3.4.17 \(\int \frac {(d \sec (e+f x))^{3/2}}{\sqrt {b \tan (e+f x)}} \, dx\) [317]

Optimal. Leaf size=131 \[ \frac {d \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{\sqrt {b} f \sqrt {b \tan (e+f x)}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{\sqrt {b} f \sqrt {b \tan (e+f x)}} \]

[Out]

d*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(d*sec(f*x+e))^(1/2)*(b*sin(f*x+e))^(1/2)/f/b^(1/2)/(b*tan(f*x+e))^(1/2
)+d*arctanh((b*sin(f*x+e))^(1/2)/b^(1/2))*(d*sec(f*x+e))^(1/2)*(b*sin(f*x+e))^(1/2)/f/b^(1/2)/(b*tan(f*x+e))^(
1/2)

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Rubi [A]
time = 0.07, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2696, 2644, 335, 218, 212, 209} \begin {gather*} \frac {d \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{\sqrt {b} f \sqrt {b \tan (e+f x)}}+\frac {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 )}{\sqrt {b} f \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(Sqrt[b]*f*Sqrt[b*Tan[e + f
*x]]) + (d*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(Sqrt[b]*f*Sqrt[b*
Tan[e + f*x]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 335

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 2644

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 2696

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]

Rubi steps

\begin {align*} \int \frac {(d \sec (e+f x))^{3/2}}{\sqrt {b \tan (e+f x)}} \, dx &=\frac {\left (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}{\sqrt {b \tan (e+f x)}}\\ &=\frac {\left (d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{b^2}\right )} \, dx,x,b \sin (e+f x)\right )}{b f \sqrt {b \tan (e+f x)}}\\ &=\frac {\left (2 d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{1-\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{b f \sqrt {b \tan (e+f x)}}\\ &=\frac {\left (d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{f \sqrt {b \tan (e+f x)}}+\frac {\left (d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{f \sqrt {b \tan (e+f x)}}\\ &=\frac {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)}}{\sqrt {b} f \sqrt {b \tan (e+f x)}}+\frac {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)}}{\sqrt {b} f \sqrt {b \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 5.01, size = 105, normalized size = 0.80 \begin {gather*} -\frac {\left (\text {ArcTan}\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 ) (d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}{b f \sec ^{\frac {3}{2}}(e+f x) \sqrt [4]{\tan ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((ArcTan[Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)] - ArcTanh[Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)])*(d
*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]])/(b*f*Sec[e + f*x]^(3/2)*(Tan[e + f*x]^2)^(1/4)))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.36, size = 344, normalized size = 2.63

method	result	size
default	\(\frac {\sqrt {2}\, \left (2 i \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \cos \left (f x +e \right ) \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}}{2 f \left (\cos \left (f x +e \right )-1\right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}\)	\(344\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2/f*2^(1/2)*(2*I*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))-I*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*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))+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))-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)*(-I*(cos(f*
x+e)-1)/sin(f*x+e))^(1/2)*(d/cos(f*x+e))^(3/2)*sin(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)/(cos(f*x+e)-1)/(b*sin(f*x+e)/cos(f*x+e))^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(3/2)/sqrt(b*tan(f*x + e)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (115) = 230\).
time = 0.57, size = 709, normalized size = 5.41 \begin {gather*} \left [-\frac {2 \, d \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (d \cos \left (f x + e\right )^{2} - {\left (d \cos \left (f x + e\right ) + d\right )} \sin \left (f x + e\right ) - d\right )}}\right ) - d \sqrt {-\frac {d}{b}} \log \left (\frac {d \cos \left (f x + e\right )^{4} - 72 \, d \cos \left (f x + e\right )^{2} + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 28 \, {\left (d \cos \left (f x + e\right )^{2} - 2 \, d\right )} \sin \left (f x + e\right ) + 72 \, d}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right )}{8 \, f}, \frac {2 \, d \sqrt {\frac {d}{b}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (d \cos \left (f x + e\right )^{2} + {\left (d \cos \left (f x + e\right ) + d\right )} \sin \left (f x + e\right ) - d\right )}}\right ) + d \sqrt {\frac {d}{b}} \log \left (\frac {d \cos \left (f x + e\right )^{4} - 72 \, d \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} - 28 \, {\left (d \cos \left (f x + e\right )^{2} - 2 \, d\right )} \sin \left (f x + e\right ) + 72 \, d}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right )}{8 \, f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(2*d*sqrt(-d/b)*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)*si
n(f*x + e) - 2*cos(f*x + e) + 4)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-d/b)*sqrt(d/cos(f*x + e))/(d*cos(f*x
+ e)^2 - (d*cos(f*x + e) + d)*sin(f*x + e) - d)) - d*sqrt(-d/b)*log((d*cos(f*x + e)^4 - 72*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*sin(f*x + e)/cos
(f*x + e))*sqrt(-d/b)*sqrt(d/cos(f*x + e)) + 28*(d*cos(f*x + e)^2 - 2*d)*sin(f*x + e) + 72*d)/(cos(f*x + e)^4
- 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)))/f, 1/8*(2*d*sqrt(d/b)*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(d/b)*sqrt(d/cos(f*x + e))/(d*cos(f*x + e)^2 + (d*cos(f*x + e) + d)*sin(f*x + e) - d)
) + d*sqrt(d/b)*log((d*cos(f*x + e)^4 - 72*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*sin(f*x + e)/cos(f*x + e))*sqrt(d/b)*sqrt(d/cos(f*x + e)) - 28*(
d*cos(f*x + e)^2 - 2*d)*sin(f*x + e) + 72*d)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - 2)*sin(f
*x + e) + 8)))/f]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {b \tan {\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(3/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Integral((d*sec(e + f*x))**(3/2)/sqrt(b*tan(e + f*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(3/2)/sqrt(b*tan(f*x + e)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(3/2)/(b*tan(e + f*x))^(1/2),x)

[Out]

int((d/cos(e + f*x))^(3/2)/(b*tan(e + f*x))^(1/2), x)

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