3.291 \(\int (d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)} \, dx\)

Optimal. Leaf size=178 \[ \frac{d^2 (b \tan (e+f x))^{3/2} \sqrt{d \sec (e+f x)}}{2 b f}-\frac{\sqrt{b} d^3 \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{\sqrt{b} d^3 \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.162123, antiderivative size = 178, 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 = {2613, 2616, 2564, 329, 298, 203, 206} \[ \frac{d^2 (b \tan (e+f x))^{3/2} \sqrt{d \sec (e+f x)}}{2 b f}-\frac{\sqrt{b} d^3 \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{\sqrt{b} d^3 \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{4 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]

[Out]

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

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && 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 (d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)} \, dx &=\frac{d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 b f}+\frac{1}{4} d^2 \int \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)} \, dx\\ &=\frac{d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 b f}+\frac{\left (d^3 \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{d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 b f}+\frac{\left (d^3 \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 b f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 b f}+\frac{\left (d^3 \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 b f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 b f}+\frac{\left (b d^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 )}{4 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}-\frac{\left (b d^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 )}{4 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=-\frac{\sqrt{b} d^3 \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{\sqrt{b} d^3 \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{d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 b f}\\ \end{align*}

Mathematica [A]  time = 1.90293, size = 174, normalized size = 0.98 \[ \frac{b (d \sec (e+f x))^{5/2} \left (4 \sec ^{\frac{5}{2}}(e+f x)-4 \sqrt{\sec (e+f x)}+2 \sqrt [4]{\tan ^2(e+f x)} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )+\sqrt [4]{\tan ^2(e+f x)} \left (\log \left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}+1\right )-\log \left (1-\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )\right )\right )}{8 f \sec ^{\frac{5}{2}}(e+f x) \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]

[Out]

(b*(d*Sec[e + f*x])^(5/2)*(-4*Sqrt[Sec[e + f*x]] + 4*Sec[e + f*x]^(5/2) + 2*ArcTan[Sqrt[Sec[e + f*x]]/(Tan[e +
f*x]^2)^(1/4)]*(Tan[e + f*x]^2)^(1/4) + (-Log[1 - Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)] + Log[1 + Sqrt[S
ec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)])*(Tan[e + f*x]^2)^(1/4)))/(8*f*Sec[e + f*x]^(5/2)*Sqrt[b*Tan[e + f*x]])

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Maple [C]  time = 0.316, size = 600, 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((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x)

[Out]

1/8/f*2^(1/2)*(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))-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))-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-s
in(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))-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))+2*cos(f*x+e)*2^(1/2)-2*2^(1/2))*cos(f*x+e)*(d/cos(f*x+e))^(5/2)*(b*
sin(f*x+e)/cos(f*x+e))^(1/2)*sin(f*x+e)/(cos(f*x+e)-1)

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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{5}{2}} \sqrt{b \tan \left (f x + e\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(5/2)*sqrt(b*tan(f*x + e)), x)

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Fricas [B]  time = 2.90462, size = 2044, normalized size = 11.48 \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))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/32*(2*sqrt(-b*d)*d^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) - sqrt(-b*d)*d^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*d^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*(2*sqr
t(b*d)*d^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) - sqrt(b*d)*d^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*d^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))]

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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))**(5/2)*(b*tan(f*x+e))**(1/2),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{5}{2}} \sqrt{b \tan \left (f x + e\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)*sqrt(b*tan(f*x + e)), x)