3.1.77 \(\int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) [77]
Optimal. Leaf size=23 \[ \frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]
[Out]
sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)
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Rubi [A]
time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps
used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {4377}
\begin {gather*} \frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sin[a + b*x]/Sin[2*a + 2*b*x]^(3/2),x]
[Out]
Sin[a + b*x]/(b*Sqrt[Sin[2*a + 2*b*x]])
Rule 4377
Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(e*Sin[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] && Eq
Q[d/b, 2] && !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
Rubi steps
\begin {align*} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx &=\frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 22, normalized size = 0.96 \begin {gather*} \frac {\sin (a+b x)}{b \sqrt {\sin (2 (a+b x))}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sin[a + b*x]/Sin[2*a + 2*b*x]^(3/2),x]
[Out]
Sin[a + b*x]/(b*Sqrt[Sin[2*(a + b*x)]])
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 10.37, size = 67736131, normalized size = 2945049.17
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| method |
result |
size |
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| default |
\(\text {Expression too large to display}\) |
\(67736131\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(b*x+a)/sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)
[Out]
result too large to display
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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(sin(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")
[Out]
integrate(sin(b*x + a)/sin(2*b*x + 2*a)^(3/2), x)
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Fricas [A]
time = 2.33, size = 39, normalized size = 1.70 \begin {gather*} \frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + \cos \left (b x + a\right )}{2 \, b \cos \left (b x + a\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")
[Out]
1/2*(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) + cos(b*x + a))/(b*cos(b*x + a))
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(b*x+a)/sin(2*b*x+2*a)**(3/2),x)
[Out]
Exception raised: SystemError >> excessive stack use: stack is 4852 deep
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 2029 vs.
\(2 (21) = 42\).
time = 12.61, size = 2029, normalized size = 88.22 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="giac")
[Out]
-1/2*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))*((2*(sqrt(2)
*tan(1/2*a)^25 + 10*sqrt(2)*tan(1/2*a)^23 + 44*sqrt(2)*tan(1/2*a)^21 + 110*sqrt(2)*tan(1/2*a)^19 + 165*sqrt(2)
*tan(1/2*a)^17 + 132*sqrt(2)*tan(1/2*a)^15 - 132*sqrt(2)*tan(1/2*a)^11 - 165*sqrt(2)*tan(1/2*a)^9 - 110*sqrt(2
)*tan(1/2*a)^7 - 44*sqrt(2)*tan(1/2*a)^5 - 10*sqrt(2)*tan(1/2*a)^3 - sqrt(2)*tan(1/2*a))*tan(1/2*b*x)/(tan(1/2
*a)^24 + 12*tan(1/2*a)^22 + 66*tan(1/2*a)^20 + 220*tan(1/2*a)^18 + 495*tan(1/2*a)^16 + 792*tan(1/2*a)^14 + 924
*tan(1/2*a)^12 + 792*tan(1/2*a)^10 + 495*tan(1/2*a)^8 + 220*tan(1/2*a)^6 + 66*tan(1/2*a)^4 + 12*tan(1/2*a)^2 +
1) + (sqrt(2)*tan(1/2*a)^26 + 5*sqrt(2)*tan(1/2*a)^24 - 10*sqrt(2)*tan(1/2*a)^22 - 154*sqrt(2)*tan(1/2*a)^20
- 605*sqrt(2)*tan(1/2*a)^18 - 1353*sqrt(2)*tan(1/2*a)^16 - 1980*sqrt(2)*tan(1/2*a)^14 - 1980*sqrt(2)*tan(1/2*a
)^12 - 1353*sqrt(2)*tan(1/2*a)^10 - 605*sqrt(2)*tan(1/2*a)^8 - 154*sqrt(2)*tan(1/2*a)^6 - 10*sqrt(2)*tan(1/2*a
)^4 + 5*sqrt(2)*tan(1/2*a)^2 + sqrt(2))/(tan(1/2*a)^24 + 12*tan(1/2*a)^22 + 66*tan(1/2*a)^20 + 220*tan(1/2*a)^
18 + 495*tan(1/2*a)^16 + 792*tan(1/2*a)^14 + 924*tan(1/2*a)^12 + 792*tan(1/2*a)^10 + 495*tan(1/2*a)^8 + 220*ta
n(1/2*a)^6 + 66*tan(1/2*a)^4 + 12*tan(1/2*a)^2 + 1))*tan(1/2*b*x) - 2*(sqrt(2)*tan(1/2*a)^25 + 10*sqrt(2)*tan(
1/2*a)^23 + 44*sqrt(2)*tan(1/2*a)^21 + 110*sqrt(2)*tan(1/2*a)^19 + 165*sqrt(2)*tan(1/2*a)^17 + 132*sqrt(2)*tan
(1/2*a)^15 - 132*sqrt(2)*tan(1/2*a)^11 - 165*sqrt(2)*tan(1/2*a)^9 - 110*sqrt(2)*tan(1/2*a)^7 - 44*sqrt(2)*tan(
1/2*a)^5 - 10*sqrt(2)*tan(1/2*a)^3 - sqrt(2)*tan(1/2*a))/(tan(1/2*a)^24 + 12*tan(1/2*a)^22 + 66*tan(1/2*a)^20
+ 220*tan(1/2*a)^18 + 495*tan(1/2*a)^16 + 792*tan(1/2*a)^14 + 924*tan(1/2*a)^12 + 792*tan(1/2*a)^10 + 495*tan(
1/2*a)^8 + 220*tan(1/2*a)^6 + 66*tan(1/2*a)^4 + 12*tan(1/2*a)^2 + 1))*cos(a)/((tan(1/2*b*x)^4*tan(1/2*a)^3 + t
an(1/2*b*x)^3*tan(1/2*a)^4 - tan(1/2*b*x)^4*tan(1/2*a) - 6*tan(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*tan(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))*b) - 1/4*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*tan(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*tan(1/2*b*x)^2*tan(1/2*a) - 6*tan(1/2*b*x)*tan(1/2*a)^2 - ta
n(1/2*a)^3 + tan(1/2*b*x) + tan(1/2*a))*(((sqrt(2)*tan(1/2*a)^26 + 5*sqrt(2)*tan(1/2*a)^24 - 10*sqrt(2)*tan(1/
2*a)^22 - 154*sqrt(2)*tan(1/2*a)^20 - 605*sqrt(2)*tan(1/2*a)^18 - 1353*sqrt(2)*tan(1/2*a)^16 - 1980*sqrt(2)*ta
n(1/2*a)^14 - 1980*sqrt(2)*tan(1/2*a)^12 - 1353*sqrt(2)*tan(1/2*a)^10 - 605*sqrt(2)*tan(1/2*a)^8 - 154*sqrt(2)
*tan(1/2*a)^6 - 10*sqrt(2)*tan(1/2*a)^4 + 5*sqrt(2)*tan(1/2*a)^2 + sqrt(2))*tan(1/2*b*x)/(tan(1/2*a)^24 + 12*t
an(1/2*a)^22 + 66*tan(1/2*a)^20 + 220*tan(1/2*a)^18 + 495*tan(1/2*a)^16 + 792*tan(1/2*a)^14 + 924*tan(1/2*a)^1
2 + 792*tan(1/2*a)^10 + 495*tan(1/2*a)^8 + 220*tan(1/2*a)^6 + 66*tan(1/2*a)^4 + 12*tan(1/2*a)^2 + 1) - 8*(sqrt
(2)*tan(1/2*a)^25 + 10*sqrt(2)*tan(1/2*a)^23 + 44*sqrt(2)*tan(1/2*a)^21 + 110*sqrt(2)*tan(1/2*a)^19 + 165*sqrt
(2)*tan(1/2*a)^17 + 132*sqrt(2)*tan(1/2*a)^15 - 132*sqrt(2)*tan(1/2*a)^11 - 165*sqrt(2)*tan(1/2*a)^9 - 110*sqr
t(2)*tan(1/2*a)^7 - 44*sqrt(2)*tan(1/2*a)^5 - 10*sqrt(2)*tan(1/2*a)^3 - sqrt(2)*tan(1/2*a))/(tan(1/2*a)^24 + 1
2*tan(1/2*a)^22 + 66*tan(1/2*a)^20 + 220*tan(1/2*a)^18 + 495*tan(1/2*a)^16 + 792*tan(1/2*a)^14 + 924*tan(1/2*a
)^12 + 792*tan(1/2*a)^10 + 495*tan(1/2*a)^8 + 220*tan(1/2*a)^6 + 66*tan(1/2*a)^4 + 12*tan(1/2*a)^2 + 1))*tan(1
/2*b*x) - (sqrt(2)*tan(1/2*a)^26 + 5*sqrt(2)*tan(1/2*a)^24 - 10*sqrt(2)*tan(1/2*a)^22 - 154*sqrt(2)*tan(1/2*a)
^20 - 605*sqrt(2)*tan(1/2*a)^18 - 1353*sqrt(2)*tan(1/2*a)^16 - 1980*sqrt(2)*tan(1/2*a)^14 - 1980*sqrt(2)*tan(1
/2*a)^12 - 1353*sqrt(2)*tan(1/2*a)^10 - 605*sqrt(2)*tan(1/2*a)^8 - 154*sqrt(2)*tan(1/2*a)^6 - 10*sqrt(2)*tan(1
/2*a)^4 + 5*sqrt(2)*tan(1/2*a)^2 + sqrt(2))/(tan(1/2*a)^24 + 12*tan(1/2*a)^22 + 66*tan(1/2*a)^20 + 220*tan(1/2
*a)^18 + 495*tan(1/2*a)^16 + 792*tan(1/2*a)^14 + 924*tan(1/2*a)^12 + 792*tan(1/2*a)^10 + 495*tan(1/2*a)^8 + 22
0*tan(1/2*a)^6 + 66*tan(1/2*a)^4 + 12*tan(1/2*a)^2 + 1))*sin(a)/((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*tan(1/2*b*x)^3*tan(1/2*a)^2 - 6*tan(1/2*b*x)^2*tan(1/2*a)^3 - ta
n(1/2*b*x)*tan(1/2*a)^4 + tan(1/2*b*x)^3 + 6*tan(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))*b)
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Mupad [B]
time = 0.28, size = 34, normalized size = 1.48 \begin {gather*} \frac {\cos \left (a+b\,x\right )\,\sqrt {\sin \left (2\,a+2\,b\,x\right )}}{b\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(a + b*x)/sin(2*a + 2*b*x)^(3/2),x)
[Out]
(cos(a + b*x)*sin(2*a + 2*b*x)^(1/2))/(b*(cos(2*a + 2*b*x) + 1))
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