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

Optimal. Leaf size=122 \[ -\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{f}-\frac {b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 f}-\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3186, 416, 523, 217, 203, 377, 206} \[ -\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{f}-\frac {b \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 f}-\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(2*f) - (a^(3/2)*ArcTanh[(S
qrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/f - (b*Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/(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 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[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 (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}}{1-x^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {b \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{2 f}+\frac {\operatorname {Subst}\left (\int \frac {-(a+b) (2 a+b)+b (3 a+b) x^2}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{2 f}\\ &=-\frac {b \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{2 f}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{f}-\frac {(b (3 a+b)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{2 f}\\ &=-\frac {b \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{2 f}-\frac {a^2 \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 )}{f}-\frac {(b (3 a+b)) \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 )}{2 f}\\ &=-\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 f}-\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{f}-\frac {b \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{2 f}\\ \end {align*}

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Mathematica [A]  time = 0.92, size = 141, normalized size = 1.16 \[ -\frac {4 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a-b \cos (2 (e+f x))+b}}\right )+\sqrt {2} b \cos (e+f x) \sqrt {2 a-b \cos (2 (e+f x))+b}-2 \sqrt {-b} (3 a+b) \log \left (\sqrt {2 a-b \cos (2 (e+f x))+b}+\sqrt {2} \sqrt {-b} \cos (e+f x)\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(4*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] + Sqrt[2]*b*Cos[e +
f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]] - 2*Sqrt[-b]*(3*a + b)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a +
b - b*Cos[2*(e + f*x)]]])/f

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fricas [B]  time = 0.88, size = 1282, normalized size = 10.51 \[ \text {result too large to display} \]

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]

[-1/16*(8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e) - (3*a + 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)) - 4*a^(3/2)*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, 1/16*(8*sqrt(-a)*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))) - 8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e) + (3*a + 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, 1/8*((3*a + 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)*b
*cos(f*x + e) + 2*a^(3/2)*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, 1/8*(4*sqrt(-a)*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))) + (3*a + 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)*b*cos(f*x + e))/f]

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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)*(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)Error: Bad Argument Type

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maple [B]  time = 2.08, size = 255, normalized size = 2.09 \[ \frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (b^{\frac {3}{2}} \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )-2 a^{\frac {3}{2}} \ln \left (\frac {-\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}-a -b}{-1+\cos ^{2}\left (f x +e \right )}\right )+3 \sqrt {b}\, a \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )-2 b \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\right )}{4 \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)*(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)*(b^(3/2)*arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)
^4+(a+b)*cos(f*x+e)^2)^(1/2))-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)/(-1+cos(f*x+e)^2))+3*b^(1/2)*a*arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*c
os(f*x+e)^2)^(1/2))-2*b*(-b*cos(f*x+e)^4+(a+b)*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 [A]  time = 0.48, size = 179, normalized size = 1.47 \[ -\frac {3 \, a \sqrt {b} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right ) + b^{\frac {3}{2}} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right ) + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right ) + a^{\frac {3}{2}} \log \left (b - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) - 1} - \frac {a}{\cos \left (f x + e\right ) - 1}\right ) - a^{\frac {3}{2}} \log \left (-b + \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) + 1} + \frac {a}{\cos \left (f x + e\right ) + 1}\right )}{2 \, f} \]

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

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

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),x)

[Out]

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

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

[Out]

Timed out

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