3.300 \(\int (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2} \, dx\)
Optimal. Leaf size=169 \[ -\frac{b^{3/2} d \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \tan (e+f x)}}-\frac{b^{3/2} d \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \tan (e+f x)}}+\frac{b \sqrt{b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 f} \]
[Out]
-(b^(3/2)*d*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(4*f*Sqrt[b*Tan[e
+ f*x]]) - (b^(3/2)*d*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(4*f*Sq
rt[b*Tan[e + f*x]]) + (b*(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]])/(2*f)
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Rubi [A] time = 0.173737, 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, 212, 206, 203} \[ -\frac{b^{3/2} d \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \tan (e+f x)}}-\frac{b^{3/2} d \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \tan (e+f x)}}+\frac{b \sqrt{b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[(d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(3/2),x]
[Out]
-(b^(3/2)*d*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(4*f*Sqrt[b*Tan[e
+ f*x]]) - (b^(3/2)*d*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(4*f*Sq
rt[b*Tan[e + f*x]]) + (b*(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]])/(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 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 (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2} \, dx &=\frac{b (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{2 f}-\frac{1}{4} b^2 \int \frac{(d \sec (e+f x))^{3/2}}{\sqrt{b \tan (e+f x)}} \, dx\\ &=\frac{b (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{2 f}-\frac{\left (b^2 d \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \int \frac{\sec (e+f x)}{\sqrt{b \sin (e+f x)}} \, dx}{4 \sqrt{b \tan (e+f x)}}\\ &=\frac{b (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{2 f}-\frac{\left (b d \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\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 )}{4 f \sqrt{b \tan (e+f x)}}\\ &=\frac{b (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{2 f}-\frac{\left (b d \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{2 f \sqrt{b \tan (e+f x)}}\\ &=\frac{b (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{2 f}-\frac{\left (b^2 d \sqrt{d \sec (e+f x)} \sqrt{b \sin (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{b \tan (e+f x)}}-\frac{\left (b^2 d \sqrt{d \sec (e+f x)} \sqrt{b \sin (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{b \tan (e+f x)}}\\ &=-\frac{b^{3/2} d \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}{4 f \sqrt{b \tan (e+f x)}}-\frac{b^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}{4 f \sqrt{b \tan (e+f x)}}+\frac{b (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{2 f}\\ \end{align*}
Mathematica [A] time = 6.29854, size = 129, normalized size = 0.76 \[ \frac{b \sqrt{b \tan (e+f x)} (d \sec (e+f x))^{3/2} \left (2 \sqrt [4]{\tan ^2(e+f x)} \sec ^{\frac{3}{2}}(e+f x)+\tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\tanh ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )\right )}{4 f \sqrt [4]{\tan ^2(e+f x)} \sec ^{\frac{3}{2}}(e+f x)} \]
Antiderivative was successfully verified.
[In]
Integrate[(d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(3/2),x]
[Out]
(b*(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]*(ArcTan[Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)] - ArcTanh[Sq
rt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)] + 2*Sec[e + f*x]^(3/2)*(Tan[e + f*x]^2)^(1/4)))/(4*f*Sec[e + f*x]^(3/
2)*(Tan[e + f*x]^2)^(1/4))
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Maple [C] time = 0.183, size = 759, normalized size = 4.5 \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))^(3/2)*(b*tan(f*x+e))^(3/2),x)
[Out]
1/8/f*2^(1/2)*(I*cos(f*x+e)^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)*(-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))+I*cos(f*x+e)^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)*(-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*I*cos(f*x+e)^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)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*Ellipti
cF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+cos(f*x+e)^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)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*E
llipticPi(((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))-cos(f*x+e)^2*sin(f*x+e)*((I*co
s(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)/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))+2*cos(f*x+e)*
2^(1/2)-2*2^(1/2))*cos(f*x+e)*(d/cos(f*x+e))^(3/2)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)/(cos(f*x+e)-1)/sin(f*x+e)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (b \tan \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((d*sec(f*x+e))^(3/2)*(b*tan(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e))^(3/2), x)
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Fricas [B] time = 2.99851, size = 2007, normalized size = 11.88 \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))^(3/2)*(b*tan(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
[1/32*(2*sqrt(-b*d)*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*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) + sqrt(-b*d)*b*d*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*d*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)), -1/32*(2*sqrt(b*d)*b*d*arc
tan(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) - sqrt(b*d)*b*d*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*d*sqrt(b*sin(f*x + e)/
cos(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e))]
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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))**(3/2)*(b*tan(f*x+e))**(3/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (b \tan \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((d*sec(f*x+e))^(3/2)*(b*tan(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e))^(3/2), x)