3.6.62 \(\int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx\) [562]

Optimal. Leaf size=187 \[ \frac {2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac {6 a \left (a^2-2 b^2\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac {2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}} \]

[Out]

2/5*b*(3*a^2-4*b^2)*(e*cos(d*x+c))^(3/2)/d/e^5+2/5*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^2/d/e/(e*cos(d*x+c))^(5/2
)-2/5*(a+b*sin(d*x+c))*(a*b-(3*a^2-4*b^2)*sin(d*x+c))/d/e^3/(e*cos(d*x+c))^(1/2)-6/5*a*(a^2-2*b^2)*(cos(1/2*d*
x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/d/e^4/cos(d*x+
c)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2770, 2940, 2748, 2721, 2719} \begin {gather*} \frac {2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac {6 a \left (a^2-2 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d e^4 \sqrt {\cos (c+d x)}}-\frac {2 \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))}{5 d e^3 \sqrt {e \cos (c+d x)}}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*b*(3*a^2 - 4*b^2)*(e*Cos[c + d*x])^(3/2))/(5*d*e^5) - (6*a*(a^2 - 2*b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[c + d*x]]) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(5*d*e*(e*Cos[c +
d*x])^(5/2)) - (2*(a + b*Sin[c + d*x])*(a*b - (3*a^2 - 4*b^2)*Sin[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c + d*x]])

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 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])

Rule 2940

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^m*((d + c*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 - 1)*Simp[a*c*(p
+ 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b
*x])

Rubi steps

\begin {align*} \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx &=\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac {2 \int \frac {(a+b \sin (c+d x)) \left (-\frac {3 a^2}{2}+2 b^2+\frac {1}{2} a b \sin (c+d x)\right )}{(e \cos (c+d x))^{3/2}} \, dx}{5 e^2}\\ &=\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac {2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}}+\frac {4 \int \sqrt {e \cos (c+d x)} \left (-\frac {3}{4} a \left (a^2-2 b^2\right )-\frac {3}{4} b \left (3 a^2-4 b^2\right ) \sin (c+d x)\right ) \, dx}{5 e^4}\\ &=\frac {2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac {2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}}-\frac {\left (3 a \left (a^2-2 b^2\right )\right ) \int \sqrt {e \cos (c+d x)} \, dx}{5 e^4}\\ &=\frac {2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac {2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}}-\frac {\left (3 a \left (a^2-2 b^2\right ) \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 e^4 \sqrt {\cos (c+d x)}}\\ &=\frac {2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac {6 a \left (a^2-2 b^2\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac {2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.76, size = 126, normalized size = 0.67 \begin {gather*} \frac {2 \left (-5 b^3-3 a \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+b \left (3 a^2+b^2\right ) \sec ^2(c+d x)+3 a^3 \sin (c+d x)-6 a b^2 \sin (c+d x)+a \left (a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*(-5*b^3 - 3*a*(a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + b*(3*a^2 + b^2)*Sec[c + d*x]^2 +
3*a^3*Sin[c + d*x] - 6*a*b^2*Sin[c + d*x] + a*(a^2 + 3*b^2)*Sec[c + d*x]*Tan[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c
+ d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(617\) vs. \(2(195)=390\).
time = 17.23, size = 618, normalized size = 3.30

method	result	size
default	\(-\frac {2 \left (12 \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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3} \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2} \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+48 a \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3} \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2} \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-48 a \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 b^{3} \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3}-6 \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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-8 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a \,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 )-20 b^{3} \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{2} b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+4 b^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \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^{3} d}\)	\(618\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*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^3*(12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^
3*sin(1/2*d*x+1/2*c)^4-24*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2))*a*b^2*sin(1/2*d*x+1/2*c)^4-24*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+48*a*b^2*cos(1/2*d*x
+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/
2*d*x+1/2*c),2^(1/2))*a^3*sin(1/2*d*x+1/2*c)^2+24*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2*sin(1/2*d*x+1/2*c)^2+24*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^4-48*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+20*b^3*sin(1/2*d*x+1/2*c)^5+3*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-6*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-8*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^2+6*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-20*b^3*sin(1/2*d*x+1/2*c)^3-3*a^2*b*sin(1/2*d*x+1/2
*c)+4*b^3*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))^3/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

e^(-7/2)*integrate((b*sin(d*x + c) + a)^3/cos(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 = 0.11, size = 181, normalized size = 0.97 \begin {gather*} -\frac {{\left (3 \, \sqrt {2} {\left (i \, a^{3} - 2 i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} {\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 ) + 3 \, \sqrt {2} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} {\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 ) + 2 \, {\left (5 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3} - {\left (a^{3} + 3 \, a b^{2} + 3 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}\right )} e^{\left (-\frac {7}{2}\right )}}{5 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/5*(3*sqrt(2)*(I*a^3 - 2*I*a*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x +
c) + I*sin(d*x + c))) + 3*sqrt(2)*(-I*a^3 + 2*I*a*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInve
rse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(5*b^3*cos(d*x + c)^2 - 3*a^2*b - b^3 - (a^3 + 3*a*b^2 + 3*(a^3
- 2*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(cos(d*x + c)))*e^(-7/2)/(d*cos(d*x + c)^3)

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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((a+b*sin(d*x+c))**3/(e*cos(d*x+c))**(7/2),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((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^3*e^(-7/2)/cos(d*x + c)^(7/2), 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 )}^3}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^3/(e*cos(c + d*x))^(7/2),x)

[Out]

int((a + b*sin(c + d*x))^3/(e*cos(c + d*x))^(7/2), x)

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