3.2.31 \(\int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx\) [131]

Optimal. Leaf size=190 \[ \frac {4 e^3}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {2 e^3 \cos (c+d x)}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {2 e^3 \cos ^3(c+d x)}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 e}{3 a^2 d (e \sin (c+d x))^{3/2}}+\frac {16 e \cos (c+d x)}{21 a^2 d (e \sin (c+d x))^{3/2}}+\frac {20 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 a^2 d \sqrt {e \sin (c+d x)}} \]

[Out]

4/7*e^3/a^2/d/(e*sin(d*x+c))^(7/2)-2/7*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(7/2)-2/7*e^3*cos(d*x+c)^3/a^2/d/(e
*sin(d*x+c))^(7/2)-4/3*e/a^2/d/(e*sin(d*x+c))^(3/2)+16/21*e*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(3/2)-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))*sin(d*x+c
)^(1/2)/a^2/d/(e*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.43, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3957, 2954, 2952, 2647, 2716, 2721, 2720, 2644, 14} \begin {gather*} \frac {4 e^3}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {2 e^3 \cos ^3(c+d x)}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {2 e^3 \cos (c+d x)}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 e}{3 a^2 d (e \sin (c+d x))^{3/2}}+\frac {16 e \cos (c+d x)}{21 a^2 d (e \sin (c+d x))^{3/2}}+\frac {20 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{21 a^2 d \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e^3)/(7*a^2*d*(e*Sin[c + d*x])^(7/2)) - (2*e^3*Cos[c + d*x])/(7*a^2*d*(e*Sin[c + d*x])^(7/2)) - (2*e^3*Cos[
c + d*x]^3)/(7*a^2*d*(e*Sin[c + d*x])^(7/2)) - (4*e)/(3*a^2*d*(e*Sin[c + d*x])^(3/2)) + (16*e*Cos[c + d*x])/(2
1*a^2*d*(e*Sin[c + d*x])^(3/2)) + (20*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*a^2*d*Sqrt[e*Si
n[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 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 2721

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

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]

Rubi steps

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

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Mathematica [A]
time = 0.94, size = 82, normalized size = 0.43 \begin {gather*} -\frac {\csc ^3(c+d x) \left (16 (8+11 \cos (c+d x)) \sin ^4\left (\frac {1}{2} (c+d x)\right )+40 F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {7}{2}}(c+d x)\right )}{42 a^2 d \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/42*(Csc[c + d*x]^3*(16*(8 + 11*Cos[c + d*x])*Sin[(c + d*x)/2]^4 + 40*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Si
n[c + d*x]^(7/2)))/(a^2*d*Sqrt[e*Sin[c + d*x]])

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Maple [A]
time = 0.22, size = 148, normalized size = 0.78

method	result	size
default	\(\frac {\frac {4 e^{3} \left (7 \left (\cos ^{2}\left (d x +c \right )\right )-4\right )}{21 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}-\frac {2 \left (5 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {9}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+11 \left (\sin ^{5}\left (d x +c \right )\right )-17 \left (\sin ^{3}\left (d x +c \right )\right )+6 \sin \left (d x +c \right )\right )}{21 a^{2} \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4/21/a^2*e^3/(e*sin(d*x+c))^(7/2)*(7*cos(d*x+c)^2-4)-2/21*(5*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin
(d*x+c)^(9/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+11*sin(d*x+c)^5-17*sin(d*x+c)^3+6*sin(d*x+c))/a^2/s
in(d*x+c)^4/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

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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(1/(a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(1/2),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 = 0.97, size = 162, normalized size = 0.85 \begin {gather*} \frac {2 \, {\left (5 \, \sqrt {-i} {\left (\sqrt {2} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {i} {\left (\sqrt {2} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \cos \left (d x + c\right ) + \sqrt {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 ) + 8\right )} \sqrt {\sin \left (d x + c\right )}\right )}}{21 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} e^{\frac {1}{2}} + 2 \, a^{2} d \cos \left (d x + c\right ) e^{\frac {1}{2}} + a^{2} d e^{\frac {1}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(1/((a*sec(d*x + c) + a)^2*sqrt(e*sin(d*x + c))), 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\,\sqrt {e\,\sin \left (c+d\,x\right )}\,{\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/((e*sin(c + d*x))^(1/2)*(a + a/cos(c + d*x))^2),x)

[Out]

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

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