\(\int \csc ^5(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [136]

Optimal result

Integrand size = 25, antiderivative size = 128 \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {3 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{8 \sqrt {a} f}-\frac {3 (a+b) \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 f}-\frac {\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 f} \]

[Out]

-1/4*(a+b-b*cos(f*x+e)^2)^(3/2)*cot(f*x+e)*csc(f*x+e)^3/f-3/8*(a+b)^2*arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*
x+e)^2)^(1/2))/f/a^(1/2)-3/8*(a+b)*cot(f*x+e)*csc(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3265, 386, 385, 212} \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {3 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{8 \sqrt {a} f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 f}-\frac {3 (a+b) \cot (e+f x) \csc (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{8 f} \]

[In]

Int[Csc[e + f*x]^5*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(-3*(a + b)^2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(8*Sqrt[a]*f) - (3*(a + b)*Sqrt[
a + b - b*Cos[e + f*x]^2]*Cot[e + f*x]*Csc[e + f*x])/(8*f) - ((a + b - b*Cos[e + f*x]^2)^(3/2)*Cot[e + f*x]*Cs
c[e + f*x]^3)/(4*f)

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 386

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*(
(c + d*x^n)^q/(a*n*(p + 1))), x] - Dist[c*(q/(a*(p + 1))), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && 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*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {\left (a+b-b x^2\right )^{3/2}}{\left (1-x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 f}-\frac {(3 (a+b)) \text {Subst}\left (\int \frac {\sqrt {a+b-b x^2}}{\left (1-x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{4 f} \\ & = -\frac {3 (a+b) \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 f}-\frac {\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 f}-\frac {\left (3 (a+b)^2\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{8 f} \\ & = -\frac {3 (a+b) \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 f}-\frac {\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 f}-\frac {\left (3 (a+b)^2\right ) \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 )}{8 f} \\ & = -\frac {3 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{8 \sqrt {a} f}-\frac {3 (a+b) \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 f}-\frac {\left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{4 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89 \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {\frac {6 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )}{\sqrt {a}}+\sqrt {2} \sqrt {2 a+b-b \cos (2 (e+f x))} \cot (e+f x) \csc (e+f x) \left (3 a+5 b+2 a \csc ^2(e+f x)\right )}{16 f} \]

[In]

Integrate[Csc[e + f*x]^5*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-1/16*((6*(a + b)^2*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/Sqrt[a] + Sqrt
[2]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*Cot[e + f*x]*Csc[e + f*x]*(3*a + 5*b + 2*a*Csc[e + f*x]^2))/f

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(375\) vs. \(2(112)=224\).

Time = 1.10 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.94

method	result	size
default	\(-\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (3 a^{2} \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right )+6 a b \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right )+3 b^{2} \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{4}\left (f x +e \right )\right )+6 a^{\frac {3}{2}} \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (\sin ^{2}\left (f x +e \right )\right )+10 b \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (f x +e \right )\right )+4 a^{\frac {3}{2}} \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\right )}{16 \sqrt {a}\, \sin \left (f x +e \right )^{4} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(376\)

[In]

int(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(3*a^2*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*c
os(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^4+6*a*b*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+
b)*cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^4+3*b^2*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4
+(a+b)*cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^4+6*a^(3/2)*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*s
in(f*x+e)^2+10*b*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*a^(1/2)*sin(f*x+e)^2+4*a^(3/2)*(cos(f*x+e)^2*(a+b*sin
(f*x+e)^2))^(1/2))/a^(1/2)/sin(f*x+e)^4/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 484, normalized size of antiderivative = 3.78 \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} - 5 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{32 \, {\left (a f \cos \left (f x + e\right )^{4} - 2 \, a f \cos \left (f x + e\right )^{2} + a f\right )}}, \frac {3 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right ) + 2 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} - 5 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a f \cos \left (f x + e\right )^{4} - 2 \, a f \cos \left (f x + e\right )^{2} + a f\right )}}\right ] \]

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/32*(3*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^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)/(cos(f*x + e)^4 - 2*c
os(f*x + e)^2 + 1)) + 4*((3*a^2 + 5*a*b)*cos(f*x + e)^3 - 5*(a^2 + a*b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 +
a + b))/(a*f*cos(f*x + e)^4 - 2*a*f*cos(f*x + e)^2 + a*f), 1/16*(3*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a
^2 + 2*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)*sq
rt(-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*((3*a^2 + 5*a*b)*c
os(f*x + e)^3 - 5*(a^2 + a*b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(a*f*cos(f*x + e)^4 - 2*a*f*cos(f
*x + e)^2 + a*f)]

Sympy [F(-1)]

Timed out. \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(csc(f*x+e)**5*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{5} \,d x } \]

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^5, x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 912 vs. \(2 (112) = 224\).

Time = 0.72 (sec) , antiderivative size = 912, normalized size of antiderivative = 7.12 \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

1/64*(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*tan(1/2*
f*x + 1/2*e)^2 + (7*a^2 + 10*a*b)/a) + 24*(a^2 + 2*a*b + b^2)*arctan(-(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))/sqrt(-a) - 1
2*(a^(5/2) + 2*a^(3/2)*b + sqrt(a)*b^2)*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 + 4*(4*(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))^3*a^2 + 12*(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))^3*a*b + 10*(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))^3*b^2 + 5*(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*a^(5/2) +
8*(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*a^(3/2)*b - 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 - 8*(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^2*b - 6*(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*
b^2 - 3*a^(7/2) - 4*a^(5/2)*b)/((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 - a)^2)/f

Mupad [F(-1)]

Timed out. \[ \int \csc ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^5} \,d x \]

[In]

int((a + b*sin(e + f*x)^2)^(3/2)/sin(e + f*x)^5,x)

[Out]

int((a + b*sin(e + f*x)^2)^(3/2)/sin(e + f*x)^5, x)