\(\int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\) [202]

Optimal result

Integrand size = 21, antiderivative size = 100 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {4 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b d^4 \sqrt {\cos (a+b x)}}+\frac {2 \sin (a+b x)}{5 b d (d \cos (a+b x))^{5/2}}-\frac {4 \sin (a+b x)}{5 b d^3 \sqrt {d \cos (a+b x)}} \] Output:

4/5*(d*cos(b*x+a))^(1/2)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))/b/d^4/cos(b 
*x+a)^(1/2)+2/5*sin(b*x+a)/b/d/(d*cos(b*x+a))^(5/2)-4/5*sin(b*x+a)/b/d^3/( 
d*cos(b*x+a))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.59 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {9}{4},\frac {5}{2},\sin ^2(a+b x)\right ) \sin ^3(2 (a+b x))}{24 b (d \cos (a+b x))^{7/2}} \] Input:

Integrate[Sin[a + b*x]^2/(d*Cos[a + b*x])^(7/2),x]
 

Output:

((Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[3/2, 9/4, 5/2, Sin[a + b*x]^2]*S 
in[2*(a + b*x)]^3)/(24*b*(d*Cos[a + b*x])^(7/2))
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3046, 3042, 3116, 3042, 3121, 3042, 3119}

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.

\(\displaystyle \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3046

\(\displaystyle \frac {2 \sin (a+b x)}{5 b d (d \cos (a+b x))^{5/2}}-\frac {2 \int \frac {1}{(d \cos (a+b x))^{3/2}}dx}{5 d^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 \sin (a+b x)}{5 b d (d \cos (a+b x))^{5/2}}-\frac {2 \left (\frac {2 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}}-\frac {2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{b d^2 \sqrt {\cos (a+b x)}}\right )}{5 d^2}\)

Input:

Int[Sin[a + b*x]^2/(d*Cos[a + b*x])^(7/2),x]
 

Output:

(2*Sin[a + b*x])/(5*b*d*(d*Cos[a + b*x])^(5/2)) - (2*((-2*Sqrt[d*Cos[a + b 
*x]]*EllipticE[(a + b*x)/2, 2])/(b*d^2*Sqrt[Cos[a + b*x]]) + (2*Sin[a + b* 
x])/(b*d*Sqrt[d*Cos[a + b*x]])))/(5*d^2)
 

Defintions of rubi rules used

rule 3042

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

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 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(364\) vs. \(2(88)=176\).

Time = 5.27 (sec) , antiderivative size = 365, normalized size of antiderivative = 3.65

method	result	size
default	\(\frac {4 \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \left (8 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{6} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-4 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+4 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4} d +\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d}}{5 d^{4} \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} \left (8 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}-12 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+6 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\)	\(365\)

Input:

int(sin(b*x+a)^2/(d*cos(b*x+a))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

4/5*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/d^4/sin(1/2* 
b*x+1/2*a)^3/(8*sin(1/2*b*x+1/2*a)^6-12*sin(1/2*b*x+1/2*a)^4+6*sin(1/2*b*x 
+1/2*a)^2-1)*(8*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)-4*(sin(1/2*b*x+1/2 
*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a) 
,2^(1/2))*sin(1/2*b*x+1/2*a)^4-8*cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^4+4 
*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticE(c 
os(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*x+1/2*a)^2+cos(1/2*b*x+1/2*a)*sin(1/2 
*b*x+1/2*a)^2-(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2 
)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2)))*(-2*sin(1/2*b*x+1/2*a)^4*d+sin(1/ 
2*b*x+1/2*a)^2*d)^(1/2)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=-\frac {2 \, {\left (-2 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 2 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + \sqrt {d \cos \left (b x + a\right )} {\left (2 \, \cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )\right )}}{5 \, b d^{4} \cos \left (b x + a\right )^{3}} \] Input:

integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")
 

Output:

-2/5*(-2*I*sqrt(1/2)*sqrt(d)*cos(b*x + a)^3*weierstrassZeta(-4, 0, weierst 
rassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) + 2*I*sqrt(1/2)*sqrt(d 
)*cos(b*x + a)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x 
 + a) - I*sin(b*x + a))) + sqrt(d*cos(b*x + a))*(2*cos(b*x + a)^2 - 1)*sin 
(b*x + a))/(b*d^4*cos(b*x + a)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(sin(b*x+a)**2/(d*cos(b*x+a))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")
 

Output:

integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(7/2), x)
 

Giac [F]

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

integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(7/2),x, algorithm="giac")
 

Output:

integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(7/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}}{\cos \left (b x +a \right )^{4}}d x \right )}{d^{4}} \] Input:

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

Output:

(sqrt(d)*int((sqrt(cos(a + b*x))*sin(a + b*x)**2)/cos(a + b*x)**4,x))/d**4