3.2.56 \(\int \frac {\csc ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [156]

Optimal. Leaf size=134 \[ -\frac {(a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 a^{5/2} f}-\frac {b (a+3 b) \cos (e+f x)}{2 a^2 (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b-b \cos ^2(e+f x)}} \]

[Out]

-1/2*(a-3*b)*arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/a^(5/2)/f-1/2*b*(a+3*b)*cos(f*x+e)/a^2/(a+
b)/f/(a+b-b*cos(f*x+e)^2)^(1/2)-1/2*cot(f*x+e)*csc(f*x+e)/a/f/(a+b-b*cos(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 134, 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 = {3265, 425, 541, 12, 385, 212} \begin {gather*} -\frac {(a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 a^{5/2} f}-\frac {b (a+3 b) \cos (e+f x)}{2 a^2 f (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a-b \cos ^2(e+f x)+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a - 3*b)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(a^(5/2)*f) - (b*(a + 3*b)*Cos
[e + f*x])/(2*a^2*(a + b)*f*Sqrt[a + b - b*Cos[e + f*x]^2]) - (Cot[e + f*x]*Csc[e + f*x])/(2*a*f*Sqrt[a + b -
b*Cos[e + f*x]^2])

Rule 12

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

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 425

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[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[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 ^3(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 )^2 \left (a+b-b x^2\right )^{3/2}} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\text {Subst}\left (\int \frac {a-b-2 b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )^{3/2}} \, dx,x,\cos (e+f x)\right )}{2 a f}\\ &=-\frac {b (a+3 b) \cos (e+f x)}{2 a^2 (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\text {Subst}\left (\int \frac {(a-3 b) (a+b)}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{2 a^2 (a+b) f}\\ &=-\frac {b (a+3 b) \cos (e+f x)}{2 a^2 (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {(a-3 b) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{2 a^2 f}\\ &=-\frac {b (a+3 b) \cos (e+f x)}{2 a^2 (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {(a-3 b) \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 )}{2 a^2 f}\\ &=-\frac {(a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 a^{5/2} f}-\frac {b (a+3 b) \cos (e+f x)}{2 a^2 (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b-b \cos ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.48, size = 134, normalized size = 1.00 \begin {gather*} \frac {-\frac {(a-3 b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )}{a^{5/2}}+\frac {\left (-2 a^2-3 a b-3 b^2+b (a+3 b) \cos (2 (e+f x))\right ) \cot (e+f x) \csc (e+f x)}{\sqrt {2} a^2 (a+b) \sqrt {2 a+b-b \cos (2 (e+f x))}}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((a - 3*b)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/a^(5/2)) + ((-2*a^2
- 3*a*b - 3*b^2 + b*(a + 3*b)*Cos[2*(e + f*x)])*Cot[e + f*x]*Csc[e + f*x])/(Sqrt[2]*a^2*(a + b)*Sqrt[2*a + b -
b*Cos[2*(e + f*x)]]))/(2*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(273\) vs. \(2(118)=236\).
time = 16.32, size = 274, normalized size = 2.04

method	result	size
default	\(\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (-\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{2 a^{2} \sin \left (f x +e \right )^{2}}-\frac {\ln \left (\frac {2 a +\left (-a +b \right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )^{2}}\right )}{4 a^{\frac {3}{2}}}+\frac {3 b \ln \left (\frac {2 a +\left (-a +b \right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )^{2}}\right )}{4 a^{\frac {5}{2}}}-\frac {b^{2} \left (\cos ^{2}\left (f x +e \right )\right )}{a^{2} \left (a +b \right ) \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(274\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3/(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^2/sin(f*x+e)^2*(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)-1/4
/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)+3/4/a^
(5/2)*b*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)-b^2/a^2
*cos(f*x+e)^2/(a+b)/(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (118) = 236\).
time = 0.69, size = 634, normalized size = 4.73 \begin {gather*} \left [-\frac {{\left ({\left (a^{2} b - 2 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + a^{3} - a^{2} b - 5 \, a b^{2} - 3 \, b^{3} - {\left (a^{3} - 7 \, a b^{2} - 6 \, b^{3}\right )} \cos \left (f x + e\right )^{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^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 2 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left ({\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}}, \frac {{\left ({\left (a^{2} b - 2 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + a^{3} - a^{2} b - 5 \, a b^{2} - 3 \, b^{3} - {\left (a^{3} - 7 \, a b^{2} - 6 \, b^{3}\right )} \cos \left (f x + e\right )^{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^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 2 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{4 \, {\left ({\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(((a^2*b - 2*a*b^2 - 3*b^3)*cos(f*x + e)^4 + a^3 - a^2*b - 5*a*b^2 - 3*b^3 - (a^3 - 7*a*b^2 - 6*b^3)*cos
(f*x + e)^2)*sqrt(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)/(co
s(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 4*((a^2*b + 3*a*b^2)*cos(f*x + e)^3 - (a^3 + 2*a^2*b + 3*a*b^2)*cos(f*
x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b + a^3*b^2)*f*cos(f*x + e)^4 - (a^5 + 3*a^4*b + 2*a^3*b^2)*f*c
os(f*x + e)^2 + (a^5 + 2*a^4*b + a^3*b^2)*f), 1/4*(((a^2*b - 2*a*b^2 - 3*b^3)*cos(f*x + e)^4 + a^3 - a^2*b - 5
*a*b^2 - 3*b^3 - (a^3 - 7*a*b^2 - 6*b^3)*cos(f*x + e)^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))) + 2*((a^2*b + 3*a*b
^2)*cos(f*x + e)^3 - (a^3 + 2*a^2*b + 3*a*b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b + a^3*b^
2)*f*cos(f*x + e)^4 - (a^5 + 3*a^4*b + 2*a^3*b^2)*f*cos(f*x + e)^2 + (a^5 + 2*a^4*b + a^3*b^2)*f)]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (125) = 250\).
time = 0.72, size = 520, normalized size = 3.88 \begin {gather*} \frac {\frac {{\left (\frac {{\left (a^{5} b + a^{4} b^{2}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{6} b + a^{5} b^{2}} + \frac {2 \, {\left (a^{5} b + 3 \, a^{4} b^{2} + 6 \, a^{3} b^{3}\right )}}{a^{6} b + a^{5} b^{2}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {a^{5} b + a^{4} b^{2} - 8 \, a^{3} b^{3}}{a^{6} b + a^{5} b^{2}}}{\sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}} - \frac {2 \, {\left (a^{\frac {3}{2}} - 3 \, \sqrt {a} b\right )} \log \left ({\left | -{\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} a - a^{\frac {3}{2}} - 2 \, \sqrt {a} b \right |}\right )}{a^{3}} + \frac {2 \, {\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} a^{\frac {3}{2}} + 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} \sqrt {a} b + a^{2}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} \sqrt {a} - a^{\frac {3}{2}}\right )} a^{2}}}{8 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((((a^5*b + a^4*b^2)*tan(1/2*f*x + 1/2*e)^2/(a^6*b + a^5*b^2) + 2*(a^5*b + 3*a^4*b^2 + 6*a^3*b^3)/(a^6*b +
a^5*b^2))*tan(1/2*f*x + 1/2*e)^2 + (a^5*b + a^4*b^2 - 8*a^3*b^3)/(a^6*b + a^5*b^2))/sqrt(a*tan(1/2*f*x + 1/2*
e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - 2*(a^(3/2) - 3*sqrt(a)*b)*log(abs(-(sqrt
(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/
2*e)^2 + a))*a - a^(3/2) - 2*sqrt(a)*b))/a^3 + 2*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e
)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^(3/2) + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^
2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a)*b +
a^2)/(((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(
1/2*f*x + 1/2*e)^2 + a))^2*sqrt(a) - a^(3/2))*a^2))/f

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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 )}^3\,{\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)^3*(a + b*sin(e + f*x)^2)^(3/2)),x)

[Out]

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

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