3.3.56 \(\int \frac {\csc ^2(e+f x)}{(a+b \sec (e+f x))^{3/2}} \, dx\) [256]

Optimal. Leaf size=318 \[ \frac {4 a \cot (e+f x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{(a-b) (a+b)^{3/2} f}-\frac {(3 a-b) \cot (e+f x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{(a-b) (a+b)^{3/2} f}-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {4 a b^2 \tan (e+f x)}{\left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}} \]

[Out]

-cot(f*x+e)/f/(a+b*sec(f*x+e))^(3/2)+4*a*cot(f*x+e)*EllipticE((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))
^(1/2))*(b*(1-sec(f*x+e))/(a+b))^(1/2)*(-b*(1+sec(f*x+e))/(a-b))^(1/2)/(a-b)/(a+b)^(3/2)/f-(3*a-b)*cot(f*x+e)*
EllipticF((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(f*x+e))/(a+b))^(1/2)*(-b*(1+sec(f*
x+e))/(a-b))^(1/2)/(a-b)/(a+b)^(3/2)/f+b^2*tan(f*x+e)/(a^2-b^2)/f/(a+b*sec(f*x+e))^(3/2)+4*a*b^2*tan(f*x+e)/(a
^2-b^2)^2/f/(a+b*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.36, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3960, 3918, 4088, 4090, 3917, 4089} \begin {gather*} \frac {4 a b^2 \tan (e+f x)}{f \left (a^2-b^2\right )^2 \sqrt {a+b \sec (e+f x)}}+\frac {b^2 \tan (e+f x)}{f \left (a^2-b^2\right ) (a+b \sec (e+f x))^{3/2}}-\frac {(3 a-b) \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{f (a-b) (a+b)^{3/2}}+\frac {4 a \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{f (a-b) (a+b)^{3/2}}-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*Cot[e + f*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e
+ f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/((a - b)*(a + b)^(3/2)*f) - ((3*a - b)*Cot[e + f*x]
*EllipticF[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]
*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/((a - b)*(a + b)^(3/2)*f) - Cot[e + f*x]/(f*(a + b*Sec[e + f*x])^(3/
2)) + (b^2*Tan[e + f*x])/((a^2 - b^2)*f*(a + b*Sec[e + f*x])^(3/2)) + (4*a*b^2*Tan[e + f*x])/((a^2 - b^2)^2*f*
Sqrt[a + b*Sec[e + f*x]])

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3918

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-b)*Cot[e + f*x]*(
(a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 -
b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 3960

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)/cos[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[Tan[e + f*x]*((a
+ b*Csc[e + f*x])^m/f), x] + Dist[b*m, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e
, f, m}, x]

Rule 4088

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a
*B)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1]

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[Csc[e + f*x]*((1 +
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rubi steps

\begin {align*} \int \frac {\csc ^2(e+f x)}{(a+b \sec (e+f x))^{3/2}} \, dx &=-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}-\frac {1}{2} (3 b) \int \frac {\sec (e+f x)}{(a+b \sec (e+f x))^{5/2}} \, dx\\ &=-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {b \int \frac {\sec (e+f x) \left (-\frac {3 a}{2}+\frac {1}{2} b \sec (e+f x)\right )}{(a+b \sec (e+f x))^{3/2}} \, dx}{a^2-b^2}\\ &=-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {4 a b^2 \tan (e+f x)}{\left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}}-\frac {(2 b) \int \frac {\sec (e+f x) \left (\frac {1}{4} \left (3 a^2+b^2\right )+a b \sec (e+f x)\right )}{\sqrt {a+b \sec (e+f x)}} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {4 a b^2 \tan (e+f x)}{\left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}}-\frac {((3 a-b) b) \int \frac {\sec (e+f x)}{\sqrt {a+b \sec (e+f x)}} \, dx}{2 (a-b) (a+b)^2}-\frac {\left (2 a b^2\right ) \int \frac {\sec (e+f x) (1+\sec (e+f x))}{\sqrt {a+b \sec (e+f x)}} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {4 a \cot (e+f x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{(a-b) (a+b)^{3/2} f}-\frac {(3 a-b) \cot (e+f x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{(a-b) (a+b)^{3/2} f}-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {4 a b^2 \tan (e+f x)}{\left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 5.07, size = 259, normalized size = 0.81 \begin {gather*} \frac {-((a-b) ((3 a-b) b+a (a-3 b) \cos (e+f x)) \csc (e+f x))+8 a b (a+b) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {b+a \cos (e+f x)}{(a+b) (1+\cos (e+f x))}} E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right ) \sec (e+f x) \sqrt {\frac {1}{1+\sec (e+f x)}}-2 b \left (3 a^2+4 a b+b^2\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {b+a \cos (e+f x)}{(a+b) (1+\cos (e+f x))}} F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right ) \sec (e+f x) \sqrt {\frac {1}{1+\sec (e+f x)}}}{\left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a - b)*((3*a - b)*b + a*(a - 3*b)*Cos[e + f*x])*Csc[e + f*x]) + 8*a*b*(a + b)*Cos[(e + f*x)/2]^2*Sqrt[(b +
a*Cos[e + f*x])/((a + b)*(1 + Cos[e + f*x]))]*EllipticE[ArcSin[Tan[(e + f*x)/2]], (a - b)/(a + b)]*Sec[e + f*
x]*Sqrt[(1 + Sec[e + f*x])^(-1)] - 2*b*(3*a^2 + 4*a*b + b^2)*Cos[(e + f*x)/2]^2*Sqrt[(b + a*Cos[e + f*x])/((a
+ b)*(1 + Cos[e + f*x]))]*EllipticF[ArcSin[Tan[(e + f*x)/2]], (a - b)/(a + b)]*Sec[e + f*x]*Sqrt[(1 + Sec[e +
f*x])^(-1)])/((a^2 - b^2)^2*f*Sqrt[a + b*Sec[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1064\) vs. \(2(294)=588\).
time = 0.20, size = 1065, normalized size = 3.35

method	result	size
default	\(\text {Expression too large to display}\)	\(1065\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(-3*cos(f*x+e)*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/
2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*b-4*cos(f*x+e)*sin(f*x+e)*(cos(f*x+e)/(cos(f*
x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b)
)^(1/2))*a*b^2-cos(f*x+e)*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))
^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^3+4*cos(f*x+e)*sin(f*x+e)*(cos(f*x+e)/(cos(
f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+
b))^(1/2))*a^2*b+4*cos(f*x+e)*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a
+b))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b^2-3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e
)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(
1/2))*a^2*b-4*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*Ellip
ticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b^2-sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*co
s(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^3+4*sin(f*
x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))
/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*b+4*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f
*x+e)+1)/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b^2+cos(f*x+e)^2*a^3-4*cos(f
*x+e)^2*a^2*b+3*cos(f*x+e)^2*a*b^2+3*cos(f*x+e)*a^2*b-4*cos(f*x+e)*a*b^2+cos(f*x+e)*b^3)*((a*cos(f*x+e)+b)/cos
(f*x+e))^(1/2)*4^(1/2)/(a*cos(f*x+e)+b)/sin(f*x+e)/(a-b)^2/(a+b)^2

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(f*x + e) + a)*csc(f*x + e)^2/(b^2*sec(f*x + e)^2 + 2*a*b*sec(f*x + e) + a^2), x)

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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