3.2.31 \(\int \frac {1}{\sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \, dx\) [131]

Optimal. Leaf size=106 \[ \frac {\text {ArcTan}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {a \sin (e+f x)}}{a f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {a \sin (e+f x)}}{a f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2681, 12, 2645, 335, 304, 209, 212} \begin {gather*} \frac {\sqrt {a \sin (e+f x)} \text {ArcTan}\left (\sqrt {\cos (e+f x)}\right )}{a f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\sqrt {a \sin (e+f x)} \tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right )}{a f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]]),x]

[Out]

(ArcTan[Sqrt[Cos[e + f*x]]]*Sqrt[a*Sin[e + f*x]])/(a*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) - (ArcTanh[Sqr
t[Cos[e + f*x]]]*Sqrt[a*Sin[e + f*x]])/(a*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 80, normalized size = 0.75 \begin {gather*} \frac {\left (\text {ArcTan}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-\tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right ) \sin (2 (e+f x))}{2 f \cos ^2(e+f x)^{3/4} \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]]),x]

[Out]

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

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Maple [A]
time = 0.33, size = 177, normalized size = 1.67

method	result	size
default	\(-\frac {\left (\cos \left (f x +e \right )-1\right ) \left (\ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )\right )}{2 f \sqrt {a \sin \left (f x +e \right )}\, \sin \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(cos(f*x+e)-1)*(ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-
cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)))/(a*sin(f
*x+e))^(1/2)/sin(f*x+e)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(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(1/(a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(a*sin(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. 223 vs. \(2 (98) = 196\).
time = 0.60, size = 453, normalized size = 4.27 \begin {gather*} \left [\frac {2 \, \sqrt {-a b} \arctan \left (\frac {2 \, \sqrt {-a b} \sqrt {a \sin \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 (a b \cos \left (f x + e\right ) + a b\right )} \sin \left (f x + e\right )}\right ) - \sqrt {-a b} \log \left (-\frac {a b \cos \left (f x + e\right )^{3} - 5 \, a b \cos \left (f x + e\right )^{2} + 4 \, \sqrt {-a b} \sqrt {a \sin \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 ) \sin \left (f x + e\right ) - 5 \, a b \cos \left (f x + e\right ) + a b}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, a b f}, -\frac {2 \, \sqrt {a b} \arctan \left (\frac {2 \, \sqrt {a b} \sqrt {a \sin \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 (a b \cos \left (f x + e\right ) - a b\right )} \sin \left (f x + e\right )}\right ) - \sqrt {a b} \log \left (\frac {4 \, \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} - {\left (a b \cos \left (f x + e\right )^{2} + 6 \, a b \cos \left (f x + e\right ) + a b\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}\right )}{4 \, a b f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(-a*b)*arctan(2*sqrt(-a*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/((a
*b*cos(f*x + e) + a*b)*sin(f*x + e))) - sqrt(-a*b)*log(-(a*b*cos(f*x + e)^3 - 5*a*b*cos(f*x + e)^2 + 4*sqrt(-a
*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 5*a*b*cos(f*x + e) + a*
b)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 3*cos(f*x + e) + 1)))/(a*b*f), -1/4*(2*sqrt(a*b)*arctan(2*sqrt(a*b)*sq
rt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/((a*b*cos(f*x + e) - a*b)*sin(f*x + e))) - s
qrt(a*b)*log((4*sqrt(a*b)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e
)) - (a*b*cos(f*x + e)^2 + 6*a*b*cos(f*x + e) + a*b)*sin(f*x + e))/((cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*sin(
f*x + e))))/(a*b*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a \sin {\left (e + f x \right )}} \sqrt {b \tan {\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(f*x+e))**(1/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

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

[Out]

integrate(1/(sqrt(a*sin(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 \frac {1}{\sqrt {a\,\sin \left (e+f\,x\right )}\,\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(1/((a*sin(e + f*x))^(1/2)*(b*tan(e + f*x))^(1/2)),x)

[Out]

int(1/((a*sin(e + f*x))^(1/2)*(b*tan(e + f*x))^(1/2)), x)

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