3.302 \(\int \frac{(b \tan (e+f x))^{3/2}}{\sqrt{d \sec (e+f x)}} \, dx\)
Optimal. Leaf size=167 \[ \frac{b^{3/2} d (b \tan (e+f x))^{3/2} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}+\frac{b^{3/2} d (b \tan (e+f x))^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}-\frac{2 d \csc (e+f x) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2}} \]
[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))
________________________________________________________________________________________
Rubi [A] time = 0.125362, antiderivative size = 167, 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 =
{2616, 2564, 321, 329, 212, 206, 203} \[ \frac{b^{3/2} d (b \tan (e+f x))^{3/2} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}+\frac{b^{3/2} d (b \tan (e+f x))^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}-\frac{2 d \csc (e+f x) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[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 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 321
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 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 212
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] &&
!GtQ[a/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])
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])
Rubi steps
\begin{align*} \int \frac{(b \tan (e+f x))^{3/2}}{\sqrt{d \sec (e+f x)}} \, dx &=\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 \tan ^{-1}\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 \tanh ^{-1}\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] time = 4.1733, size = 64, normalized size = 0.38 \[ \frac{2 (b \tan (e+f x))^{5/2} \, _2F_1\left (-\frac{1}{4},-\frac{1}{4};\frac{3}{4};\sec ^2(e+f x)\right )}{b f \left (-\tan ^2(e+f x)\right )^{5/4} \sqrt{d \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(b*Tan[e + f*x])^(3/2)/Sqrt[d*Sec[e + f*x]],x]
[Out]
(2*Hypergeometric2F1[-1/4, -1/4, 3/4, Sec[e + f*x]^2]*(b*Tan[e + f*x])^(5/2))/(b*f*Sqrt[d*Sec[e + f*x]]*(-Tan[
e + f*x]^2)^(5/4))
________________________________________________________________________________________
Maple [C] time = 0.227, size = 719, normalized size = 4.3 \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))^(3/2)/(d*sec(f*x+e))^(1/2),x)
[Out]
1/2/f*2^(1/2)*(2*I*sin(f*x+e)*((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)*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-I*(cos(f*x+e)-1)/sin(f*
x+e))^(1/2)-I*sin(f*x+e)*((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))*(-I*(cos(f*x+e)-1)/
sin(f*x+e))^(1/2)-I*sin(f*x+e)*((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))/si
n(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))*(-I*(cos(f*x+
e)-1)/sin(f*x+e))^(1/2)+sin(f*x+e)*((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))*(-I*(cos(
f*x+e)-1)/sin(f*x+e))^(1/2)-sin(f*x+e)*((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))*(-I*(
cos(f*x+e)-1)/sin(f*x+e))^(1/2)-2*cos(f*x+e)*2^(1/2)+2*2^(1/2))*(b*sin(f*x+e)/cos(f*x+e))^(3/2)*cos(f*x+e)/(co
s(f*x+e)-1)/(d/cos(f*x+e))^(1/2)/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{3}{2}}}{\sqrt{d \sec \left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
Fricas [B] time = 5.35585, size = 1886, normalized size = 11.29 \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))^(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(-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))**(3/2)/(d*sec(f*x+e))**(1/2),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{3}{2}}}{\sqrt{d \sec \left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)