3.4.2 \(\int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx\) [302]

Optimal. Leaf size=201 \[ \frac {16 \cot (c+d x) \sqrt {e \csc (c+d x)}}{21 a^2 d}-\frac {2 \cot ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {4 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 a^2 d}-\frac {2 \cot (c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}+\frac {4 \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}+\frac {20 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 a^2 d} \]

[Out]

16/21*cot(d*x+c)*(e*csc(d*x+c))^(1/2)/a^2/d-2/7*cot(d*x+c)^3*(e*csc(d*x+c))^(1/2)/a^2/d-4/3*csc(d*x+c)*(e*csc(
d*x+c))^(1/2)/a^2/d-2/7*cot(d*x+c)*csc(d*x+c)^2*(e*csc(d*x+c))^(1/2)/a^2/d+4/7*csc(d*x+c)^3*(e*csc(d*x+c))^(1/
2)/a^2/d-20/21*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*
x),2^(1/2))*(e*csc(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/a^2/d

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Rubi [A]
time = 0.31, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3963, 3957, 2954, 2952, 2647, 2716, 2720, 2644, 14} \begin {gather*} \frac {4 \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {4 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 a^2 d}-\frac {2 \cot ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {2 \cot (c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}+\frac {16 \cot (c+d x) \sqrt {e \csc (c+d x)}}{21 a^2 d}+\frac {20 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{21 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Csc[c + d*x]]/(a + a*Sec[c + d*x])^2,x]

[Out]

(16*Cot[c + d*x]*Sqrt[e*Csc[c + d*x]])/(21*a^2*d) - (2*Cot[c + d*x]^3*Sqrt[e*Csc[c + d*x]])/(7*a^2*d) - (4*Csc
[c + d*x]*Sqrt[e*Csc[c + d*x]])/(3*a^2*d) - (2*Cot[c + d*x]*Csc[c + d*x]^2*Sqrt[e*Csc[c + d*x]])/(7*a^2*d) + (
4*Csc[c + d*x]^3*Sqrt[e*Csc[c + d*x]])/(7*a^2*d) + (20*Sqrt[e*Csc[c + d*x]]*EllipticF[(c - Pi/2 + d*x)/2, 2]*S
qrt[Sin[c + d*x]])/(21*a^2*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2647

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a*Cos[e +
f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Dist[a^2*((m - 1)/(b^2*(n + 1))), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2954

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e +
f*x])^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 3963

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_), x_Symbol] :> Dist[g^Int
Part[p]*(g*Sec[e + f*x])^FracPart[p]*Cos[e + f*x]^FracPart[p], Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x],
x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sqrt {e \csc (c+d x)}}{(a+a \sec (c+d x))^2} \, dx &=\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {\sin (c+d x)}} \, dx\\ &=\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^2(c+d x)}{(-a-a \cos (c+d x))^2 \sqrt {\sin (c+d x)}} \, dx\\ &=\frac {\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{\sin ^{\frac {9}{2}}(c+d x)} \, dx}{a^4}\\ &=\frac {\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \left (\frac {a^2 \cos ^2(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}-\frac {2 a^2 \cos ^3(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}+\frac {a^2 \cos ^4(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}\right ) \, dx}{a^4}\\ &=\frac {\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^2(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)} \, dx}{a^2}+\frac {\left (\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^4(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)} \, dx}{a^2}-\frac {\left (2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^3(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac {2 \cot ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {2 \cot (c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {\left (2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sin ^{\frac {5}{2}}(c+d x)} \, dx}{7 a^2}-\frac {\left (6 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {\cos ^2(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)} \, dx}{7 a^2}-\frac {\left (2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1-x^2}{x^{9/2}} \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=\frac {16 \cot (c+d x) \sqrt {e \csc (c+d x)}}{21 a^2 d}-\frac {2 \cot ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {2 \cot (c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {\left (2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{21 a^2}+\frac {\left (4 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{7 a^2}-\frac {\left (2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {1}{x^{9/2}}-\frac {1}{x^{5/2}}\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=\frac {16 \cot (c+d x) \sqrt {e \csc (c+d x)}}{21 a^2 d}-\frac {2 \cot ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {4 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 a^2 d}-\frac {2 \cot (c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}+\frac {4 \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}+\frac {20 \sqrt {e \csc (c+d x)} F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.45, size = 82, normalized size = 0.41 \begin {gather*} -\frac {4 \csc ^3(c+d x) \sqrt {e \csc (c+d x)} \left (2 (8+11 \cos (c+d x)) \sin ^4\left (\frac {1}{2} (c+d x)\right )+5 F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {7}{2}}(c+d x)\right )}{21 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Csc[c + d*x]]/(a + a*Sec[c + d*x])^2,x]

[Out]

(-4*Csc[c + d*x]^3*Sqrt[e*Csc[c + d*x]]*(2*(8 + 11*Cos[c + d*x])*Sin[(c + d*x)/2]^4 + 5*EllipticF[(-2*c + Pi -
2*d*x)/4, 2]*Sin[c + d*x]^(7/2)))/(21*a^2*d)

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Maple [C] Result contains complex when optimal does not.
time = 0.20, size = 474, normalized size = 2.36

method	result	size
default	\(-\frac {\sqrt {\frac {e}{\sin \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{3} \left (10 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+20 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}+10 i \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )+11 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-3 \sqrt {2}\, \cos \left (d x +c \right )-8 \sqrt {2}\right ) \sqrt {2}}{21 a^{2} d \sin \left (d x +c \right )^{7}}\)	\(474\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/21/a^2/d*(e/sin(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^3*(10*I*sin(d*x+c)*cos(d*x+c)^2*(-I*(-1+cos(
d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x
+c))^(1/2)*EllipticF(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))+20*I*cos(d*x+c)*sin(d*x+c)*((
I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-
I)/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))+10*I*EllipticF(((I*
cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-I*(-1
+cos(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*sin(d*x+c)+11*cos(d*x+c)^2*2^(1
/2)-3*2^(1/2)*cos(d*x+c)-8*2^(1/2))/sin(d*x+c)^7*2^(1/2)

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.47, size = 160, normalized size = 0.80 \begin {gather*} -\frac {2 \, {\left (5 \, \sqrt {2 i} {\left (i \, \cos \left (d x + c\right )^{2} e^{\frac {1}{2}} + 2 i \, \cos \left (d x + c\right ) e^{\frac {1}{2}} + i \, e^{\frac {1}{2}}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {-2 i} {\left (-i \, \cos \left (d x + c\right )^{2} e^{\frac {1}{2}} - 2 i \, \cos \left (d x + c\right ) e^{\frac {1}{2}} - i \, e^{\frac {1}{2}}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (11 \, \cos \left (d x + c\right ) e^{\frac {1}{2}} + 8 \, e^{\frac {1}{2}}\right )} \sqrt {\sin \left (d x + c\right )}\right )}}{21 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/21*(5*sqrt(2*I)*(I*cos(d*x + c)^2*e^(1/2) + 2*I*cos(d*x + c)*e^(1/2) + I*e^(1/2))*weierstrassPInverse(4, 0,
cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(-2*I)*(-I*cos(d*x + c)^2*e^(1/2) - 2*I*cos(d*x + c)*e^(1/2) - I*e^(1/
2))*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + (11*cos(d*x + c)*e^(1/2) + 8*e^(1/2))*sqrt(sin(
d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sqrt {e \csc {\left (c + d x \right )}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*csc(d*x+c))**(1/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(sqrt(e*csc(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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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((e*csc(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(sqrt(e*csc(d*x + c))/(a*sec(d*x + c) + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/sin(c + d*x))^(1/2)/(a + a/cos(c + d*x))^2,x)

[Out]

int((cos(c + d*x)^2*(e/sin(c + d*x))^(1/2))/(a^2*(cos(c + d*x) + 1)^2), x)

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