\(\int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\) [218]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=-\frac {8 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}+\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{3 b d^3}+\frac {2 \sin ^3(a+b x)}{3 b d (d \cos (a+b x))^{3/2}} \] Output:

-8/3*cos(b*x+a)^(1/2)*InverseJacobiAM(1/2*a+1/2*b*x,2^(1/2))/b/d^2/(d*cos( 
b*x+a))^(1/2)+4/3*(d*cos(b*x+a))^(1/2)*sin(b*x+a)/b/d^3+2/3*sin(b*x+a)^3/b 
/d/(d*cos(b*x+a))^(3/2)
 

Mathematica [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.59 \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {5}{2},\frac {7}{2},\sin ^2(a+b x)\right ) \sin ^5(a+b x)}{5 b d (d \cos (a+b x))^{3/2}} \] Input:

Integrate[Sin[a + b*x]^4/(d*Cos[a + b*x])^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03, 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, 3048, 3042, 3121, 3042, 3120}

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.

\(\displaystyle \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (a+b x)^4}{(d \cos (a+b x))^{5/2}}dx\)

\(\Big \downarrow \) 3046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3048

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 \sin ^3(a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \left (\frac {4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {d \cos (a+b x)}}-\frac {2 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d}\right )}{d^2}\)

Input:

Int[Sin[a + b*x]^4/(d*Cos[a + b*x])^(5/2),x]
 

Output:

(2*Sin[a + b*x]^3)/(3*b*d*(d*Cos[a + b*x])^(3/2)) - (2*((4*Sqrt[Cos[a + b* 
x]]*EllipticF[(a + b*x)/2, 2])/(3*b*Sqrt[d*Cos[a + b*x]]) - (2*Sqrt[d*Cos[ 
a + b*x]]*Sin[a + b*x])/(3*b*d)))/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 3048

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

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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. \(285\) vs. \(2(89)=178\).

Time = 4.75 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.80

method	result	size
default	\(-\frac {8 \left (-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{6} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+2 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-\cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}}{3 d^{2} \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sqrt {-d \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\)	\(286\)

Input:

int(sin(b*x+a)^4/(d*cos(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-8/3*(-2*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)+2*cos(1/2*b*x+1/2*a)*sin( 
1/2*b*x+1/2*a)^4+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)*EllipticF(cos(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)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2)))/d^2*(d*(2*cos(1 
/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/(2*cos(1/2*b*x+1/2*a)^2-1)/ 
(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a 
)/(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 = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.11 \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=-\frac {2 \, {\left (-4 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 4 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - \sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )\right )}}{3 \, b d^{3} \cos \left (b x + a\right )^{2}} \] Input:

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(sin(b*x+a)**4/(d*cos(b*x+a))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")
 

Output:

integrate(sin(b*x + a)^4/(d*cos(b*x + a))^(5/2), x)
 

Giac [F]

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

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(5/2),x, algorithm="giac")
 

Output:

integrate(sin(b*x + a)^4/(d*cos(b*x + a))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^4}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \] Input:

int(sin(a + b*x)^4/(d*cos(a + b*x))^(5/2),x)
 

Output:

int(sin(a + b*x)^4/(d*cos(a + b*x))^(5/2), x)
 

Reduce [F]

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

int(sin(b*x+a)^4/(d*cos(b*x+a))^(5/2),x)
 

Output:

(sqrt(d)*int((sqrt(cos(a + b*x))*sin(a + b*x)**4)/cos(a + b*x)**3,x))/d**3