\(\int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx\) [285]

Optimal result

Integrand size = 25, antiderivative size = 106 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {6 c (c \sin (a+b x))^{3/2}}{77 b d^3 (d \cos (a+b x))^{7/2}}-\frac {8 c (c \sin (a+b x))^{3/2}}{77 b d^5 (d \cos (a+b x))^{3/2}} \] Output:

2/11*c*(c*sin(b*x+a))^(3/2)/b/d/(d*cos(b*x+a))^(11/2)-6/77*c*(c*sin(b*x+a) 
)^(3/2)/b/d^3/(d*cos(b*x+a))^(7/2)-8/77*c*(c*sin(b*x+a))^(3/2)/b/d^5/(d*co 
s(b*x+a))^(3/2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.54 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {2 c^4 (9+2 \cos (2 (a+b x))) \tan ^5(a+b x)}{77 b d^6 \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}} \] Input:

Integrate[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(13/2),x]
 

Output:

(2*c^4*(9 + 2*Cos[2*(a + b*x)])*Tan[a + b*x]^5)/(77*b*d^6*Sqrt[d*Cos[a + b 
*x]]*(c*Sin[a + b*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3046, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3051

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3043

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

Input:

Int[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(13/2),x]
 

Output:

(2*c*(c*Sin[a + b*x])^(3/2))/(11*b*d*(d*Cos[a + b*x])^(11/2)) - (3*c^2*((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 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 = 5.47 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {2 \sqrt {c \sin \left (b x +a \right )}\, c^{2} \left (4 \tan \left (b x +a \right )^{3}+7 \sec \left (b x +a \right )^{2} \tan \left (b x +a \right )^{3}\right )}{77 b \sqrt {d \cos \left (b x +a \right )}\, d^{6}}\)	\(61\)

Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x,method=_RETURNVERBOSE)
 

Output:

2/77/b*(c*sin(b*x+a))^(1/2)*c^2/(d*cos(b*x+a))^(1/2)/d^6*(4*tan(b*x+a)^3+7 
*sec(b*x+a)^2*tan(b*x+a)^3)
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=-\frac {2 \, {\left (4 \, c^{2} \cos \left (b x + a\right )^{4} + 3 \, c^{2} \cos \left (b x + a\right )^{2} - 7 \, c^{2}\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{77 \, b d^{7} \cos \left (b x + a\right )^{6}} \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x, algorithm="fricas" 
)
 

Output:

-2/77*(4*c^2*cos(b*x + a)^4 + 3*c^2*cos(b*x + a)^2 - 7*c^2)*sqrt(d*cos(b*x 
 + a))*sqrt(c*sin(b*x + a))*sin(b*x + a)/(b*d^7*cos(b*x + a)^6)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\text {Timed out} \] Input:

integrate((c*sin(b*x+a))**(5/2)/(d*cos(b*x+a))**(13/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x, algorithm="maxima" 
)
 

Output:

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(13/2), x)
 

Giac [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x, algorithm="giac")
 

Output:

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(13/2), x)
 

Mupad [B] (verification not implemented)

Time = 33.06 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.66 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=-\frac {{\mathrm {e}}^{-a\,5{}\mathrm {i}-b\,x\,5{}\mathrm {i}}\,\sqrt {c\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,\left (\frac {96\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (3\,a+3\,b\,x\right )}{77\,b\,d^6}+\frac {16\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (5\,a+5\,b\,x\right )}{77\,b\,d^6}-\frac {368\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (a+b\,x\right )}{77\,b\,d^6}\right )}{32\,{\cos \left (a+b\,x\right )}^5\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}} \] Input:

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(13/2),x)
 

Output:

-(exp(- a*5i - b*x*5i)*(c*((exp(- a*1i - b*x*1i)*1i)/2 - (exp(a*1i + b*x*1 
i)*1i)/2))^(1/2)*((96*c^2*exp(a*5i + b*x*5i)*sin(3*a + 3*b*x))/(77*b*d^6) 
+ (16*c^2*exp(a*5i + b*x*5i)*sin(5*a + 5*b*x))/(77*b*d^6) - (368*c^2*exp(a 
*5i + b*x*5i)*sin(a + b*x))/(77*b*d^6)))/(32*cos(a + b*x)^5*(d*(exp(- a*1i 
 - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2))
 

Reduce [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}}{\cos \left (b x +a \right )^{7}}d x \right ) c^{2}}{d^{7}} \] Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x)
 

Output:

(sqrt(d)*sqrt(c)*int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x))*sin(a + b*x)** 
2)/cos(a + b*x)**7,x)*c**2)/d**7