3.6.53 \(\int \frac {(a+b \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx\) [553]

Optimal. Leaf size=119 \[ \frac {2 a b \sqrt {e \cos (c+d x)}}{3 d e^3}+\frac {2 \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d e (e \cos (c+d x))^{3/2}} \]

[Out]

2/3*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))/d/e/(e*cos(d*x+c))^(3/2)+2/3*(a^2-2*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co
s(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/e^2/(e*cos(d*x+c))^(1/2)+2/3*a*b*(e*
cos(d*x+c))^(1/2)/d/e^3

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Rubi [A]
time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2770, 2748, 2721, 2720} \begin {gather*} \frac {2 \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 a b \sqrt {e \cos (c+d x)}}{3 d e^3}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))}{3 d e (e \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^2/(e*Cos[c + d*x])^(5/2),x]

[Out]

(2*a*b*Sqrt[e*Cos[c + d*x]])/(3*d*e^3) + (2*(a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*e
^2*Sqrt[e*Cos[c + d*x]]) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x]))/(3*d*e*(e*Cos[c + d*x])^(3/2))

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 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2770

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*C
os[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Dist[1/(g^2*(p +
1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*S
in[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (Integers
Q[2*m, 2*p] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {(a+b \sin (c+d x))^2}{(e \cos (c+d x))^{5/2}} \, dx &=\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d e (e \cos (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {a^2}{2}+b^2+\frac {1}{2} a b \sin (c+d x)}{\sqrt {e \cos (c+d x)}} \, dx}{3 e^2}\\ &=\frac {2 a b \sqrt {e \cos (c+d x)}}{3 d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d e (e \cos (c+d x))^{3/2}}+\frac {\left (a^2-2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{3 e^2}\\ &=\frac {2 a b \sqrt {e \cos (c+d x)}}{3 d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d e (e \cos (c+d x))^{3/2}}+\frac {\left (\left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 e^2 \sqrt {e \cos (c+d x)}}\\ &=\frac {2 a b \sqrt {e \cos (c+d x)}}{3 d e^3}+\frac {2 \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d e (e \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 72, normalized size = 0.61 \begin {gather*} \frac {2 \left (2 a b+\left (a^2-2 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (a^2+b^2\right ) \sin (c+d x)\right )}{3 d e (e \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^2/(e*Cos[c + d*x])^(5/2),x]

[Out]

(2*(2*a*b + (a^2 - 2*b^2)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + (a^2 + b^2)*Sin[c + d*x]))/(3*d*e*(e*
Cos[c + d*x])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(332\) vs. \(2(131)=262\).
time = 7.82, size = 333, normalized size = 2.80

method	result	size
default	\(-\frac {2 \left (2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a^{2} \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, b^{2} \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+2 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\)	\(333\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))^2/(e*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^2*(2*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*sin(1/2*d*x+1/2*c)^2-4
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*sin(1
/2*d*x+1/2*c)^2-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))*a^2+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)*b^2+2*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*a*b*sin(1/
2*d*x+1/2*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((a+b*sin(d*x+c))^2/(e*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate((b*sin(d*x + c) + a)^2/cos(d*x + c)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 127, normalized size = 1.07 \begin {gather*} \frac {{\left (\sqrt {2} {\left (-i \, a^{2} + 2 i \, b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, a^{2} - 2 i \, b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}\right )} e^{\left (-\frac {5}{2}\right )}}{3 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^2/(e*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(-I*a^2 + 2*I*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqr
t(2)*(I*a^2 - 2*I*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(2*a*b + (
a^2 + b^2)*sin(d*x + c))*sqrt(cos(d*x + c)))*e^(-5/2)/(d*cos(d*x + c)^2)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))**2/(e*cos(d*x+c))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3881 deep

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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((a+b*sin(d*x+c))^2/(e*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^2*e^(-5/2)/cos(d*x + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^2/(e*cos(c + d*x))^(5/2),x)

[Out]

int((a + b*sin(c + d*x))^2/(e*cos(c + d*x))^(5/2), x)

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