3.311 \(\int \frac{(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=168 \[ -\frac{b^{5/2} \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{d f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{b^{5/2} \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{d f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}} \]
[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))
________________________________________________________________________________________
Rubi [A] time = 0.165321, antiderivative size = 168, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used =
{2610, 2616, 2564, 329, 298, 203, 206} \[ -\frac{b^{5/2} \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{d f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{b^{5/2} \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{d f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{2 b (b \tan (e+f x))^{3/2}}{3 f (d \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[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 2610
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 2616
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]
Rule 2564
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 329
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 298
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 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{3/2}} \, dx &=-\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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} \tan ^{-1}\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} \tanh ^{-1}\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] time = 1.05884, size = 181, normalized size = 1.08 \[ \frac{\csc ^3(e+f x) (b \tan (e+f x))^{5/2} \sqrt{d \sec (e+f x)} \left (-4 \sin ^2(e+f x) \sqrt{\sec (e+f x)}+6 \sqrt [4]{\tan ^2(e+f x)} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )+3 \sqrt [4]{\tan ^2(e+f x)} \left (\log \left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}+1\right )-\log \left (1-\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )\right )\right )}{6 d^2 f \sec ^{\frac{7}{2}}(e+f x)} \]
Antiderivative was successfully verified.
[In]
Integrate[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(3/2),x]
[Out]
(Csc[e + f*x]^3*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(5/2)*(-4*Sqrt[Sec[e + f*x]]*Sin[e + f*x]^2 + 6*ArcTan[S
qrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)]*(Tan[e + f*x]^2)^(1/4) + 3*(-Log[1 - Sqrt[Sec[e + f*x]]/(Tan[e + f*x
]^2)^(1/4)] + Log[1 + Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)])*(Tan[e + f*x]^2)^(1/4)))/(6*d^2*f*Sec[e + f*
x]^(7/2))
________________________________________________________________________________________
Maple [C] time = 0.211, size = 570, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(3/2),x)
[Out]
-1/6/f*2^(1/2)*(3*I*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*c
os(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,
1/2*2^(1/2))-3*I*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(
f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2
*2^(1/2))+3*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e
)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1
/2))+3*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-s
in(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))+
2*cos(f*x+e)*2^(1/2)-2*2^(1/2))*(b*sin(f*x+e)/cos(f*x+e))^(5/2)*cos(f*x+e)/(cos(f*x+e)-1)/(d/cos(f*x+e))^(3/2)
/sin(f*x+e)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
Fricas [B] time = 5.52909, size = 1952, normalized size = 11.62 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*tan(f*x+e))**(5/2)/(d*sec(f*x+e))**(3/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)