\(\int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx\) [266]

Optimal result

Integrand size = 25, antiderivative size = 112 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}+\frac {16 (c \sin (a+b x))^{3/2}}{77 b c d^3 (d \cos (a+b x))^{7/2}}+\frac {64 (c \sin (a+b x))^{3/2}}{231 b c d^5 (d \cos (a+b x))^{3/2}} \] Output:

2/11*(c*sin(b*x+a))^(3/2)/b/c/d/(d*cos(b*x+a))^(11/2)+16/77*(c*sin(b*x+a)) 
^(3/2)/b/c/d^3/(d*cos(b*x+a))^(7/2)+64/231*(c*sin(b*x+a))^(3/2)/b/c/d^5/(d 
*cos(b*x+a))^(3/2)
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {2 \sqrt {d \cos (a+b x)} (45+28 \cos (2 (a+b x))+4 \cos (4 (a+b x))) \sec ^6(a+b x) (c \sin (a+b x))^{3/2}}{231 b c d^7} \] Input:

Integrate[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(13/2),x]
 

Output:

(2*Sqrt[d*Cos[a + b*x]]*(45 + 28*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*Se 
c[a + b*x]^6*(c*Sin[a + b*x])^(3/2))/(231*b*c*d^7)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3051, 3042, 3051, 3042, 3043}

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.

\(\displaystyle \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}}dx\)

\(\Big \downarrow \) 3051

\(\displaystyle \frac {8 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}}dx}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}}dx}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\)

\(\Big \downarrow \) 3051

\(\displaystyle \frac {8 \left (\frac {4 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}}dx}{7 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 \left (\frac {4 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}}dx}{7 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\)

\(\Big \downarrow \) 3043

\(\displaystyle \frac {8 \left (\frac {8 (c \sin (a+b x))^{3/2}}{21 b c d^3 (d \cos (a+b x))^{3/2}}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\)

Input:

Int[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(13/2),x]
 

Output:

(2*(c*Sin[a + b*x])^(3/2))/(11*b*c*d*(d*Cos[a + b*x])^(11/2)) + (8*((2*(c* 
Sin[a + b*x])^(3/2))/(7*b*c*d*(d*Cos[a + b*x])^(7/2)) + (8*(c*Sin[a + b*x] 
)^(3/2))/(21*b*c*d^3*(d*Cos[a + b*x])^(3/2))))/(11*d^2)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3043

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^( 
m_.), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Cos[e + f*x])^(n + 1)/ 
(a*b*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2, 0] & 
& NeQ[m, -1]
 

rule 3051

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(-(b*Sin[e + f*x])^(n + 1))*((a*Cos[e + f*x])^(m + 1) 
/(a*b*f*(m + 1))), x] + Simp[(m + n + 2)/(a^2*(m + 1))   Int[(b*Sin[e + f*x 
])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m 
, -1] && IntegersQ[2*m, 2*n]
 

Maple [A] (verified)

Time = 12.33 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {2 \left (32 \cos \left (b x +a \right )^{4}+24 \cos \left (b x +a \right )^{2}+21\right ) \sqrt {c \sin \left (b x +a \right )}\, \tan \left (b x +a \right ) \sec \left (b x +a \right )^{4}}{231 b \sqrt {d \cos \left (b x +a \right )}\, d^{6}}\)	\(65\)

Input:

int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x,method=_RETURNVERBOSE)
 

Output:

2/231/b*(32*cos(b*x+a)^4+24*cos(b*x+a)^2+21)*(c*sin(b*x+a))^(1/2)/(d*cos(b 
*x+a))^(1/2)/d^6*tan(b*x+a)*sec(b*x+a)^4
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {2 \, {\left (32 \, \cos \left (b x + a\right )^{4} + 24 \, \cos \left (b x + a\right )^{2} + 21\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{231 \, b d^{7} \cos \left (b x + a\right )^{6}} \] Input:

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x, algorithm="fricas" 
)
 

Output:

2/231*(32*cos(b*x + a)^4 + 24*cos(b*x + a)^2 + 21)*sqrt(d*cos(b*x + a))*sq 
rt(c*sin(b*x + a))*sin(b*x + a)/(b*d^7*cos(b*x + a)^6)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx=\text {Timed out} \] Input:

integrate((c*sin(b*x+a))**(1/2)/(d*cos(b*x+a))**(13/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x, algorithm="maxima" 
)
 

Output:

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(13/2), x)
 

Giac [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(13/2), x)
 

Mupad [B] (verification not implemented)

Time = 32.91 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.93 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx=-\frac {\sqrt {c\,\sin \left (a+b\,x\right )}\,\left (2\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )\,\left (2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2+\sin \left (5\,a+5\,b\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {1984\,\sin \left (a+b\,x\right )\,\left (-2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2+\sin \left (5\,a+5\,b\,x\right )\,1{}\mathrm {i}+1\right )}{231\,b\,d^6}+\frac {256\,\sin \left (3\,a+3\,b\,x\right )\,\left (-2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2+\sin \left (5\,a+5\,b\,x\right )\,1{}\mathrm {i}+1\right )}{77\,b\,d^6}+\frac {128\,\sin \left (5\,a+5\,b\,x\right )\,\left (-2\,{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2+\sin \left (5\,a+5\,b\,x\right )\,1{}\mathrm {i}+1\right )}{231\,b\,d^6}\right )}{32\,{\left ({\sin \left (a+b\,x\right )}^2-1\right )}^3\,\sqrt {-d\,\left (2\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )}} \] Input:

int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(13/2),x)
 

Output:

-((c*sin(a + b*x))^(1/2)*(2*sin(a/2 + (b*x)/2)^2 - 1)*(sin(5*a + 5*b*x)*1i 
 + 2*sin((5*a)/2 + (5*b*x)/2)^2 - 1)*((1984*sin(a + b*x)*(sin(5*a + 5*b*x) 
*1i - 2*sin((5*a)/2 + (5*b*x)/2)^2 + 1))/(231*b*d^6) + (256*sin(3*a + 3*b* 
x)*(sin(5*a + 5*b*x)*1i - 2*sin((5*a)/2 + (5*b*x)/2)^2 + 1))/(77*b*d^6) + 
(128*sin(5*a + 5*b*x)*(sin(5*a + 5*b*x)*1i - 2*sin((5*a)/2 + (5*b*x)/2)^2 
+ 1))/(231*b*d^6)))/(32*(sin(a + b*x)^2 - 1)^3*(-d*(2*sin(a/2 + (b*x)/2)^2 
 - 1))^(1/2))
 

Reduce [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}}{\cos \left (b x +a \right )^{7}}d x \right )}{d^{7}} \] Input:

int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(13/2),x)
 

Output:

(sqrt(d)*sqrt(c)*int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x)))/cos(a + b*x)* 
*7,x))/d**7