3.4.15 \(\int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx\) [315]

3.4.15.1 Optimal result

Integrand size = 25, antiderivative size = 178 \[ \int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\frac {3 d^3 \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{4 \sqrt {b} f \sqrt {b \tan (e+f x)}}+\frac {3 d^3 \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{4 \sqrt {b} f \sqrt {b \tan (e+f x)}}+\frac {d^2 (d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}{2 b f} \]

output

3/4*d^3*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(d*sec(f*x+e))^(1/2)*(b*sin(f 
*x+e))^(1/2)/f/b^(1/2)/(b*tan(f*x+e))^(1/2)+3/4*d^3*arctanh((b*sin(f*x+e)) 
^(1/2)/b^(1/2))*(d*sec(f*x+e))^(1/2)*(b*sin(f*x+e))^(1/2)/f/b^(1/2)/(b*tan 
(f*x+e))^(1/2)+1/2*d^2*(d*sec(f*x+e))^(3/2)*(b*tan(f*x+e))^(1/2)/b/f
 

3.4.15.2 Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.75 \[ \int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\frac {d^2 (d \sec (e+f x))^{3/2} \left (3 \arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+3 \text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+2 \sec ^2(e+f x)^{3/4} \sqrt {\tan (e+f x)}\right ) \sqrt {\tan (e+f x)}}{4 f \sec ^2(e+f x)^{3/4} \sqrt {b \tan (e+f x)}} \]

input

Integrate[(d*Sec[e + f*x])^(7/2)/Sqrt[b*Tan[e + f*x]],x]
 

output

(d^2*(d*Sec[e + f*x])^(3/2)*(3*ArcTan[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^ 
(1/4)] + 3*ArcTanh[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^(1/4)] + 2*(Sec[e + 
 f*x]^2)^(3/4)*Sqrt[Tan[e + f*x]])*Sqrt[Tan[e + f*x]])/(4*f*(Sec[e + f*x]^ 
2)^(3/4)*Sqrt[b*Tan[e + f*x]])
 

3.4.15.3 Rubi [A] (warning: unable to verify)

Time = 0.55 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.73, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3093, 3042, 3096, 3042, 3044, 27, 266, 756, 216, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}}dx\)

\(\Big \downarrow \) 3093

\(\displaystyle \frac {3}{4} d^2 \int \frac {(d \sec (e+f x))^{3/2}}{\sqrt {b \tan (e+f x)}}dx+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{4} d^2 \int \frac {(d \sec (e+f x))^{3/2}}{\sqrt {b \tan (e+f x)}}dx+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 3096

\(\displaystyle \frac {3 d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {\sec (e+f x)}{\sqrt {b \sin (e+f x)}}dx}{4 \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\cos (e+f x) \sqrt {b \sin (e+f x)}}dx}{4 \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 3044

\(\displaystyle \frac {3 d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {b^2}{\sqrt {b \sin (e+f x)} \left (b^2-b^2 \sin ^2(e+f x)\right )}d(b \sin (e+f x))}{4 b f \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 b d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {b \sin (e+f x)} \left (b^2-b^2 \sin ^2(e+f x)\right )}d(b \sin (e+f x))}{4 f \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 b d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}}{2 f \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {3 b d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \left (\frac {\int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}}{2 b}+\frac {\int \frac {1}{b^2 \sin ^2(e+f x)+b}d\sqrt {b \sin (e+f x)}}{2 b}\right )}{2 f \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {3 b d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \left (\frac {\int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 b^{3/2}}\right )}{2 f \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3 b d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \left (\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \sin (e+f x)\right )}{2 b^{3/2}}\right )}{2 f \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 b f}\)

input

Int[(d*Sec[e + f*x])^(7/2)/Sqrt[b*Tan[e + f*x]],x]
 

output

(3*b*d^3*(ArcTan[Sqrt[b]*Sin[e + f*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Sin[e 
 + f*x]]/(2*b^(3/2)))*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(2*f*Sqrt 
[b*Tan[e + f*x]]) + (d^2*(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]])/(2*b 
*f)
 

3.4.15.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3093

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*(m + n - 1))), x] + Simp[a^2*((m - 2)/(m + n - 1))   Int[(a*Sec[e + 
 f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && ( 
GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[ 
2*m, 2*n]
 

rule 3096

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a^(m + n)*((b*Tan[e + f*x])^n/((a*Sec[e + f*x])^n*(b* 
Sin[e + f*x])^n))   Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x] /; 
FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]
 

3.4.15.4 Maple [A] (verified)

Time = 12.21 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.14

method	result	size
default	\(-\frac {\sqrt {d \sec \left (f x +e \right )}\, d^{3} \left (3 \sin \left (f x +e \right ) \arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )-3 \sin \left (f x +e \right ) \operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )-2 \tan \left (f x +e \right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \tan \left (f x +e \right ) \sec \left (f x +e \right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right )}{4 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {b \tan \left (f x +e \right )}}\)	\(203\)

input

int((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

output

-1/4/f*(d*sec(f*x+e))^(1/2)*d^3/(cos(f*x+e)+1)/(sin(f*x+e)/(cos(f*x+e)+1)^ 
2)^(1/2)/(b*tan(f*x+e))^(1/2)*(3*sin(f*x+e)*arctan((sin(f*x+e)/(cos(f*x+e) 
+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x+e)))-3*sin(f*x+e)*arctanh((sin(f*x+e)/(co 
s(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x+e)))-2*tan(f*x+e)*(sin(f*x+e)/(co 
s(f*x+e)+1)^2)^(1/2)-2*tan(f*x+e)*sec(f*x+e)*(sin(f*x+e)/(cos(f*x+e)+1)^2) 
^(1/2))
 

3.4.15.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (144) = 288\).

Time = 0.42 (sec) , antiderivative size = 782, normalized size of antiderivative = 4.39 \[ \int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\left [-\frac {6 \, b d^{3} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (d \cos \left (f x + e\right )^{2} - {\left (d \cos \left (f x + e\right ) + d\right )} \sin \left (f x + e\right ) - d\right )}}\right ) \cos \left (f x + e\right ) - 3 \, b d^{3} \sqrt {-\frac {d}{b}} \cos \left (f x + e\right ) \log \left (\frac {d \cos \left (f x + e\right )^{4} - 72 \, d \cos \left (f x + e\right )^{2} + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 28 \, {\left (d \cos \left (f x + e\right )^{2} - 2 \, d\right )} \sin \left (f x + e\right ) + 72 \, d}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) - 16 \, d^{3} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{32 \, b f \cos \left (f x + e\right )}, \frac {6 \, b d^{3} \sqrt {\frac {d}{b}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (d \cos \left (f x + e\right )^{2} + {\left (d \cos \left (f x + e\right ) + d\right )} \sin \left (f x + e\right ) - d\right )}}\right ) \cos \left (f x + e\right ) + 3 \, b d^{3} \sqrt {\frac {d}{b}} \cos \left (f x + e\right ) \log \left (\frac {d \cos \left (f x + e\right )^{4} - 72 \, d \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} - 28 \, {\left (d \cos \left (f x + e\right )^{2} - 2 \, d\right )} \sin \left (f x + e\right ) + 72 \, d}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) + 16 \, d^{3} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{32 \, b f \cos \left (f x + e\right )}\right ] \]

input

integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas")
 

output

[-1/32*(6*b*d^3*sqrt(-d/b)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 - 
 (cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)* 
sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-d/b)*sqrt(d/cos(f*x + e))/(d*cos(f 
*x + e)^2 - (d*cos(f*x + e) + d)*sin(f*x + e) - d))*cos(f*x + e) - 3*b*d^3 
*sqrt(-d/b)*cos(f*x + e)*log((d*cos(f*x + e)^4 - 72*d*cos(f*x + e)^2 + 8*( 
7*cos(f*x + e)^3 - (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos( 
f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-d/b)*sqrt(d/cos(f*x + e) 
) + 28*(d*cos(f*x + e)^2 - 2*d)*sin(f*x + e) + 72*d)/(cos(f*x + e)^4 - 8*c 
os(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)) - 16*d^3*sqrt(b* 
sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)))/(b*f*cos(f*x + e)), 1/32* 
(6*b*d^3*sqrt(d/b)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 + (cos(f* 
x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*sqrt(b*s 
in(f*x + e)/cos(f*x + e))*sqrt(d/b)*sqrt(d/cos(f*x + e))/(d*cos(f*x + e)^2 
 + (d*cos(f*x + e) + d)*sin(f*x + e) - d))*cos(f*x + e) + 3*b*d^3*sqrt(d/b 
)*cos(f*x + e)*log((d*cos(f*x + e)^4 - 72*d*cos(f*x + e)^2 - 8*(7*cos(f*x 
+ e)^3 + (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))* 
sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/b)*sqrt(d/cos(f*x + e)) - 28*(d*c 
os(f*x + e)^2 - 2*d)*sin(f*x + e) + 72*d)/(cos(f*x + e)^4 - 8*cos(f*x + e) 
^2 + 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)) + 16*d^3*sqrt(b*sin(f*x + e 
)/cos(f*x + e))*sqrt(d/cos(f*x + e)))/(b*f*cos(f*x + e))]
 

3.4.15.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\text {Timed out} \]

input

integrate((d*sec(f*x+e))**(7/2)/(b*tan(f*x+e))**(1/2),x)
 

output

Timed out
 

3.4.15.7 Maxima [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]

input

integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")
 

output

integrate((d*sec(f*x + e))^(7/2)/sqrt(b*tan(f*x + e)), x)
 

3.4.15.8 Giac [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]

input

integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")
 

output

integrate((d*sec(f*x + e))^(7/2)/sqrt(b*tan(f*x + e)), x)
 

3.4.15.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{\sqrt {b \tan (e+f x)}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]

input

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

output

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