\(\int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx\) [274]

Optimal result

Integrand size = 25, antiderivative size = 106 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\frac {2 c \sqrt {c \sin (a+b x)}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {2 c \sqrt {c \sin (a+b x)}}{45 b d^3 (d \cos (a+b x))^{5/2}}-\frac {8 c \sqrt {c \sin (a+b x)}}{45 b d^5 \sqrt {d \cos (a+b x)}} \] Output:

2/9*c*(c*sin(b*x+a))^(1/2)/b/d/(d*cos(b*x+a))^(9/2)-2/45*c*(c*sin(b*x+a))^ 
(1/2)/b/d^3/(d*cos(b*x+a))^(5/2)-8/45*c*(c*sin(b*x+a))^(1/2)/b/d^5/(d*cos( 
b*x+a))^(1/2)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.54 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\frac {2 \sqrt {d \cos (a+b x)} (7+2 \cos (2 (a+b x))) \sec ^5(a+b x) (c \sin (a+b x))^{5/2}}{45 b c d^6} \] Input:

Integrate[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(11/2),x]
 

Output:

(2*Sqrt[d*Cos[a + b*x]]*(7 + 2*Cos[2*(a + b*x)])*Sec[a + b*x]^5*(c*Sin[a + 
 b*x])^(5/2))/(45*b*c*d^6)
 

Rubi [A] (verified)

Time = 0.49 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3051

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3043

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

Input:

Int[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(11/2),x]
 

Output:

(2*c*Sqrt[c*Sin[a + b*x]])/(9*b*d*(d*Cos[a + b*x])^(9/2)) - (c^2*((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)
 

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 = 5.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.56

method	result	size
default	\(\frac {2 \sqrt {c \sin \left (b x +a \right )}\, c \left (4 \tan \left (b x +a \right )^{2}+5 \sec \left (b x +a \right )^{2} \tan \left (b x +a \right )^{2}\right )}{45 b \sqrt {d \cos \left (b x +a \right )}\, d^{5}}\)	\(59\)

Input:

int((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(11/2),x,method=_RETURNVERBOSE)
 

Output:

2/45/b*(c*sin(b*x+a))^(1/2)*c/(d*cos(b*x+a))^(1/2)/d^5*(4*tan(b*x+a)^2+5*s 
ec(b*x+a)^2*tan(b*x+a)^2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.58 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=-\frac {2 \, {\left (4 \, c \cos \left (b x + a\right )^{4} + c \cos \left (b x + a\right )^{2} - 5 \, c\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )}}{45 \, b d^{6} \cos \left (b x + a\right )^{5}} \] Input:

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(11/2),x, algorithm="fricas" 
)
 

Output:

-2/45*(4*c*cos(b*x + a)^4 + c*cos(b*x + a)^2 - 5*c)*sqrt(d*cos(b*x + a))*s 
qrt(c*sin(b*x + a))/(b*d^6*cos(b*x + a)^5)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\text {Timed out} \] Input:

integrate((c*sin(b*x+a))**(3/2)/(d*cos(b*x+a))**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(11/2),x, algorithm="maxima" 
)
 

Output:

integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(11/2), x)
 

Giac [F]

\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(11/2),x, algorithm="giac")
 

Output:

integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(11/2), x)
 

Mupad [B] (verification not implemented)

Time = 32.15 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.95 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=-\frac {\sqrt {c\,\sin \left (a+b\,x\right )}\,\left (2\,{\sin \left (2\,a+2\,b\,x\right )}^2+\sin \left (4\,a+4\,b\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {32\,c\,\left (-2\,{\sin \left (2\,a+2\,b\,x\right )}^2+\sin \left (4\,a+4\,b\,x\right )\,1{}\mathrm {i}+1\right )}{15\,b\,d^5}+\frac {16\,c\,\left (2\,{\sin \left (2\,a+2\,b\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,a+2\,b\,x\right )}^2+\sin \left (4\,a+4\,b\,x\right )\,1{}\mathrm {i}+1\right )}{45\,b\,d^5}+\frac {16\,c\,\left (2\,{\sin \left (a+b\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,a+2\,b\,x\right )}^2+\sin \left (4\,a+4\,b\,x\right )\,1{}\mathrm {i}+1\right )}{9\,b\,d^5}\right )}{16\,{\left ({\sin \left (a+b\,x\right )}^2-1\right )}^2\,\sqrt {-d\,\left (2\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )}} \] Input:

int((c*sin(a + b*x))^(3/2)/(d*cos(a + b*x))^(11/2),x)
 

Output:

-((c*sin(a + b*x))^(1/2)*(sin(4*a + 4*b*x)*1i + 2*sin(2*a + 2*b*x)^2 - 1)* 
((32*c*(sin(4*a + 4*b*x)*1i - 2*sin(2*a + 2*b*x)^2 + 1))/(15*b*d^5) + (16* 
c*(2*sin(2*a + 2*b*x)^2 - 1)*(sin(4*a + 4*b*x)*1i - 2*sin(2*a + 2*b*x)^2 + 
 1))/(45*b*d^5) + (16*c*(2*sin(a + b*x)^2 - 1)*(sin(4*a + 4*b*x)*1i - 2*si 
n(2*a + 2*b*x)^2 + 1))/(9*b*d^5)))/(16*(sin(a + b*x)^2 - 1)^2*(-d*(2*sin(a 
/2 + (b*x)/2)^2 - 1))^(1/2))
 

Reduce [F]

\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{11/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, c \left (-\cos \left (b x +a \right )^{5} \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}}{\cos \left (b x +a \right )^{4} \sin \left (b x +a \right )}d x \right ) b +2 \sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\right )}{9 \cos \left (b x +a \right )^{5} b \,d^{6}} \] Input:

int((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(11/2),x)
 

Output:

(sqrt(d)*sqrt(c)*c*( - cos(a + b*x)**5*int((sqrt(sin(a + b*x))*sqrt(cos(a 
+ b*x)))/(cos(a + b*x)**4*sin(a + b*x)),x)*b + 2*sqrt(sin(a + b*x))*sqrt(c 
os(a + b*x))))/(9*cos(a + b*x)**5*b*d**6)