3.309 \(\int \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx\)
Optimal. Leaf size=169 \[ \frac{3 b^{5/2} d \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{3 b^{5/2} d \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{b (b \tan (e+f x))^{3/2} \sqrt{d \sec (e+f x)}}{2 f} \]
[Out]
(3*b^(5/2)*d*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(4*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e
+ f*x]]) - (3*b^(5/2)*d*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(4*f*Sqrt[d*Sec[e + f*x]]
*Sqrt[b*Sin[e + f*x]]) + (b*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2))/(2*f)
________________________________________________________________________________________
Rubi [A] time = 0.152934, antiderivative size = 169, 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 =
{2611, 2616, 2564, 329, 298, 203, 206} \[ \frac{3 b^{5/2} d \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{3 b^{5/2} d \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{b (b \tan (e+f x))^{3/2} \sqrt{d \sec (e+f x)}}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(5/2),x]
[Out]
(3*b^(5/2)*d*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(4*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e
+ f*x]]) - (3*b^(5/2)*d*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(4*f*Sqrt[d*Sec[e + f*x]]
*Sqrt[b*Sin[e + f*x]]) + (b*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2))/(2*f)
Rule 2611
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 + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[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 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx &=\frac{b \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac{1}{4} \left (3 b^2\right ) \int \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)} \, dx\\ &=\frac{b \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac{\left (3 b^2 d \sqrt{b \tan (e+f x)}\right ) \int \sec (e+f x) \sqrt{b \sin (e+f x)} \, dx}{4 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{b \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac{\left (3 b d \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 )}{4 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{b \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac{\left (3 b d \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 )}{2 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{b \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac{\left (3 b^3 d \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 )}{4 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}+\frac{\left (3 b^3 d \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 )}{4 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{3 b^{5/2} d \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{b \tan (e+f x)}}{4 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}-\frac{3 b^{5/2} d \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{b \tan (e+f x)}}{4 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}+\frac{b \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}\\ \end{align*}
Mathematica [A] time = 1.67164, size = 182, normalized size = 1.08 \[ \frac{\csc ^3(e+f x) (b \tan (e+f x))^{5/2} \sqrt{d \sec (e+f x)} \left (4 \sec ^{\frac{5}{2}}(e+f x)-4 \sqrt{\sec (e+f x)}-6 \sqrt [4]{\tan ^2(e+f x)} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )+3 \sqrt [4]{\tan ^2(e+f x)} \left (\log \left (1-\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\log \left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}+1\right )\right )\right )}{8 f \sec ^{\frac{7}{2}}(e+f x)} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(5/2),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]] + 4*Sec[e + f*x]^(5/2) - 6*
ArcTan[Sqrt[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)))/(8*f*Sec[e +
f*x]^(7/2))
________________________________________________________________________________________
Maple [C] time = 0.196, size = 602, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(5/2),x)
[Out]
1/8/f*2^(1/2)*(b*sin(f*x+e)/cos(f*x+e))^(5/2)*(d/cos(f*x+e))^(1/2)*cos(f*x+e)*(3*I*cos(f*x+e)^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)*(-I*(cos(f*x+e)-1)/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)^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)*(-I*(cos(f*x+e)-1)/
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*cos(f*x+e
)^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)*(-I*(cos(f*
x+e)-1)/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*c
os(f*x+e)^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)*(-I
*(cos(f*x+e)-1)/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))/(cos(f*x+e)-1)/sin(f*x+e)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
integrate(sqrt(d*sec(f*x + e))*(b*tan(f*x + e))^(5/2), x)
________________________________________________________________________________________
Fricas [B] time = 2.9858, size = 2047, normalized size = 12.11 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[1/32*(6*sqrt(-b*d)*b^2*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*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*cos(
f*x + e)^2 - b*d - (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*cos(f*x + e) + 3*sqrt(-b*d)*b^2*cos(f*x + e)*log((b
*d*cos(f*x + e)^4 - 72*b*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*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)) + 72*b*d + 28*(b*d*cos(
f*x + e)^2 - 2*b*d)*sin(f*x + e))/(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^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)), 1/32*(6*sqr
t(b*d)*b^2*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*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*cos(f*x + e)^2 - b
*d + (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*cos(f*x + e) + 3*sqrt(b*d)*b^2*cos(f*x + e)*log((b*d*cos(f*x + e)
^4 - 72*b*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*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)) + 72*b*d - 28*(b*d*cos(f*x + e)^2 - 2*b
*d)*sin(f*x + e))/(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^2*sqrt
(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e))]
________________________________________________________________________________________
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((d*sec(f*x+e))**(1/2)*(b*tan(f*x+e))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(5/2),x, algorithm="giac")
[Out]
integrate(sqrt(d*sec(f*x + e))*(b*tan(f*x + e))^(5/2), x)