\(\int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx\) [1511]

Optimal result

Integrand size = 35, antiderivative size = 502 \[ \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx=-\frac {3 a b \left (-2 a^2+b^2\right ) \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{5 f \sqrt {d \sin (e+f x)}}+\frac {\sec ^5(e+f x) \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{9/2}}{5 d f}-\frac {3 a \sec ^3(e+f x) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)} \left (-a \left (7 a^2+b^2\right )+2 b \left (-7 a^2+b^2\right ) \sin (e+f x)+5 a \left (a^2-b^2\right ) \sin ^2(e+f x)+\left (8 a^2 b-4 b^3\right ) \sin ^3(e+f x)\right )}{20 d f}-\frac {3 a (a+b)^{3/2} \left (5 a^2+3 a b-4 b^2\right ) \sqrt {-\frac {a (-1+\csc (e+f x))}{a+b}} \sqrt {\frac {a (1+\csc (e+f x))}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right ),-\frac {a+b}{a-b}\right ) \tan (e+f x)}{20 \sqrt {d} f}-\frac {3 b \left (2 a^4-3 a^2 b^2+b^4\right ) \sqrt {-\frac {a (-1+\csc (e+f x))}{a+b}} E\left (\arcsin \left (\sqrt {-\frac {b+a \csc (e+f x)}{a-b}}\right )|1-\frac {2 a}{a+b}\right ) \sqrt {d \sin (e+f x)} \sqrt {-\frac {a \csc ^2(e+f x) (1+\sin (e+f x)) (a+b \sin (e+f x))}{(a-b)^2}} \tan (e+f x)}{5 d f \sqrt {a+b \sin (e+f x)}} \] Output:

-3/5*a*b*(-2*a^2+b^2)*cos(f*x+e)*(a+b*sin(f*x+e))^(1/2)/f/(d*sin(f*x+e))^( 
1/2)+1/5*sec(f*x+e)^5*(d*sin(f*x+e))^(1/2)*(a+b*sin(f*x+e))^(9/2)/d/f-3/20 
*a*sec(f*x+e)^3*(d*sin(f*x+e))^(1/2)*(a+b*sin(f*x+e))^(1/2)*(-a*(7*a^2+b^2 
)+2*b*(-7*a^2+b^2)*sin(f*x+e)+5*a*(a^2-b^2)*sin(f*x+e)^2+(8*a^2*b-4*b^3)*s 
in(f*x+e)^3)/d/f-3/20*a*(a+b)^(3/2)*(5*a^2+3*a*b-4*b^2)*(-a*(-1+csc(f*x+e) 
)/(a+b))^(1/2)*(a*(1+csc(f*x+e))/(a-b))^(1/2)*EllipticF(d^(1/2)*(a+b*sin(f 
*x+e))^(1/2)/(a+b)^(1/2)/(d*sin(f*x+e))^(1/2),(-(a+b)/(a-b))^(1/2))*tan(f* 
x+e)/d^(1/2)/f-3/5*b*(2*a^4-3*a^2*b^2+b^4)*(-a*(-1+csc(f*x+e))/(a+b))^(1/2 
)*EllipticE((-(b+a*csc(f*x+e))/(a-b))^(1/2),(1-2*a/(a+b))^(1/2))*(d*sin(f* 
x+e))^(1/2)*(-a*csc(f*x+e)^2*(1+sin(f*x+e))*(a+b*sin(f*x+e))/(a-b)^2)^(1/2 
)*tan(f*x+e)/d/f/(a+b*sin(f*x+e))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 21.41 (sec) , antiderivative size = 1600, normalized size of antiderivative = 3.19 \[ \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx =\text {Too large to display} \] Input:

Integrate[(Sec[e + f*x]^6*(a + b*Sin[e + f*x])^(9/2))/Sqrt[d*Sin[e + f*x]] 
,x]
 

Output:

(Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]]*((Sec[e + f*x]*(15*a^4 - 15*a^2*b^2 
 + 4*b^4 + 24*a^3*b*Sin[e + f*x] - 12*a*b^3*Sin[e + f*x]))/20 + (Sec[e + f 
*x]^3*(3*a^4 - 3*a^2*b^2 - 4*b^4 + 9*a^3*b*Sin[e + f*x] - 5*a*b^3*Sin[e + 
f*x]))/10 + (Sec[e + f*x]^5*(a^4 + 6*a^2*b^2 + b^4 + 4*a^3*b*Sin[e + f*x] 
+ 4*a*b^3*Sin[e + f*x]))/5))/(f*Sqrt[d*Sin[e + f*x]]) + (3*a*Sqrt[Sin[e + 
f*x]]*((4*a*(5*a^4 - 9*a^2*b^2 + 4*b^4)*Sqrt[((a + b)*Cot[(-e + Pi/2 - f*x 
)/2]^2)/(-a + b)]*EllipticF[ArcSin[Sqrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b 
*Sin[e + f*x]))/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sec[e + f*x]*Sin[(-e + Pi/2 
- f*x)/2]^4*Sqrt[-(((a + b)*Csc[(-e + Pi/2 - f*x)/2]^2*Sin[e + f*x])/a)]*S 
qrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a])/((a + b)*Sqrt[Si 
n[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]) + 4*a*(-8*a^3*b + 4*a*b^3)*((Sqrt[(( 
a + b)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-a + b)]*EllipticF[ArcSin[Sqrt[(Csc[(- 
e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a]/Sqrt[2]], (-2*a)/(-a + b)]*S 
ec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[-(((a + b)*Csc[(-e + Pi/2 - f* 
x)/2]^2*Sin[e + f*x])/a)]*Sqrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + 
f*x]))/a])/((a + b)*Sqrt[Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]) - (Sqrt[( 
(a + b)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-a + b)]*EllipticPi[-(a/b), ArcSin[Sq 
rt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a]/Sqrt[2]], (-2*a)/( 
-a + b)]*Sec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[-(((a + b)*Csc[(-e + 
 Pi/2 - f*x)/2]^2*Sin[e + f*x])/a)]*Sqrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a...
 

Rubi [F]

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.

\(\displaystyle \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \sin (e+f x))^{9/2}}{\cos (e+f x)^6 \sqrt {d \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3367

\(\displaystyle \frac {9}{10} a \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{7/2}}{\sqrt {d \sin (e+f x)}}dx+\frac {\sec ^5(e+f x) \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{9/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{10} a \int \frac {(a+b \sin (e+f x))^{7/2}}{\cos (e+f x)^4 \sqrt {d \sin (e+f x)}}dx+\frac {\sec ^5(e+f x) \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{9/2}}{5 d f}\)

\(\Big \downarrow \) 3404

\(\displaystyle \frac {9}{10} a \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{7/2}}{\sqrt {d \sin (e+f x)}}dx+\frac {\sec ^5(e+f x) \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{9/2}}{5 d f}\)

Input:

Int[(Sec[e + f*x]^6*(a + b*Sin[e + f*x])^(9/2))/Sqrt[d*Sin[e + f*x]],x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 3042

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

rule 3367

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[2*(g*Cos 
[e + f*x])^(p + 1)*Sqrt[d*Sin[e + f*x]]*((a + b*Sin[e + f*x])^m/(d*f*g*(2*m 
 + 1))), x] + Simp[2*a*(m/(g^2*(2*m + 1)))   Int[(g*Cos[e + f*x])^(p + 2)*( 
(a + b*Sin[e + f*x])^(m - 1)/Sqrt[d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, 
e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && EqQ[m + p + 3/2, 0]
 

rule 3404

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Unin 
tegrable[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, 
x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2375\) vs. \(2(456)=912\).

Time = 9.91 (sec) , antiderivative size = 2376, normalized size of antiderivative = 4.73

method	result	size
default	\(\text {Expression too large to display}\)	\(2376\)

Input:

int(sec(f*x+e)^6*(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

-1/40/f/(a+b*sin(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)*(tan(f*x+e)*sec(f*x+e) 
^4*(-15*cos(f*x+e)^4-6*cos(f*x+e)^2-4)*2^(1/2)*a^5+2^(1/2)*b*a^4*(15*cos(f 
*x+e)+24-15*sec(f*x+e)-4*sec(f*x+e)^3-20*sec(f*x+e)^5)+tan(f*x+e)*sec(f*x+ 
e)^4*(24*cos(f*x+e)^5+9*cos(f*x+e)^4+4*cos(f*x+e)^2-40)*2^(1/2)*a^3*b^2-2^ 
(1/2)*(15*cos(f*x+e)+12-11*sec(f*x+e)-56*sec(f*x+e)^3+40*sec(f*x+e)^5)*a^2 
*b^3+tan(f*x+e)*sec(f*x+e)^4*(-12*cos(f*x+e)^5-2*cos(f*x+e)^4+34*cos(f*x+e 
)^2-20)*2^(1/2)*a*b^4-4*sin(f*x+e)*tan(f*x+e)^5*2^(1/2)*b^5+(48*cos(f*x+e) 
+48)*(-a^2+b^2)^(1/2)*((-a*cot(f*x+e)+(-a^2+b^2)^(1/2)+b+a*csc(f*x+e))/(b+ 
(-a^2+b^2)^(1/2)))^(1/2)*(1/(-a^2+b^2)^(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2 
)-b-a*csc(f*x+e)))^(1/2)*(-(csc(f*x+e)-cot(f*x+e))/(b+(-a^2+b^2)^(1/2))*a) 
^(1/2)*a^2*b^2*EllipticE(((-a*cot(f*x+e)+(-a^2+b^2)^(1/2)+b+a*csc(f*x+e))/ 
(b+(-a^2+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^( 
1/2))^(1/2))+(-24*cos(f*x+e)-24)*(-a^2+b^2)^(1/2)*((-a*cot(f*x+e)+(-a^2+b^ 
2)^(1/2)+b+a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2)*(1/(-a^2+b^2)^(1/2)*( 
a*cot(f*x+e)+(-a^2+b^2)^(1/2)-b-a*csc(f*x+e)))^(1/2)*(-(csc(f*x+e)-cot(f*x 
+e))/(b+(-a^2+b^2)^(1/2))*a)^(1/2)*b^4*EllipticE(((-a*cot(f*x+e)+(-a^2+b^2 
)^(1/2)+b+a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-a^2+ 
b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))+(-48*cos(f*x+e)-48)*((-a*cot(f*x+e)+( 
-a^2+b^2)^(1/2)+b+a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2)*(1/(-a^2+b^2)^ 
(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2)-b-a*csc(f*x+e)))^(1/2)*(-(csc(f*x+...
 

Fricas [F]

\[ \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx=\int { \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {9}{2}} \sec \left (f x + e\right )^{6}}{\sqrt {d \sin \left (f x + e\right )}} \,d x } \] Input:

integrate(sec(f*x+e)^6*(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x, algo 
rithm="fricas")
 

Output:

integral(-(4*(a*b^3*cos(f*x + e)^2 - a^3*b - a*b^3)*sec(f*x + e)^6*sin(f*x 
 + e) - (b^4*cos(f*x + e)^4 + a^4 + 6*a^2*b^2 + b^4 - 2*(3*a^2*b^2 + b^4)* 
cos(f*x + e)^2)*sec(f*x + e)^6)*sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + 
e))/(d*sin(f*x + e)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**6*(a+b*sin(f*x+e))**(9/2)/(d*sin(f*x+e))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx=\int { \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {9}{2}} \sec \left (f x + e\right )^{6}}{\sqrt {d \sin \left (f x + e\right )}} \,d x } \] Input:

integrate(sec(f*x+e)^6*(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x, algo 
rithm="maxima")
 

Output:

integrate((b*sin(f*x + e) + a)^(9/2)*sec(f*x + e)^6/sqrt(d*sin(f*x + e)), 
x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)^6*(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x, algo 
rithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx=\int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^{9/2}}{{\cos \left (e+f\,x\right )}^6\,\sqrt {d\,\sin \left (e+f\,x\right )}} \,d x \] Input:

int((a + b*sin(e + f*x))^(9/2)/(cos(e + f*x)^6*(d*sin(e + f*x))^(1/2)),x)
 

Output:

int((a + b*sin(e + f*x))^(9/2)/(cos(e + f*x)^6*(d*sin(e + f*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^6(e+f x) (a+b \sin (e+f x))^{9/2}}{\sqrt {d \sin (e+f x)}} \, dx=\frac {\sqrt {d}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right ) b +a}\, \sec \left (f x +e \right )^{6}}{\sin \left (f x +e \right )}d x \right ) a^{4}+\left (\int \sqrt {\sin \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right ) b +a}\, \sec \left (f x +e \right )^{6} \sin \left (f x +e \right )^{3}d x \right ) b^{4}+4 \left (\int \sqrt {\sin \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right ) b +a}\, \sec \left (f x +e \right )^{6} \sin \left (f x +e \right )^{2}d x \right ) a \,b^{3}+6 \left (\int \sqrt {\sin \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right ) b +a}\, \sec \left (f x +e \right )^{6} \sin \left (f x +e \right )d x \right ) a^{2} b^{2}+4 \left (\int \sqrt {\sin \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right ) b +a}\, \sec \left (f x +e \right )^{6}d x \right ) a^{3} b \right )}{d} \] Input:

int(sec(f*x+e)^6*(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x)
 

Output:

(sqrt(d)*(int((sqrt(sin(e + f*x))*sqrt(sin(e + f*x)*b + a)*sec(e + f*x)**6 
)/sin(e + f*x),x)*a**4 + int(sqrt(sin(e + f*x))*sqrt(sin(e + f*x)*b + a)*s 
ec(e + f*x)**6*sin(e + f*x)**3,x)*b**4 + 4*int(sqrt(sin(e + f*x))*sqrt(sin 
(e + f*x)*b + a)*sec(e + f*x)**6*sin(e + f*x)**2,x)*a*b**3 + 6*int(sqrt(si 
n(e + f*x))*sqrt(sin(e + f*x)*b + a)*sec(e + f*x)**6*sin(e + f*x),x)*a**2* 
b**2 + 4*int(sqrt(sin(e + f*x))*sqrt(sin(e + f*x)*b + a)*sec(e + f*x)**6,x 
)*a**3*b))/d