3.2.55 \(\int \frac {\csc (e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [155]

Optimal. Leaf size=79 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{a^{3/2} f}+\frac {b \cos (e+f x)}{a (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3265, 390, 385, 212} \begin {gather*} \frac {b \cos (e+f x)}{a f (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{a^{3/2} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]]/(a^(3/2)*f)) + (b*Cos[e + f*x])/(a*(a + b)*f*
Sqrt[a + b - b*Cos[e + f*x]^2])

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 3265

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, Dist[-ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.23, size = 93, normalized size = 1.18 \begin {gather*} \frac {-\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+\frac {\sqrt {2} \sqrt {a} b \cos (e+f x)}{(a+b) \sqrt {2 a+b-b \cos (2 (e+f x))}}}{a^{3/2} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] + (Sqrt[2]*Sqrt[a]*b*Cos[e + f*x]
)/((a + b)*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]))/(a^(3/2)*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(164\) vs. \(2(71)=142\).
time = 12.07, size = 165, normalized size = 2.09

method	result	size
default	\(\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (-\frac {\ln \left (\frac {2 a +\left (-a +b \right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )^{2}}\right )}{2 a^{\frac {3}{2}}}+\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a \left (a +b \right ) \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (75) = 150\).
time = 0.54, size = 174, normalized size = 2.20 \begin {gather*} \frac {\frac {2 \, b^{2} \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{2} b + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b^{2}} - \frac {\log \left (b - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) - 1} - \frac {a}{\cos \left (f x + e\right ) - 1}\right )}{a^{\frac {3}{2}}} + \frac {\log \left (-b + \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) + 1} + \frac {a}{\cos \left (f x + e\right ) + 1}\right )}{a^{\frac {3}{2}}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (71) = 142\).
time = 0.50, size = 422, normalized size = 5.34 \begin {gather*} \left [-\frac {4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) - {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}, -\frac {2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) - {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right )}{2 \, {\left ({\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(e + f*x)/(a + b*sin(e + f*x)**2)**(3/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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