3.135 \(\int \csc ^3(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\)
Optimal. Leaf size=128 \[ -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{f}-\frac {\sqrt {a} (a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 f}-\frac {a \cot (e+f x) \csc (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 f} \]
[Out]
-b^(3/2)*arctan(cos(f*x+e)*b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/f-1/2*(a+3*b)*arctanh(cos(f*x+e)*a^(1/2)/(a+b-b
*cos(f*x+e)^2)^(1/2))*a^(1/2)/f-1/2*a*cot(f*x+e)*csc(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/f
________________________________________________________________________________________
Rubi [A] time = 0.15, antiderivative size = 128, 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 =
{3186, 413, 523, 217, 203, 377, 206} \[ -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{f}-\frac {\sqrt {a} (a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 f}-\frac {a \cot (e+f x) \csc (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^3*(a + b*Sin[e + f*x]^2)^(3/2),x]
[Out]
-((b^(3/2)*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/f) - (Sqrt[a]*(a + 3*b)*ArcTanh[(Sqr
t[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(2*f) - (a*Sqrt[a + b - b*Cos[e + f*x]^2]*Cot[e + f*x]*Csc
[e + f*x])/(2*f)
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 377
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 413
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
Rule 3186
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 \csc ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b-b x^2\right )^{3/2}}{\left (1-x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {a \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{2 f}+\frac {\operatorname {Subst}\left (\int \frac {-(a+b) (a+2 b)+2 b^2 x^2}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{2 f}\\ &=-\frac {a \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{2 f}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{f}-\frac {(a (a+3 b)) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{2 f}\\ &=-\frac {a \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{2 f}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{f}-\frac {(a (a+3 b)) \operatorname {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 f}\\ &=-\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{f}-\frac {\sqrt {a} (a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 f}-\frac {a \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{2 f}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 1.22, size = 147, normalized size = 1.15 \[ -\frac {4 (-b)^{3/2} \log \left (\sqrt {2 a-b \cos (2 (e+f x))+b}+\sqrt {2} \sqrt {-b} \cos (e+f x)\right )+2 \sqrt {a} (a+3 b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a-b \cos (2 (e+f x))+b}}\right )+\sqrt {2} a \cot (e+f x) \csc (e+f x) \sqrt {2 a-b \cos (2 (e+f x))+b}}{4 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[e + f*x]^3*(a + b*Sin[e + f*x]^2)^(3/2),x]
[Out]
-1/4*(2*Sqrt[a]*(a + 3*b)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] + Sqrt[2]
*a*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*Cot[e + f*x]*Csc[e + f*x] + 4*(-b)^(3/2)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*
x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/f
________________________________________________________________________________________
fricas [B] time = 0.94, size = 1449, normalized size = 11.32 \[ \text {result too large to display} \]
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*(4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x + e) + (b*cos(f*x + e)^2 - b)*sqrt(-b)*log(128*b^4*cos(f*x +
e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^
2*b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^7 - 24*
(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos
(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)) + ((a + 3*b)*cos(f*x + e)^2 - a - 3*b)*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*cos(f*x + e)^2
+ 1)))/(f*cos(f*x + e)^2 - f), 1/8*(2*((a + 3*b)*cos(f*x + e)^2 - a - 3*b)*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)))
+ 4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x + e) + (b*cos(f*x + e)^2 - b)*sqrt(-b)*log(128*b^4*cos(f*x + e)^
8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^
2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b
^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x
+ e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)))/(f*cos(f*x + e)^2 - f), 1/8*(2*(b*cos(f*x + e)^2 - b)*sqrt(b
)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2
+ a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 - 3*(a*b^2 + b^3)*cos(f*x + e)^3 + (a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)
)) + 4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x + e) + ((a + 3*b)*cos(f*x + e)^2 - a - 3*b)*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*cos(f*x + e)^
2 + 1)))/(f*cos(f*x + e)^2 - f), 1/4*(((a + 3*b)*cos(f*x + e)^2 - a - 3*b)*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))) +
(b*cos(f*x + e)^2 - b)*sqrt(b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b
+ b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 - 3*(a*b^2 + b^3)*cos(f*x + e)^3 + (a^2*b
+ 2*a*b^2 + b^3)*cos(f*x + e))) + 2*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x + e))/(f*cos(f*x + e)^2 - f)]
________________________________________________________________________________________
giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/
2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_n
ostep/2)Error: Bad Argument Type
________________________________________________________________________________________
maple [B] time = 2.36, size = 287, normalized size = 2.24 \[ \frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (2 b^{\frac {3}{2}} \arctan \left (\frac {2 b \left (\sin ^{2}\left (f x +e \right )\right )+a -b}{2 \sqrt {b}\, \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}}\right ) \left (\sin ^{2}\left (f x +e \right )\right )-a^{\frac {3}{2}} \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{2}\left (f x +e \right )\right )-3 \sqrt {a}\, b \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \left (\sin ^{2}\left (f x +e \right )\right )-2 a \sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\right )}{4 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]
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)
[Out]
1/4*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(2*b^(3/2)*arctan(1/2/b^(1/2)*(2*b*sin(f*x+e)^2+a-b)/(cos(f*x+e)^2
*(a+b*sin(f*x+e)^2))^(1/2))*sin(f*x+e)^2-a^(3/2)*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)^2-3*a^(1/2)*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)^2-2*a*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2))/sin(f*x+
e)^2/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{3}\,{d x} \]
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((b*sin(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^3, x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*sin(e + f*x)^2)^(3/2)/sin(e + f*x)^3,x)
[Out]
int((a + b*sin(e + f*x)^2)^(3/2)/sin(e + f*x)^3, x)
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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]
Timed out
________________________________________________________________________________________