\(\int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx\) [286]

Optimal result

Integrand size = 25, antiderivative size = 141 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx=\frac {2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac {2 c (c \sin (a+b x))^{3/2}}{55 b d^3 (d \cos (a+b x))^{11/2}}-\frac {16 c (c \sin (a+b x))^{3/2}}{385 b d^5 (d \cos (a+b x))^{7/2}}-\frac {64 c (c \sin (a+b x))^{3/2}}{1155 b d^7 (d \cos (a+b x))^{3/2}} \] Output:

2/15*c*(c*sin(b*x+a))^(3/2)/b/d/(d*cos(b*x+a))^(15/2)-2/55*c*(c*sin(b*x+a) 
)^(3/2)/b/d^3/(d*cos(b*x+a))^(11/2)-16/385*c*(c*sin(b*x+a))^(3/2)/b/d^5/(d 
*cos(b*x+a))^(7/2)-64/1155*c*(c*sin(b*x+a))^(3/2)/b/d^7/(d*cos(b*x+a))^(3/ 
2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.48 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx=\frac {2 \sqrt {d \cos (a+b x)} (117+44 \cos (2 (a+b x))+4 \cos (4 (a+b x))) \sec ^8(a+b x) (c \sin (a+b x))^{7/2}}{1155 b c d^9} \] Input:

Integrate[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(17/2),x]
 

Output:

(2*Sqrt[d*Cos[a + b*x]]*(117 + 44*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*S 
ec[a + b*x]^8*(c*Sin[a + b*x])^(7/2))/(1155*b*c*d^9)
 

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.

\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}}dx\)

\(\Big \downarrow \) 3046

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac {c^2 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}}dx}{5 d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac {c^2 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{13/2}}dx}{5 d^2}\)

\(\Big \downarrow \) 3051

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac {c^2 \left (\frac {8 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}}dx}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\right )}{5 d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac {c^2 \left (\frac {8 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}}dx}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\right )}{5 d^2}\)

\(\Big \downarrow \) 3051

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac {c^2 \left (\frac {8 \left (\frac {4 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}}dx}{7 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\right )}{5 d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac {c^2 \left (\frac {8 \left (\frac {4 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}}dx}{7 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\right )}{5 d^2}\)

\(\Big \downarrow \) 3043

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{15 b d (d \cos (a+b x))^{15/2}}-\frac {c^2 \left (\frac {8 \left (\frac {8 (c \sin (a+b x))^{3/2}}{21 b c d^3 (d \cos (a+b x))^{3/2}}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}\right )}{11 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{11 b c d (d \cos (a+b x))^{11/2}}\right )}{5 d^2}\)

Input:

Int[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(17/2),x]
 

Output:

(2*c*(c*Sin[a + b*x])^(3/2))/(15*b*d*(d*Cos[a + b*x])^(15/2)) - (c^2*((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)))/(5*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 = 5.84 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.50

method	result	size
default	\(\frac {2 \sqrt {c \sin \left (b x +a \right )}\, \left (32 \cos \left (b x +a \right )^{4}+56 \cos \left (b x +a \right )^{2}+77\right ) c^{2} \tan \left (b x +a \right )^{3} \sec \left (b x +a \right )^{4}}{1155 b \sqrt {d \cos \left (b x +a \right )}\, d^{8}}\)	\(70\)

Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x,method=_RETURNVERBOSE)
 

Output:

2/1155/b*(c*sin(b*x+a))^(1/2)*(32*cos(b*x+a)^4+56*cos(b*x+a)^2+77)*c^2/(d* 
cos(b*x+a))^(1/2)/d^8*tan(b*x+a)^3*sec(b*x+a)^4
 

Fricas [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.62 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx=-\frac {2 \, {\left (32 \, c^{2} \cos \left (b x + a\right )^{6} + 24 \, c^{2} \cos \left (b x + a\right )^{4} + 21 \, c^{2} \cos \left (b x + a\right )^{2} - 77 \, c^{2}\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{1155 \, b d^{9} \cos \left (b x + a\right )^{8}} \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x, algorithm="fricas" 
)
 

Output:

-2/1155*(32*c^2*cos(b*x + a)^6 + 24*c^2*cos(b*x + a)^4 + 21*c^2*cos(b*x + 
a)^2 - 77*c^2)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*sin(b*x + a)/(b*d 
^9*cos(b*x + a)^8)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx=\text {Timed out} \] Input:

integrate((c*sin(b*x+a))**(5/2)/(d*cos(b*x+a))**(17/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {17}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x, algorithm="maxima" 
)
 

Output:

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(17/2), x)
 

Giac [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {17}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x, algorithm="giac")
 

Output:

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(17/2), x)
 

Mupad [B] (verification not implemented)

Time = 33.99 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.47 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx=-\frac {{\mathrm {e}}^{-a\,7{}\mathrm {i}-b\,x\,7{}\mathrm {i}}\,\sqrt {c\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,\left (\frac {1216\,c^2\,{\mathrm {e}}^{a\,7{}\mathrm {i}+b\,x\,7{}\mathrm {i}}\,\sin \left (3\,a+3\,b\,x\right )}{385\,b\,d^8}+\frac {1024\,c^2\,{\mathrm {e}}^{a\,7{}\mathrm {i}+b\,x\,7{}\mathrm {i}}\,\sin \left (5\,a+5\,b\,x\right )}{1155\,b\,d^8}+\frac {128\,c^2\,{\mathrm {e}}^{a\,7{}\mathrm {i}+b\,x\,7{}\mathrm {i}}\,\sin \left (7\,a+7\,b\,x\right )}{1155\,b\,d^8}-\frac {3392\,c^2\,{\mathrm {e}}^{a\,7{}\mathrm {i}+b\,x\,7{}\mathrm {i}}\,\sin \left (a+b\,x\right )}{231\,b\,d^8}\right )}{128\,{\cos \left (a+b\,x\right )}^7\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}} \] Input:

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(17/2),x)
 

Output:

-(exp(- a*7i - b*x*7i)*(c*((exp(- a*1i - b*x*1i)*1i)/2 - (exp(a*1i + b*x*1 
i)*1i)/2))^(1/2)*((1216*c^2*exp(a*7i + b*x*7i)*sin(3*a + 3*b*x))/(385*b*d^ 
8) + (1024*c^2*exp(a*7i + b*x*7i)*sin(5*a + 5*b*x))/(1155*b*d^8) + (128*c^ 
2*exp(a*7i + b*x*7i)*sin(7*a + 7*b*x))/(1155*b*d^8) - (3392*c^2*exp(a*7i + 
 b*x*7i)*sin(a + b*x))/(231*b*d^8)))/(128*cos(a + b*x)^7*(d*(exp(- a*1i - 
b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2))
 

Reduce [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{17/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}}{\cos \left (b x +a \right )^{9}}d x \right ) c^{2}}{d^{9}} \] Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(17/2),x)
 

Output:

(sqrt(d)*sqrt(c)*int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x))*sin(a + b*x)** 
2)/cos(a + b*x)**9,x)*c**2)/d**9