\(\int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx\) [302]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=-\frac {2 d \csc (e+f x) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2}}+\frac {b^{3/2} d \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2} (b \sin (e+f x))^{3/2}}+\frac {b^{3/2} d \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2} (b \sin (e+f x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 167, 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 = {2696, 2644, 327, 335, 218, 212, 209} \[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {b^{3/2} d (b \tan (e+f x))^{3/2} \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}+\frac {b^{3/2} d (b \tan (e+f x))^{3/2} \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}-\frac {2 d \csc (e+f x) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

(-2*d*Csc[e + f*x]*(b*Tan[e + f*x])^(3/2))/(f*(d*Sec[e + f*x])^(3/2)) + (b^(3/2)*d*ArcTan[Sqrt[b*Sin[e + f*x]]
/Sqrt[b]]*(b*Tan[e + f*x])^(3/2))/(f*(d*Sec[e + f*x])^(3/2)*(b*Sin[e + f*x])^(3/2)) + (b^(3/2)*d*ArcTanh[Sqrt[
b*Sin[e + f*x]]/Sqrt[b]]*(b*Tan[e + f*x])^(3/2))/(f*(d*Sec[e + f*x])^(3/2)*(b*Sin[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 218

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

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.70 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.40 \[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {5}{4},\frac {9}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} (b \tan (e+f x))^{5/2}}{5 b f \sqrt {d \sec (e+f x)}} \]

[In]

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

[Out]

(2*Hypergeometric2F1[5/4, 5/4, 9/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(1/4)*(b*Tan[e + f*x])^(5/2))/(5*b*f*Sqr
t[d*Sec[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(291\) vs. \(2(139)=278\).

Time = 19.50 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.75

method	result	size
default	\(-\frac {\sin \left (f x +e \right ) \left (\operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right ) \cos \left (f x +e \right )+\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )-\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )-2 \sin \left (f x +e \right )\right ) \sqrt {b \tan \left (f x +e \right )}\, b}{f \left (\cos \left (f x +e \right )-1\right ) \sqrt {d \sec \left (f x +e \right )}\, \left (\cos \left (f x +e \right )+1\right )}\)	\(292\)

[In]

int((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (139) = 278\).

Time = 0.55 (sec) , antiderivative size = 741, normalized size of antiderivative = 4.44 \[ \int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx=\left [-\frac {2 \, b 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 ) - b 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 ) + 16 \, b \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 )}{8 \, d f}, \frac {2 \, b 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 ) + b 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 ) - 16 \, b \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 )}{8 \, d f}\right ] \]

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(2*b*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)) - b*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)) + 16*b*sqrt(b*sin(f*x + e)/cos(f*x + e))*sq
rt(d/cos(f*x + e))*cos(f*x + e))/(d*f), 1/8*(2*b*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)) + b*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)) - 16*b*sq
rt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e))/(d*f)]

Sympy [F]

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

[In]

integrate((b*tan(f*x+e))**(3/2)/(d*sec(f*x+e))**(1/2),x)

[Out]

Integral((b*tan(e + f*x))**(3/2)/sqrt(d*sec(e + f*x)), x)

Maxima [F]

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

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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