3.4.6 \(\int \frac {1}{(e \csc (c+d x))^{7/2} (a+a \sec (c+d x))^2} \, dx\) [306]

Optimal. Leaf size=172 \[ -\frac {4}{a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {26 \cos (c+d x)}{21 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {2 \cos ^3(c+d x)}{7 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {52 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{21 a^2 d e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {4 \sin ^2(c+d x)}{5 a^2 d e^3 \sqrt {e \csc (c+d x)}} \]

[Out]

-4/a^2/d/e^3/(e*csc(d*x+c))^(1/2)+26/21*cos(d*x+c)/a^2/d/e^3/(e*csc(d*x+c))^(1/2)+2/7*cos(d*x+c)^3/a^2/d/e^3/(
e*csc(d*x+c))^(1/2)+4/5*sin(d*x+c)^2/a^2/d/e^3/(e*csc(d*x+c))^(1/2)-52/21*(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))/a^2/d/e^3/(e*csc(d*x+c))^(1/2)/sin(d*x+
c)^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3963, 3957, 2954, 2952, 2649, 2720, 2644, 14} \begin {gather*} -\frac {4}{a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {2 \cos ^3(c+d x)}{7 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {26 \cos (c+d x)}{21 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {4 \sin ^2(c+d x)}{5 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {52 F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{21 a^2 d e^3 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Csc[c + d*x])^(7/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

-4/(a^2*d*e^3*Sqrt[e*Csc[c + d*x]]) + (26*Cos[c + d*x])/(21*a^2*d*e^3*Sqrt[e*Csc[c + d*x]]) + (2*Cos[c + d*x]^
3)/(7*a^2*d*e^3*Sqrt[e*Csc[c + d*x]]) + (52*EllipticF[(c - Pi/2 + d*x)/2, 2])/(21*a^2*d*e^3*Sqrt[e*Csc[c + d*x
]]*Sqrt[Sin[c + d*x]]) + (4*Sin[c + d*x]^2)/(5*a^2*d*e^3*Sqrt[e*Csc[c + d*x]])

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 2649

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(b*Sin[e +
f*x])^(n + 1)*((a*Cos[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Sin[e + f*x])^
n*(a*Cos[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m
, 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 {1}{(e \csc (c+d x))^{7/2} (a+a \sec (c+d x))^2} \, dx &=\frac {\int \frac {\sin ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx}{e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {\int \frac {\cos ^2(c+d x) \sin ^{\frac {7}{2}}(c+d x)}{(-a-a \cos (c+d x))^2} \, dx}{e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {\int \frac {\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{\sqrt {\sin (c+d x)}} \, dx}{a^4 e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {\int \left (\frac {a^2 \cos ^2(c+d x)}{\sqrt {\sin (c+d x)}}-\frac {2 a^2 \cos ^3(c+d x)}{\sqrt {\sin (c+d x)}}+\frac {a^2 \cos ^4(c+d x)}{\sqrt {\sin (c+d x)}}\right ) \, dx}{a^4 e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {\int \frac {\cos ^2(c+d x)}{\sqrt {\sin (c+d x)}} \, dx}{a^2 e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {\int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x)}} \, dx}{a^2 e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {2 \int \frac {\cos ^3(c+d x)}{\sqrt {\sin (c+d x)}} \, dx}{a^2 e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {2 \cos (c+d x)}{3 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {2 \cos ^3(c+d x)}{7 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {2 \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 a^2 e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {6 \int \frac {\cos ^2(c+d x)}{\sqrt {\sin (c+d x)}} \, dx}{7 a^2 e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {2 \text {Subst}\left (\int \frac {1-x^2}{\sqrt {x}} \, dx,x,\sin (c+d x)\right )}{a^2 d e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {26 \cos (c+d x)}{21 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {2 \cos ^3(c+d x)}{7 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{3 a^2 d e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {4 \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{7 a^2 e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {2 \text {Subst}\left (\int \left (\frac {1}{\sqrt {x}}-x^{3/2}\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=-\frac {4}{a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {26 \cos (c+d x)}{21 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {2 \cos ^3(c+d x)}{7 a^2 d e^3 \sqrt {e \csc (c+d x)}}+\frac {52 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{21 a^2 d e^3 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {4 \sin ^2(c+d x)}{5 a^2 d e^3 \sqrt {e \csc (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.43, size = 94, normalized size = 0.55 \begin {gather*} \frac {\sqrt {e \csc (c+d x)} \left (-520 F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+(-756+305 \cos (c+d x)-84 \cos (2 (c+d x))+15 \cos (3 (c+d x))) \sqrt {\sin (c+d x)}\right ) \sqrt {\sin (c+d x)}}{210 a^2 d e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Csc[c + d*x])^(7/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(Sqrt[e*Csc[c + d*x]]*(-520*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + (-756 + 305*Cos[c + d*x] - 84*Cos[2*(c + d*x
)] + 15*Cos[3*(c + d*x)])*Sqrt[Sin[c + d*x]])*Sqrt[Sin[c + d*x]])/(210*a^2*d*e^4)

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Maple [C] Result contains complex when optimal does not.
time = 0.17, size = 221, normalized size = 1.28

method	result	size
default	\(-\frac {\left (130 i \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right ) \sqrt {\frac {-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i}{\sin \left (d x +c \right )}}\, \sqrt {\frac {i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}-15 \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}+57 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}-107 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+233 \sqrt {2}\, \cos \left (d x +c \right )-168 \sqrt {2}\right ) \sqrt {2}}{105 a^{2} d \left (-1+\cos \left (d x +c \right )\right ) \left (\frac {e}{\sin \left (d x +c \right )}\right )^{\frac {7}{2}} \sin \left (d x +c \right )^{3}}\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*csc(d*x+c))^(7/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/105/a^2/d*(130*I*sin(d*x+c)*((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)/si
n(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))*(-I*(-1+cos(d*x+c))/sin(
d*x+c))^(1/2)-15*cos(d*x+c)^4*2^(1/2)+57*cos(d*x+c)^3*2^(1/2)-107*cos(d*x+c)^2*2^(1/2)+233*2^(1/2)*cos(d*x+c)-
168*2^(1/2))/(-1+cos(d*x+c))/(e/sin(d*x+c))^(7/2)/sin(d*x+c)^3*2^(1/2)

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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(1/(e*csc(d*x+c))^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

e^(-7/2)*integrate(1/((a*sec(d*x + c) + a)^2*csc(d*x + c)^(7/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.73, size = 96, normalized size = 0.56 \begin {gather*} \frac {2 \, {\left ({\left (15 \, \cos \left (d x + c\right )^{3} - 42 \, \cos \left (d x + c\right )^{2} + 65 \, \cos \left (d x + c\right ) - 168\right )} \sqrt {\sin \left (d x + c\right )} - 65 i \, \sqrt {2 i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 65 i \, \sqrt {-2 i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )} e^{\left (-\frac {7}{2}\right )}}{105 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*csc(d*x+c))^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

2/105*((15*cos(d*x + c)^3 - 42*cos(d*x + c)^2 + 65*cos(d*x + c) - 168)*sqrt(sin(d*x + c)) - 65*I*sqrt(2*I)*wei
erstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 65*I*sqrt(-2*I)*weierstrassPInverse(4, 0, cos(d*x + c)
- I*sin(d*x + c)))*e^(-7/2)/(a^2*d)

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Sympy [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(1/(e*csc(d*x+c))**(7/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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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(1/(e*csc(d*x+c))^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*csc(d*x + c))^(7/2)*(a*sec(d*x + c) + a)^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{7/2}\,{\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(1/((a + a/cos(c + d*x))^2*(e/sin(c + d*x))^(7/2)),x)

[Out]

int(cos(c + d*x)^2/(a^2*(e/sin(c + d*x))^(7/2)*(cos(c + d*x) + 1)^2), x)

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