\(\int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx\) [270]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx=-\frac {2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}}-\frac {4 \cos (a+b x)}{5 b d^3 \sqrt {d \tan (a+b x)}}-\frac {4 \cos (a+b x) E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {d \tan (a+b x)}}{5 b d^4 \sqrt {\sin (2 a+2 b x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 1.86 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx=-\frac {2 \left (4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sec ^2(a+b x)+3 \left (-2+\csc ^2(a+b x)+\csc ^4(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sin (a+b x) \sqrt {d \tan (a+b x)}}{15 b d^4 \sqrt {\sec ^2(a+b x)}} \] Input:

Integrate[Sec[a + b*x]^3/(d*Tan[a + b*x])^(7/2),x]
 

Output:

(-2*(4*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x]^2 + 
3*(-2 + Csc[a + b*x]^2 + Csc[a + b*x]^4)*Sqrt[Sec[a + b*x]^2])*Sin[a + b*x 
]*Sqrt[d*Tan[a + b*x]])/(15*b*d^4*Sqrt[Sec[a + b*x]^2])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3088, 3042, 3088, 3042, 3095, 3042, 3052, 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.

\(\displaystyle \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (a+b x)^3}{(d \tan (a+b x))^{7/2}}dx\)

\(\Big \downarrow \) 3088

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3088

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3095

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 \left (-\frac {2 \cos (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{b d^2 \sqrt {\sin (2 a+2 b x)}}-\frac {2 \cos (a+b x)}{b d \sqrt {d \tan (a+b x)}}\right )}{5 d^2}-\frac {2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}}\)

Input:

Int[Sec[a + b*x]^3/(d*Tan[a + b*x])^(7/2),x]
 

Output:

(-2*Sec[a + b*x])/(5*b*d*(d*Tan[a + b*x])^(5/2)) + (2*((-2*Cos[a + b*x])/( 
b*d*Sqrt[d*Tan[a + b*x]]) - (2*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*S 
qrt[d*Tan[a + b*x]])/(b*d^2*Sqrt[Sin[2*a + 2*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 3052

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] 
, x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e 
 + 2*f*x]])   Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
 

rule 3088

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 
1)/(b*f*(n + 1))), x] - Simp[a^2*((m - 2)/(b^2*(n + 1)))   Int[(a*Sec[e + f 
*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && 
LtQ[n, -1] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, -3/2])) && IntegersQ[2*m, 
2*n]
 

rule 3095

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] 
:> Simp[Sqrt[Cos[e + f*x]]*(Sqrt[b*Tan[e + f*x]]/Sqrt[Sin[e + f*x]])   Int[ 
Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(266\) vs. \(2(97)=194\).

Time = 1.10 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.43

method	result	size
default	\(-\frac {\sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-2-2 \sec \left (b x +a \right )\right )+\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (1+\sec \left (b x +a \right )\right )+2+\cot \left (b x +a \right ) \csc \left (b x +a \right )\right ) \sqrt {2}}{5 b \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {d \tan \left (b x +a \right )}\, d^{3}}\)	\(267\)

Input:

int(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-1/5/b*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)/(-sin(b*x+a)*cos( 
b*x+a)/(cos(b*x+a)+1)^2)^(1/2)/(d*tan(b*x+a))^(1/2)/d^3*((csc(b*x+a)-cot(b 
*x+a)+1)^(1/2)*(-2*csc(b*x+a)+2*cot(b*x+a)+2)^(1/2)*(-csc(b*x+a)+cot(b*x+a 
))^(1/2)*EllipticE((csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))*(-2-2*sec( 
b*x+a))+(csc(b*x+a)-cot(b*x+a)+1)^(1/2)*(-2*csc(b*x+a)+2*cot(b*x+a)+2)^(1/ 
2)*(-csc(b*x+a)+cot(b*x+a))^(1/2)*EllipticF((csc(b*x+a)-cot(b*x+a)+1)^(1/2 
),1/2*2^(1/2))*(1+sec(b*x+a))+2+cot(b*x+a)*csc(b*x+a))*2^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.19 \[ \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx=-\frac {2 \, {\left ({\left (i \, \cos \left (b x + a\right )^{2} - i\right )} \sqrt {i \, d} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + {\left (-i \, \cos \left (b x + a\right )^{2} + i\right )} \sqrt {-i \, d} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + {\left (-i \, \cos \left (b x + a\right )^{2} + i\right )} \sqrt {i \, d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + {\left (i \, \cos \left (b x + a\right )^{2} - i\right )} \sqrt {-i \, d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + {\left (2 \, \cos \left (b x + a\right )^{4} - 3 \, \cos \left (b x + a\right )^{2}\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{5 \, {\left (b d^{4} \cos \left (b x + a\right )^{2} - b d^{4}\right )} \sin \left (b x + a\right )} \] Input:

integrate(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x, algorithm="fricas")
 

Output:

-2/5*((I*cos(b*x + a)^2 - I)*sqrt(I*d)*elliptic_e(arcsin(cos(b*x + a) + I* 
sin(b*x + a)), -1)*sin(b*x + a) + (-I*cos(b*x + a)^2 + I)*sqrt(-I*d)*ellip 
tic_e(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1)*sin(b*x + a) + (-I*cos(b* 
x + a)^2 + I)*sqrt(I*d)*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), 
-1)*sin(b*x + a) + (I*cos(b*x + a)^2 - I)*sqrt(-I*d)*elliptic_f(arcsin(cos 
(b*x + a) - I*sin(b*x + a)), -1)*sin(b*x + a) + (2*cos(b*x + a)^4 - 3*cos( 
b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))/((b*d^4*cos(b*x + a)^2 - b* 
d^4)*sin(b*x + a))
 

Sympy [F]

\[ \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx=\int \frac {\sec ^{3}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(sec(b*x+a)**3/(d*tan(b*x+a))**(7/2),x)
 

Output:

Integral(sec(a + b*x)**3/(d*tan(a + b*x))**(7/2), x)
 

Maxima [F]

\[ \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx=\int { \frac {\sec \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x, algorithm="maxima")
 

Output:

integrate(sec(b*x + a)^3/(d*tan(b*x + a))^(7/2), x)
 

Giac [F]

\[ \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx=\int { \frac {\sec \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x, algorithm="giac")
 

Output:

integrate(sec(b*x + a)^3/(d*tan(b*x + a))^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx=\int \frac {1}{{\cos \left (a+b\,x\right )}^3\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{7/2}} \,d x \] Input:

int(1/(cos(a + b*x)^3*(d*tan(a + b*x))^(7/2)),x)
 

Output:

int(1/(cos(a + b*x)^3*(d*tan(a + b*x))^(7/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (b x +a \right )}\, \sec \left (b x +a \right )^{3}}{\tan \left (b x +a \right )^{4}}d x \right )}{d^{4}} \] Input:

int(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x)
 

Output:

(sqrt(d)*int((sqrt(tan(a + b*x))*sec(a + b*x)**3)/tan(a + b*x)**4,x))/d**4