3.2.27 \(\int \frac {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx\) [127]

Optimal. Leaf size=162 \[ \frac {52 e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 a^2 d \sqrt {e \sin (c+d x)}}-\frac {4 e^3 \sqrt {e \sin (c+d x)}}{a^2 d}+\frac {26 e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 a^2 d}+\frac {2 e^3 \cos ^3(c+d x) \sqrt {e \sin (c+d x)}}{7 a^2 d}+\frac {4 e (e \sin (c+d x))^{5/2}}{5 a^2 d} \]

[Out]

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

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Rubi [A]
time = 0.38, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3957, 2954, 2952, 2649, 2721, 2720, 2644, 14} \begin {gather*} \frac {52 e^4 \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)}}-\frac {4 e^3 \sqrt {e \sin (c+d x)}}{a^2 d}+\frac {2 e^3 \cos ^3(c+d x) \sqrt {e \sin (c+d x)}}{7 a^2 d}+\frac {26 e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 a^2 d}+\frac {4 e (e \sin (c+d x))^{5/2}}{5 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(52*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*a^2*d*Sqrt[e*Sin[c + d*x]]) - (4*e^3*Sqrt[e*S
in[c + d*x]])/(a^2*d) + (26*e^3*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(21*a^2*d) + (2*e^3*Cos[c + d*x]^3*Sqrt[e*S
in[c + d*x]])/(7*a^2*d) + (4*e*(e*Sin[c + d*x])^(5/2))/(5*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 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 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 {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) (e \sin (c+d x))^{7/2}}{(-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}{\sqrt {e \sin (c+d x)}} \, dx}{a^4}\\ &=\frac {e^4 \int \left (\frac {a^2 \cos ^2(c+d x)}{\sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos ^3(c+d x)}{\sqrt {e \sin (c+d x)}}+\frac {a^2 \cos ^4(c+d x)}{\sqrt {e \sin (c+d x)}}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x)}{\sqrt {e \sin (c+d x)}} \, dx}{a^2}+\frac {e^4 \int \frac {\cos ^4(c+d x)}{\sqrt {e \sin (c+d x)}} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \frac {\cos ^3(c+d x)}{\sqrt {e \sin (c+d x)}} \, dx}{a^2}\\ &=\frac {2 e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d}+\frac {2 e^3 \cos ^3(c+d x) \sqrt {e \sin (c+d x)}}{7 a^2 d}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {1-\frac {x^2}{e^2}}{\sqrt {x}} \, dx,x,e \sin (c+d x)\right )}{a^2 d}+\frac {\left (2 e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 a^2}+\frac {\left (6 e^4\right ) \int \frac {\cos ^2(c+d x)}{\sqrt {e \sin (c+d x)}} \, dx}{7 a^2}\\ &=\frac {26 e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 a^2 d}+\frac {2 e^3 \cos ^3(c+d x) \sqrt {e \sin (c+d x)}}{7 a^2 d}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt {x}}-\frac {x^{3/2}}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}+\frac {\left (4 e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{7 a^2}+\frac {\left (2 e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 a^2 \sqrt {e \sin (c+d x)}}\\ &=\frac {4 e^4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 a^2 d \sqrt {e \sin (c+d x)}}-\frac {4 e^3 \sqrt {e \sin (c+d x)}}{a^2 d}+\frac {26 e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 a^2 d}+\frac {2 e^3 \cos ^3(c+d x) \sqrt {e \sin (c+d x)}}{7 a^2 d}+\frac {4 e (e \sin (c+d x))^{5/2}}{5 a^2 d}+\frac {\left (4 e^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 {52 e^4 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)}}-\frac {4 e^3 \sqrt {e \sin (c+d x)}}{a^2 d}+\frac {26 e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 a^2 d}+\frac {2 e^3 \cos ^3(c+d x) \sqrt {e \sin (c+d x)}}{7 a^2 d}+\frac {4 e (e \sin (c+d x))^{5/2}}{5 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 1.03, size = 94, normalized size = 0.58 \begin {gather*} -\frac {e^3 \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 {e \sin (c+d x)}}{210 a^2 d \sqrt {\sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/210*(e^3*(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[e*Sin[c + d*x]])/(a^2*d*Sqrt[Sin[c + d*x]])

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Maple [A]
time = 0.23, size = 145, normalized size = 0.90

method	result	size
default	\(-\frac {2 e^{4} \left (-15 \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+65 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+42 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-65 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+168 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{105 a^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/a^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^4*(-15*cos(d*x+c)^4*sin(d*x+c)+65*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*
x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+42*cos(d*x+c)^3*sin(d*x+c)-65*cos(
d*x+c)^2*sin(d*x+c)+168*cos(d*x+c)*sin(d*x+c))/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))^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

e^(7/2)*integrate(sin(d*x + c)^(7/2)/(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 = 0.92, size = 113, normalized size = 0.70 \begin {gather*} \frac {2 \, {\left (65 \, \sqrt {2} \sqrt {-i} e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 65 \, \sqrt {2} \sqrt {i} e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (15 \, \cos \left (d x + c\right )^{3} e^{\frac {7}{2}} - 42 \, \cos \left (d x + c\right )^{2} e^{\frac {7}{2}} + 65 \, \cos \left (d x + c\right ) e^{\frac {7}{2}} - 168 \, e^{\frac {7}{2}}\right )} \sqrt {\sin \left (d x + c\right )}\right )}}{105 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((e*sin(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\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}}{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))^(7/2)/(a + a/cos(c + d*x))^2,x)

[Out]

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

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