\(\int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) [165]

Optimal result

Integrand size = 20, antiderivative size = 24 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, 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 = {4376} \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]

[In]

Int[Cos[a + b*x]/Sin[2*a + 2*b*x]^(3/2),x]

[Out]

-(Cos[a + b*x]/(b*Sqrt[Sin[2*a + 2*b*x]]))

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {\cos (a+b x)}{b \sqrt {\sin (2 (a+b x))}} \]

[In]

Integrate[Cos[a + b*x]/Sin[2*a + 2*b*x]^(3/2),x]

[Out]

-(Cos[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 = 6.42 (sec) , antiderivative size = 57905011, normalized size of antiderivative = 2412708.79

method	result	size
default	\(\text {Expression too large to display}\)	\(57905011\)

[In]

int(cos(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 = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + \sin \left (b x + a\right )}{2 \, b \sin \left (b x + a\right )} \]

[In]

integrate(cos(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)) + sin(b*x + a))/(b*sin(b*x + a))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cos(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(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. \(2 (22) = 44\).

Time = 13.87 (sec) , antiderivative size = 2029, normalized size of antiderivative = 84.54 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Too large to display} \]

[In]

integrate(cos(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="giac")

[Out]

-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*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)*t
an(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)*ta
n(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*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) - 8*(sqrt(2)*tan(1/2*a)^25 + 10*sqrt(2)*tan(1/2*a)^23 + 44*sq
rt(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*sqr
t(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*ta
n(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 + 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 + 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 + tan(1/2*a)^3 - tan(1/2*b*x) - tan(1/2*a))*b) + 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*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))*((2*(sqrt(2)*tan(1/2*a)^2
5 + 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)^1
7 + 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*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) + (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*s
qrt(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*tan(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 + 4
4*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 + 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 = 20.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {\sin \left (2\,a+2\,b\,x\right )}}{2\,b\,\sin \left (a+b\,x\right )} \]

[In]

int(cos(a + b*x)/sin(2*a + 2*b*x)^(3/2),x)

[Out]

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