3.2.18 \(\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx\) [118]

Optimal. Leaf size=107 \[ -\frac {\text {ArcTan}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}}-\frac {\tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 107, 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, 218, 212, 209} \begin {gather*} -\frac {\sqrt {\cos (e+f x)} \text {ArcTan}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}}-\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right )}{a f \sqrt {a \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((ArcTan[Sqrt[Cos[e + f*x]]]*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(a*f*Sqrt[a*Sin[e + f*x]])) - (ArcTanh[
Sqrt[Cos[e + f*x]]]*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(a*f*Sqrt[a*Sin[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 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 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 {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx &=\frac {\left (\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \int \frac {\csc (e+f x)}{a \sqrt {\cos (e+f x)}} \, dx}{\sqrt {a \sin (e+f x)}}\\ &=\frac {\left (\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \int \frac {\csc (e+f x)}{\sqrt {\cos (e+f x)}} \, dx}{a \sqrt {a \sin (e+f x)}}\\ &=-\frac {\left (\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-x^2\right )} \, dx,x,\cos (e+f x)\right )}{a f \sqrt {a \sin (e+f x)}}\\ &=-\frac {\left (2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\sqrt {\cos (e+f x)}\right )}{a f \sqrt {a \sin (e+f x)}}\\ &=-\frac {\left (\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\cos (e+f x)}\right )}{a f \sqrt {a \sin (e+f x)}}-\frac {\left (\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\cos (e+f x)}\right )}{a f \sqrt {a \sin (e+f x)}}\\ &=-\frac {\tan ^{-1}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}}-\frac {\tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{a f \sqrt {a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 72, normalized size = 0.67 \begin {gather*} -\frac {b \left (\text {ArcTan}\left (\sqrt [4]{\cos ^2(e+f x)}\right )+\tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(184\) vs. \(2(91)=182\).
time = 0.34, size = 185, normalized size = 1.73

method	result	size
default	\(\frac {\left (\cos \left (f x +e \right )-1\right ) \left (\ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right )-\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )\right ) \cos \left (f x +e \right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}{2 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sin \left (f x +e \right )}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(cos(f*x+e)-1)*(ln(-2*(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)))*cos(f*x
+e)*(b*sin(f*x+e)/cos(f*x+e))^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(a*sin(f*x+e))^(3/2)/sin(f*x+e)

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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((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (99) = 198\).
time = 0.66, size = 447, normalized size = 4.18 \begin {gather*} \left [\frac {2 \, \sqrt {-\frac {b}{a}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{a}} \cos \left (f x + e\right )}{{\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right )}\right ) + \sqrt {-\frac {b}{a}} \log \left (-\frac {b \cos \left (f x + e\right )^{3} + 4 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, b \cos \left (f x + e\right )^{2} - 5 \, b \cos \left (f x + e\right ) + 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 f}, \frac {2 \, \sqrt {\frac {b}{a}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{{\left (b \cos \left (f x + e\right ) - b\right )} \sin \left (f x + e\right )}\right ) + \sqrt {\frac {b}{a}} \log \left (\frac {4 \, {\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 )}} \sqrt {\frac {b}{a}} - {\left (b \cos \left (f x + e\right )^{2} + 6 \, b \cos \left (f x + e\right ) + 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 f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))**(1/2)/(a*sin(f*x+e))**(3/2),x)

[Out]

Integral(sqrt(b*tan(e + f*x))/(a*sin(e + f*x))**(3/2), 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((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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