3.3.93 \(\int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx\) [293]

Optimal. Leaf size=132 \[ -\frac {\sqrt {b} d \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {\sqrt {b} d \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 132, 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, 304, 209, 212} \begin {gather*} \frac {\sqrt {b} d \sqrt {b \tan (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {\sqrt {b} d \sqrt {b \tan (e+f x)} \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]],x]

[Out]

-((Sqrt[b]*d*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e +
f*x]])) + (Sqrt[b]*d*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(f*Sqrt[d*Sec[e + f*x]]*Sqrt
[b*Sin[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 304

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

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Mathematica [A]
time = 0.79, size = 136, normalized size = 1.03 \begin {gather*} \frac {b \left (2 \text {ArcTan}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\log \left (1-\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )+\log \left (1+\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )\right ) \sqrt {d \sec (e+f x)} \sqrt [4]{\tan ^2(e+f x)}}{2 f \sqrt {\sec (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]],x]

[Out]

(b*(2*ArcTan[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)] +
Log[1 + Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)])*Sqrt[d*Sec[e + f*x]]*(Tan[e + f*x]^2)^(1/4))/(2*f*Sqrt[Se
c[e + f*x]]*Sqrt[b*Tan[e + f*x]])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.40, size = 306, normalized size = 2.32

method	result	size
default	\(\frac {\sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {2}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (i \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right )}{2 f \left (\cos \left (f x +e \right )-1\right )}\)	\(306\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(sqrt(d*sec(f*x + e))*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. 351 vs. \(2 (116) = 232\).
time = 0.52, size = 710, normalized size = 5.38 \begin {gather*} \left [-\frac {2 \, \sqrt {-b d} \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 {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d - {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) - \sqrt {-b d} \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b 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 {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d + 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\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 \, \sqrt {b d} \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 {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d + {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) - \sqrt {b d} \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b 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 {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d - 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\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))^(1/2)*(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(2*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*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))) - sqrt(-b*d)*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))/(c
os(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*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*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))) - sqrt(b*d)*log((b*d*cos(f*x + e)^4 - 72*b*d*cos(f*x + e)^2 - 8*(7*cos(f*x + e)^3 + (co
s(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)))/f]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {b \tan {\left (e + f x \right )}} \sqrt {d \sec {\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))**(1/2)*(b*tan(f*x+e))**(1/2),x)

[Out]

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

[Out]

integrate(sqrt(d*sec(f*x + e))*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 \sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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