\(\int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx\) [275]

Optimal result

Integrand size = 25, antiderivative size = 141 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx=\frac {2 c \sqrt {c \sin (a+b x)}}{13 b d (d \cos (a+b x))^{13/2}}-\frac {2 c \sqrt {c \sin (a+b x)}}{117 b d^3 (d \cos (a+b x))^{9/2}}-\frac {16 c \sqrt {c \sin (a+b x)}}{585 b d^5 (d \cos (a+b x))^{5/2}}-\frac {64 c \sqrt {c \sin (a+b x)}}{585 b d^7 \sqrt {d \cos (a+b x)}} \] Output:

2/13*c*(c*sin(b*x+a))^(1/2)/b/d/(d*cos(b*x+a))^(13/2)-2/117*c*(c*sin(b*x+a 
))^(1/2)/b/d^3/(d*cos(b*x+a))^(9/2)-16/585*c*(c*sin(b*x+a))^(1/2)/b/d^5/(d 
*cos(b*x+a))^(5/2)-64/585*c*(c*sin(b*x+a))^(1/2)/b/d^7/(d*cos(b*x+a))^(1/2 
)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.48 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx=\frac {2 \sqrt {d \cos (a+b x)} (77+36 \cos (2 (a+b x))+4 \cos (4 (a+b x))) \sec ^7(a+b x) (c \sin (a+b x))^{5/2}}{585 b c d^8} \] Input:

Integrate[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(15/2),x]
 

Output:

(2*Sqrt[d*Cos[a + b*x]]*(77 + 36*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*Se 
c[a + b*x]^7*(c*Sin[a + b*x])^(5/2))/(585*b*c*d^8)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3046, 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.

\(\displaystyle \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}}dx\)

\(\Big \downarrow \) 3046

\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{13 b d (d \cos (a+b x))^{13/2}}-\frac {c^2 \int \frac {1}{(d \cos (a+b x))^{11/2} \sqrt {c \sin (a+b x)}}dx}{13 d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{13 b d (d \cos (a+b x))^{13/2}}-\frac {c^2 \int \frac {1}{(d \cos (a+b x))^{11/2} \sqrt {c \sin (a+b x)}}dx}{13 d^2}\)

\(\Big \downarrow \) 3051

\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{13 b d (d \cos (a+b x))^{13/2}}-\frac {c^2 \left (\frac {8 \int \frac {1}{(d \cos (a+b x))^{7/2} \sqrt {c \sin (a+b x)}}dx}{9 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{9 b c d (d \cos (a+b x))^{9/2}}\right )}{13 d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{13 b d (d \cos (a+b x))^{13/2}}-\frac {c^2 \left (\frac {8 \int \frac {1}{(d \cos (a+b x))^{7/2} \sqrt {c \sin (a+b x)}}dx}{9 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{9 b c d (d \cos (a+b x))^{9/2}}\right )}{13 d^2}\)

\(\Big \downarrow \) 3051

\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{13 b d (d \cos (a+b x))^{13/2}}-\frac {c^2 \left (\frac {8 \left (\frac {4 \int \frac {1}{(d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}dx}{5 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{5 b c d (d \cos (a+b x))^{5/2}}\right )}{9 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{9 b c d (d \cos (a+b x))^{9/2}}\right )}{13 d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{13 b d (d \cos (a+b x))^{13/2}}-\frac {c^2 \left (\frac {8 \left (\frac {4 \int \frac {1}{(d \cos (a+b x))^{3/2} \sqrt {c \sin (a+b x)}}dx}{5 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{5 b c d (d \cos (a+b x))^{5/2}}\right )}{9 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{9 b c d (d \cos (a+b x))^{9/2}}\right )}{13 d^2}\)

\(\Big \downarrow \) 3043

\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{13 b d (d \cos (a+b x))^{13/2}}-\frac {c^2 \left (\frac {8 \left (\frac {8 \sqrt {c \sin (a+b x)}}{5 b c d^3 \sqrt {d \cos (a+b x)}}+\frac {2 \sqrt {c \sin (a+b x)}}{5 b c d (d \cos (a+b x))^{5/2}}\right )}{9 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{9 b c d (d \cos (a+b x))^{9/2}}\right )}{13 d^2}\)

Input:

Int[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(15/2),x]
 

Output:

(2*c*Sqrt[c*Sin[a + b*x]])/(13*b*d*(d*Cos[a + b*x])^(13/2)) - (c^2*((2*Sqr 
t[c*Sin[a + b*x]])/(9*b*c*d*(d*Cos[a + b*x])^(9/2)) + (8*((2*Sqrt[c*Sin[a 
+ b*x]])/(5*b*c*d*(d*Cos[a + b*x])^(5/2)) + (8*Sqrt[c*Sin[a + b*x]])/(5*b* 
c*d^3*Sqrt[d*Cos[a + b*x]])))/(9*d^2)))/(13*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 3046

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

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 = 5.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.48

method	result	size
default	\(\frac {2 \sqrt {c \sin \left (b x +a \right )}\, \left (32 \cos \left (b x +a \right )^{4}+40 \cos \left (b x +a \right )^{2}+45\right ) c \tan \left (b x +a \right )^{2} \sec \left (b x +a \right )^{4}}{585 b \sqrt {d \cos \left (b x +a \right )}\, d^{7}}\)	\(68\)

Input:

int((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(15/2),x,method=_RETURNVERBOSE)
 

Output:

2/585/b*(c*sin(b*x+a))^(1/2)*(32*cos(b*x+a)^4+40*cos(b*x+a)^2+45)*c/(d*cos 
(b*x+a))^(1/2)/d^7*tan(b*x+a)^2*sec(b*x+a)^4
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.52 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx=-\frac {2 \, {\left (32 \, c \cos \left (b x + a\right )^{6} + 8 \, c \cos \left (b x + a\right )^{4} + 5 \, c \cos \left (b x + a\right )^{2} - 45 \, c\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )}}{585 \, b d^{8} \cos \left (b x + a\right )^{7}} \] Input:

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(15/2),x, algorithm="fricas" 
)
 

Output:

-2/585*(32*c*cos(b*x + a)^6 + 8*c*cos(b*x + a)^4 + 5*c*cos(b*x + a)^2 - 45 
*c)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))/(b*d^8*cos(b*x + a)^7)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx=\text {Timed out} \] Input:

integrate((c*sin(b*x+a))**(3/2)/(d*cos(b*x+a))**(15/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {15}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(15/2),x, algorithm="maxima" 
)
 

Output:

integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(15/2), x)
 

Giac [F]

\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {15}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(15/2),x, algorithm="giac")
 

Output:

integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(15/2), x)
 

Mupad [B] (verification not implemented)

Time = 35.05 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.37 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx=-\frac {{\mathrm {e}}^{-a\,6{}\mathrm {i}-b\,x\,6{}\mathrm {i}}\,\sqrt {c\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,\left (-\frac {3776\,c\,{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}}{585\,b\,d^7}+\frac {2752\,c\,{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,\cos \left (2\,a+2\,b\,x\right )}{585\,b\,d^7}+\frac {896\,c\,{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,\cos \left (4\,a+4\,b\,x\right )}{585\,b\,d^7}+\frac {128\,c\,{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,\cos \left (6\,a+6\,b\,x\right )}{585\,b\,d^7}\right )}{64\,{\cos \left (a+b\,x\right )}^6\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}} \] Input:

int((c*sin(a + b*x))^(3/2)/(d*cos(a + b*x))^(15/2),x)
 

Output:

-(exp(- a*6i - b*x*6i)*(c*((exp(- a*1i - b*x*1i)*1i)/2 - (exp(a*1i + b*x*1 
i)*1i)/2))^(1/2)*((2752*c*exp(a*6i + b*x*6i)*cos(2*a + 2*b*x))/(585*b*d^7) 
 - (3776*c*exp(a*6i + b*x*6i))/(585*b*d^7) + (896*c*exp(a*6i + b*x*6i)*cos 
(4*a + 4*b*x))/(585*b*d^7) + (128*c*exp(a*6i + b*x*6i)*cos(6*a + 6*b*x))/( 
585*b*d^7)))/(64*cos(a + b*x)^6*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b* 
x*1i)/2))^(1/2))
 

Reduce [F]

\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{15/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, c \left (-\cos \left (b x +a \right )^{7} \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}}{\cos \left (b x +a \right )^{6} \sin \left (b x +a \right )}d x \right ) b +2 \sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\right )}{13 \cos \left (b x +a \right )^{7} b \,d^{8}} \] Input:

int((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(15/2),x)
 

Output:

(sqrt(d)*sqrt(c)*c*( - cos(a + b*x)**7*int((sqrt(sin(a + b*x))*sqrt(cos(a 
+ b*x)))/(cos(a + b*x)**6*sin(a + b*x)),x)*b + 2*sqrt(sin(a + b*x))*sqrt(c 
os(a + b*x))))/(13*cos(a + b*x)**7*b*d**8)