3.293 \(\int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx\)

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

[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 = 0.10, 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 = {2616, 2564, 329, 298, 203, 206} \[ \frac {\sqrt {b} d \sqrt {b \tan (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {\sqrt {b} d \sqrt {b \tan (e+f x)} \tan ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}} \]

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 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])

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 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 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 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]

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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.73, size = 136, normalized size = 1.03 \[ \frac {b \sqrt [4]{\tan ^2(e+f x)} \sqrt {d \sec (e+f x)} \left (2 \tan ^{-1}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\log \left (1-\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )+\log \left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}+1\right )\right )}{2 f \sqrt {\sec (e+f x)} \sqrt {b \tan (e+f x)}} \]

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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fricas [B]  time = 0.80, size = 654, normalized size = 4.95 \[ \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 ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )}\,{d x} \]

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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maple [C]  time = 0.67, size = 302, normalized size = 2.29 \[ -\frac {\sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (i \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right )}{2 f \left (-1+\cos \left (f x +e \right )\right )} \]

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)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan {\left (e + f x \right )}} \sqrt {d \sec {\left (e + f x \right )}}\, dx \]

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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