\(\int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx\) [311]

Optimal result

Integrand size = 25, antiderivative size = 168 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}} \]

[Out]

-b^(5/2)*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2)/d/f/(d*sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1/
2)+b^(5/2)*arctanh((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(1/2)/d/f/(d*sec(f*x+e))^(1/2)/(b*sin(f*x+e))^
(1/2)-2/3*b*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2690, 2696, 2644, 335, 304, 209, 212} \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {b^{5/2} \sqrt {b \tan (e+f x)} \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{d f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {b^{5/2} \sqrt {b \tan (e+f x)} \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{d f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}} \]

[In]

Int[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(3/2),x]

[Out]

-((b^(5/2)*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(d*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e +
f*x]])) + (b^(5/2)*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(d*f*Sqrt[d*Sec[e + f*x]]*Sqrt
[b*Sin[e + f*x]]) - (2*b*(b*Tan[e + f*x])^(3/2))/(3*f*(d*Sec[e + f*x])^(3/2))

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[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] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2690

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

Rule 2696

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}}+\frac {b^2 \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx}{d^2} \\ & = -\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}}+\frac {\left (b^2 \sqrt {b \tan (e+f x)}\right ) \int \sec (e+f x) \sqrt {b \sin (e+f x)} \, dx}{d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \\ & = -\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}}+\frac {\left (b \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{b^2}} \, dx,x,b \sin (e+f x)\right )}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \\ & = -\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \\ & = -\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}}+\frac {\left (b^3 \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {\left (b^3 \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \\ & = -\frac {b^{5/2} \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{d f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.83 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {\sqrt {d \sec (e+f x)} \left (3 \arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )-3 \text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+\sqrt [4]{\sec ^2(e+f x)} \sin (2 (e+f x)) \sqrt {\tan (e+f x)}\right ) (b \tan (e+f x))^{5/2}}{3 d^2 f \sqrt [4]{\sec ^2(e+f x)} \tan ^{\frac {5}{2}}(e+f x)} \]

[In]

Integrate[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(3/2),x]

[Out]

-1/3*(Sqrt[d*Sec[e + f*x]]*(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)] + (Sec[e + f*x]^2)^(1/4)*Sin[2*(e + f*x)]*Sqrt[Tan[e + f*x]])*(b*Tan[e + f*x])^(5/2))
/(d^2*f*(Sec[e + f*x]^2)^(1/4)*Tan[e + f*x]^(5/2))

Maple [A] (verified)

Time = 10.71 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {\sin \left (f x +e \right ) \sqrt {b \tan \left (f x +e \right )}\, \left (3 \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )+3 \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )+2 \cos \left (f x +e \right )-2\right ) b^{2}}{3 f \left (\cos \left (f x +e \right )-1\right ) \sqrt {d \sec \left (f x +e \right )}\, d}\)	\(168\)

[In]

int((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (138) = 276\).

Time = 0.59 (sec) , antiderivative size = 766, normalized size of antiderivative = 4.56 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=\left [-\frac {16 \, b^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 6 \, b^{2} d \sqrt {-\frac {b}{d}} \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 {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b \cos \left (f x + e\right )^{2} - {\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - b\right )}}\right ) - 3 \, b^{2} d \sqrt {-\frac {b}{d}} \log \left (\frac {b \cos \left (f x + e\right )^{4} - 72 \, b \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 {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 28 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right ) + 72 \, b}{\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 )}{24 \, d^{2} f}, -\frac {16 \, b^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 6 \, b^{2} d \sqrt {\frac {b}{d}} \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 {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b \cos \left (f x + e\right )^{2} + {\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - b\right )}}\right ) - 3 \, b^{2} d \sqrt {\frac {b}{d}} \log \left (\frac {b \cos \left (f x + e\right )^{4} - 72 \, b \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 {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} - 28 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right ) + 72 \, b}{\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 )}{24 \, d^{2} f}\right ] \]

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/24*(16*b^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + 6*b^2*d*sqrt
(-b/d)*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(-b/d)*sqrt(d/cos(f*x + e))/(b*cos(f*x + e)^2 - (b*cos
(f*x + e) + b)*sin(f*x + e) - b)) - 3*b^2*d*sqrt(-b/d)*log((b*cos(f*x + e)^4 - 72*b*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(-b/d)*sqrt(d/cos(f*x + e)) + 28*(b*cos(f*x + e)^2 - 2*b)*sin(f*x + e) + 72*b)/(cos(f*x + e)^4 - 8*cos(f
*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)))/(d^2*f), -1/24*(16*b^2*sqrt(b*sin(f*x + e)/cos(f*x + e)
)*sqrt(d/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + 6*b^2*d*sqrt(b/d)*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(b/d)*sqrt(d/cos(f*x + e))/(b*cos(f*x + e)^2 + (b*cos(f*x + e) + b)*sin(f*x + e) - b)) - 3*b^2*d*sqrt(
b/d)*log((b*cos(f*x + e)^4 - 72*b*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(b/d)*sqrt(d/cos(f*x + e)) - 28*(b*cos(f*x +
e)^2 - 2*b)*sin(f*x + e) + 72*b)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8
)))/(d^2*f)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((b*tan(f*x+e))**(5/2)/(d*sec(f*x+e))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(3/2), x)

Giac [F]

\[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((b*tan(e + f*x))^(5/2)/(d/cos(e + f*x))^(3/2),x)

[Out]

int((b*tan(e + f*x))^(5/2)/(d/cos(e + f*x))^(3/2), x)