+2\,{\sin \left (3\,a+3\,b\,x\right )}^2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(a + b*x)^3/sin(2*a + 2*b*x)^(5/2),x)
[Out]
(sin(2*a + 2*b*x)^(1/2)*((2*sin(a/2 + (b*x)/2)^2)/3 - sin((3*a)/2 + (3*b*x)/2)^2 + sin((5*a)/2 + (5*b*x)/2)^2/
3))/(b*(2*sin(3*a + 3*b*x)^2 - 12*sin(2*a + 2*b*x)^2 + 30*sin(a + b*x)^2))
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