Integrand size = 23, antiderivative size = 79 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {2 \left (a^2-b^2\right )}{3 b^3 d (a+b \sin (c+d x))^{3/2}}-\frac {4 a}{b^3 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \sqrt {a+b \sin (c+d x)}}{b^3 d} \] Output:
2/3*(a^2-b^2)/b^3/d/(a+b*sin(d*x+c))^(3/2)-4*a/b^3/d/(a+b*sin(d*x+c))^(1/2 )-2*(a+b*sin(d*x+c))^(1/2)/b^3/d
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {2 \left (8 a^2+b^2+12 a b \sin (c+d x)+3 b^2 \sin ^2(c+d x)\right )}{3 b^3 d (a+b \sin (c+d x))^{3/2}} \] Input:
Integrate[Cos[c + d*x]^3/(a + b*Sin[c + d*x])^(5/2),x]
Output:
(-2*(8*a^2 + b^2 + 12*a*b*Sin[c + d*x] + 3*b^2*Sin[c + d*x]^2))/(3*b^3*d*( a + b*Sin[c + d*x])^(3/2))
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3147, 476, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3}{(a+b \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \frac {b^2-b^2 \sin ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}}d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (\frac {2 a}{(a+b \sin (c+d x))^{3/2}}-\frac {1}{\sqrt {a+b \sin (c+d x)}}+\frac {b^2-a^2}{(a+b \sin (c+d x))^{5/2}}\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 \left (a^2-b^2\right )}{3 (a+b \sin (c+d x))^{3/2}}-\frac {4 a}{\sqrt {a+b \sin (c+d x)}}-2 \sqrt {a+b \sin (c+d x)}}{b^3 d}\) |
Input:
Int[Cos[c + d*x]^3/(a + b*Sin[c + d*x])^(5/2),x]
Output:
((2*(a^2 - b^2))/(3*(a + b*Sin[c + d*x])^(3/2)) - (4*a)/Sqrt[a + b*Sin[c + d*x]] - 2*Sqrt[a + b*Sin[c + d*x]])/(b^3*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(-\frac {2 \left (\sqrt {a +b \sin \left (d x +c \right )}+\frac {2 a}{\sqrt {a +b \sin \left (d x +c \right )}}-\frac {a^{2}-b^{2}}{3 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,b^{3}}\) | \(60\) |
default | \(-\frac {2 \left (\sqrt {a +b \sin \left (d x +c \right )}+\frac {2 a}{\sqrt {a +b \sin \left (d x +c \right )}}-\frac {a^{2}-b^{2}}{3 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,b^{3}}\) | \(60\) |
Input:
int(cos(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/d/b^3*((a+b*sin(d*x+c))^(1/2)+2/(a+b*sin(d*x+c))^(1/2)*a-1/3*(a^2-b^2)/ (a+b*sin(d*x+c))^(3/2))
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b^{2} \cos \left (d x + c\right )^{2} - 12 \, a b \sin \left (d x + c\right ) - 8 \, a^{2} - 4 \, b^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{3 \, {\left (b^{5} d \cos \left (d x + c\right )^{2} - 2 \, a b^{4} d \sin \left (d x + c\right ) - {\left (a^{2} b^{3} + b^{5}\right )} d\right )}} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-2/3*(3*b^2*cos(d*x + c)^2 - 12*a*b*sin(d*x + c) - 8*a^2 - 4*b^2)*sqrt(b*s in(d*x + c) + a)/(b^5*d*cos(d*x + c)^2 - 2*a*b^4*d*sin(d*x + c) - (a^2*b^3 + b^5)*d)
Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (73) = 146\).
Time = 2.39 (sec) , antiderivative size = 304, normalized size of antiderivative = 3.85 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\begin {cases} \frac {x \cos ^{3}{\left (c \right )}}{a^{\frac {5}{2}}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\frac {2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}}{a^{\frac {5}{2}}} & \text {for}\: b = 0 \\\frac {x \cos ^{3}{\left (c \right )}}{\left (a + b \sin {\left (c \right )}\right )^{\frac {5}{2}}} & \text {for}\: d = 0 \\- \frac {16 a^{2}}{3 a b^{3} d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{4} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} - \frac {24 a b \sin {\left (c + d x \right )}}{3 a b^{3} d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{4} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} - \frac {8 b^{2} \sin ^{2}{\left (c + d x \right )}}{3 a b^{3} d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{4} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} - \frac {2 b^{2} \cos ^{2}{\left (c + d x \right )}}{3 a b^{3} d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{4} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3/(a+b*sin(d*x+c))**(5/2),x)
Output:
Piecewise((x*cos(c)**3/a**(5/2), Eq(b, 0) & Eq(d, 0)), ((2*sin(c + d*x)**3 /(3*d) + sin(c + d*x)*cos(c + d*x)**2/d)/a**(5/2), Eq(b, 0)), (x*cos(c)**3 /(a + b*sin(c))**(5/2), Eq(d, 0)), (-16*a**2/(3*a*b**3*d*sqrt(a + b*sin(c + d*x)) + 3*b**4*d*sqrt(a + b*sin(c + d*x))*sin(c + d*x)) - 24*a*b*sin(c + d*x)/(3*a*b**3*d*sqrt(a + b*sin(c + d*x)) + 3*b**4*d*sqrt(a + b*sin(c + d *x))*sin(c + d*x)) - 8*b**2*sin(c + d*x)**2/(3*a*b**3*d*sqrt(a + b*sin(c + d*x)) + 3*b**4*d*sqrt(a + b*sin(c + d*x))*sin(c + d*x)) - 2*b**2*cos(c + d*x)**2/(3*a*b**3*d*sqrt(a + b*sin(c + d*x)) + 3*b**4*d*sqrt(a + b*sin(c + d*x))*sin(c + d*x)), True))
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (\frac {3 \, \sqrt {b \sin \left (d x + c\right ) + a}}{b^{2}} + \frac {6 \, {\left (b \sin \left (d x + c\right ) + a\right )} a - a^{2} + b^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{3 \, b d} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
-2/3*(3*sqrt(b*sin(d*x + c) + a)/b^2 + (6*(b*sin(d*x + c) + a)*a - a^2 + b ^2)/((b*sin(d*x + c) + a)^(3/2)*b^2))/(b*d)
Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
Timed out
Time = 21.89 (sec) , antiderivative size = 1402, normalized size of antiderivative = 17.75 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:
int(cos(c + d*x)^3/(a + b*sin(c + d*x))^(5/2),x)
Output:
(16*a^2*b^2*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i)*(a + (b* (cos(d*x) - sin(d*x)*1i)*(cos(c) - sin(c)*1i)*1i)/2 - (b*(cos(d*x) + sin(d *x)*1i)*(cos(c) + sin(c)*1i)*1i)/2)^(1/2))/(3*(a^2*b^5*d - b^7*d + 2*b^7*d *(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - b^7*d*(cos(4*d*x) + sin(4*d*x)*1i)*(cos(4*c) + sin(4*c)*1i) - a*b^6*d*(cos(3*d*x) + sin(3*d *x)*1i)*(cos(3*c) + sin(3*c)*1i)*4i + a*b^6*d*(cos(d*x) + sin(d*x)*1i)*(co s(c) + sin(c)*1i)*4i + 2*a^2*b^5*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - 4*a^4*b^3*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin( 2*c)*1i) + a^3*b^4*d*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i) *4i + a^2*b^5*d*(cos(4*d*x) + sin(4*d*x)*1i)*(cos(4*c) + sin(4*c)*1i) - a^ 3*b^4*d*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*4i)) - (8*a^4*(cos(2 *d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i)*(a + (b*(cos(d*x) - sin(d* x)*1i)*(cos(c) - sin(c)*1i)*1i)/2 - (b*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*1i)/2)^(1/2))/(3*(a^2*b^5*d - b^7*d + 2*b^7*d*(cos(2*d*x) + sin (2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - b^7*d*(cos(4*d*x) + sin(4*d*x)*1i)* (cos(4*c) + sin(4*c)*1i) - a*b^6*d*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i)*4i + a*b^6*d*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)* 4i + 2*a^2*b^5*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - 4 *a^4*b^3*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) + a^3*b^4 *d*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i)*4i + a^2*b^5*d...
Time = 0.17 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {2 \sqrt {\sin \left (d x +c \right ) b +a}\, \left (-\cos \left (d x +c \right )^{2} b^{2}-4 \sin \left (d x +c \right )^{2} b^{2}-12 \sin \left (d x +c \right ) a b -8 a^{2}\right )}{3 b^{3} d \left (\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +a^{2}\right )} \] Input:
int(cos(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x)
Output:
(2*sqrt(sin(c + d*x)*b + a)*( - cos(c + d*x)**2*b**2 - 4*sin(c + d*x)**2*b **2 - 12*sin(c + d*x)*a*b - 8*a**2))/(3*b**3*d*(sin(c + d*x)**2*b**2 + 2*s in(c + d*x)*a*b + a**2))