3.2.30 \(\int \frac {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx\) [130]

Optimal. Leaf size=188 \[ \frac {4 e^3}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos (c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos ^3(c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {4 e}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {16 e \cos (c+d x)}{5 a^2 d \sqrt {e \sin (c+d x)}}+\frac {28 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}} \]

[Out]

4/5*e^3/a^2/d/(e*sin(d*x+c))^(5/2)-2/5*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(5/2)-2/5*e^3*cos(d*x+c)^3/a^2/d/(e
*sin(d*x+c))^(5/2)-4*e/a^2/d/(e*sin(d*x+c))^(1/2)+16/5*e*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(1/2)-28/5*(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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c)
)^(1/2)/a^2/d/sin(d*x+c)^(1/2)

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Rubi [A]
time = 0.40, antiderivative size = 188, 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, 2719, 2644, 14} \begin {gather*} \frac {4 e^3}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos ^3(c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos (c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {4 e}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {16 e \cos (c+d x)}{5 a^2 d \sqrt {e \sin (c+d x)}}+\frac {28 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e^3)/(5*a^2*d*(e*Sin[c + d*x])^(5/2)) - (2*e^3*Cos[c + d*x])/(5*a^2*d*(e*Sin[c + d*x])^(5/2)) - (2*e^3*Cos[
c + d*x]^3)/(5*a^2*d*(e*Sin[c + d*x])^(5/2)) - (4*e)/(a^2*d*Sqrt[e*Sin[c + d*x]]) + (16*e*Cos[c + d*x])/(5*a^2
*d*Sqrt[e*Sin[c + d*x]]) + (28*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a^2*d*Sqrt[Sin[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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 {\sqrt {e \sin (c+d x)}}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sqrt {e \sin (c+d x)}}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx}{a^4}\\ &=\frac {e^4 \int \left (\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{7/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{7/2}}+\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{7/2}}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{7/2}} \, dx}{a^2}+\frac {e^4 \int \frac {\cos ^4(c+d x)}{(e \sin (c+d x))^{7/2}} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \frac {\cos ^3(c+d x)}{(e \sin (c+d x))^{7/2}} \, dx}{a^2}\\ &=-\frac {2 e^3 \cos (c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos ^3(c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {\left (2 e^2\right ) \int \frac {1}{(e \sin (c+d x))^{3/2}} \, dx}{5 a^2}-\frac {\left (6 e^2\right ) \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx}{5 a^2}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {1-\frac {x^2}{e^2}}{x^{7/2}} \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=-\frac {2 e^3 \cos (c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos ^3(c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac {16 e \cos (c+d x)}{5 a^2 d \sqrt {e \sin (c+d x)}}+\frac {2 \int \sqrt {e \sin (c+d x)} \, dx}{5 a^2}+\frac {12 \int \sqrt {e \sin (c+d x)} \, dx}{5 a^2}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \left (\frac {1}{x^{7/2}}-\frac {1}{e^2 x^{3/2}}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=\frac {4 e^3}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos (c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos ^3(c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {4 e}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {16 e \cos (c+d x)}{5 a^2 d \sqrt {e \sin (c+d x)}}+\frac {\left (2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 a^2 \sqrt {\sin (c+d x)}}+\frac {\left (12 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 a^2 \sqrt {\sin (c+d x)}}\\ &=\frac {4 e^3}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos (c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {2 e^3 \cos ^3(c+d x)}{5 a^2 d (e \sin (c+d x))^{5/2}}-\frac {4 e}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {16 e \cos (c+d x)}{5 a^2 d \sqrt {e \sin (c+d x)}}+\frac {28 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.90, size = 222, normalized size = 1.18 \begin {gather*} \frac {4 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {e \sin (c+d x)} \left (\frac {56 i e^{2 i c} \left (3 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};e^{2 i (c+d x)}\right )+e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{2 i (c+d x)}\right )\right )}{\left (1+e^{2 i c}\right ) \sqrt {1-e^{2 i (c+d x)}}}+\frac {3}{4} \sec (c) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (49 \sin \left (\frac {1}{2} (c-d x)\right )+35 \sin \left (\frac {1}{2} (3 c+d x)\right )-23 \sin \left (\frac {1}{2} (c+3 d x)\right )+5 \sin \left (\frac {1}{2} (5 c+3 d x)\right )\right )\right )}{15 a^2 d (1+\sec (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*Sqrt[e*Sin[c + d*x]]*(((56*I)*E^((2*I)*c)*(3*Hypergeometric2F1[-1/4, 1/2,
3/4, E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))]))/((1 + E^((2
*I)*c))*Sqrt[1 - E^((2*I)*(c + d*x))]) + (3*Sec[c]*Sec[(c + d*x)/2]^3*(49*Sin[(c - d*x)/2] + 35*Sin[(3*c + d*x
)/2] - 23*Sin[(c + 3*d*x)/2] + 5*Sin[(5*c + 3*d*x)/2]))/4))/(15*a^2*d*(1 + Sec[c + d*x])^2)

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Maple [A]
time = 0.24, size = 205, normalized size = 1.09

method	result	size
default	\(\frac {-\frac {2 e \left (-\frac {2 e^{2}}{5 \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {2}{\sqrt {e \sin \left (d x +c \right )}}\right )}{a^{2}}-\frac {2 e \left (14 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-7 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+9 \left (\sin ^{5}\left (d x +c \right )\right )-11 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{5 a^{2} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(205\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*e/a^2*(-2/5*e^2/(e*sin(d*x+c))^(5/2)+2/(e*sin(d*x+c))^(1/2))-2/5*e*(14*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)
+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-7*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+
2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+9*sin(d*x+c)^5-11*sin(d*x+c)^3+2*sin(d*
x+c))/a^2/sin(d*x+c)^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.72, size = 184, normalized size = 0.98 \begin {gather*} -\frac {2 \, {\left ({\left (9 \, \cos \left (d x + c\right ) e^{\frac {1}{2}} + 8 \, e^{\frac {1}{2}}\right )} \sin \left (d x + c\right )^{\frac {3}{2}} + 7 \, \sqrt {-i} {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} e^{\frac {1}{2}} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {1}{2}} - i \, \sqrt {2} e^{\frac {1}{2}}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 7 \, \sqrt {i} {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} e^{\frac {1}{2}} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {1}{2}} + i \, \sqrt {2} e^{\frac {1}{2}}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )\right )}}{5 \, {\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*sin(d*x+c))^(1/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/5*((9*cos(d*x + c)*e^(1/2) + 8*e^(1/2))*sin(d*x + c)^(3/2) + 7*sqrt(-I)*(-I*sqrt(2)*cos(d*x + c)^2*e^(1/2)
- 2*I*sqrt(2)*cos(d*x + c)*e^(1/2) - I*sqrt(2)*e^(1/2))*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*
x + c) + I*sin(d*x + c))) + 7*sqrt(I)*(I*sqrt(2)*cos(d*x + c)^2*e^(1/2) + 2*I*sqrt(2)*cos(d*x + c)*e^(1/2) + I
*sqrt(2)*e^(1/2))*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*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 \sin {\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*sin(d*x+c))**(1/2)/(a+a*sec(d*x+c))**2,x)

[Out]

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

[Out]

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

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