\(\int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) [77]

Optimal result

Integrand size = 20, antiderivative size = 23 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {4377} \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]

[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*} \text {integral}& = \frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\sin (a+b x)}{b \sqrt {\sin (2 (a+b x))}} \]

[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)]])

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 12.36 (sec) , antiderivative size = 65687946, normalized size of antiderivative = 2855997.65

method	result	size
default	\(\text {Expression too large to display}\)	\(65687946\)

[In]

int(sin(b*x+a)/sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + \cos \left (b x + a\right )}{2 \, b \cos \left (b x + a\right )} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(b*x+a)/sin(2*b*x+2*a)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\sin \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2029 vs. \(2 (21) = 42\).

Time = 14.02 (sec) , antiderivative size = 2029, normalized size of antiderivative = 88.22 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Too large to display} \]

[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)

Mupad [B] (verification not implemented)

Time = 19.95 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\cos \left (a+b\,x\right )\,\sqrt {\sin \left (2\,a+2\,b\,x\right )}}{b\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )} \]

[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))