3.580 \(\int \frac{(a+b \cos (c+d x))^3}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=149 \[ \frac{2 b \left (a^2+b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}-\frac{6 a \left (a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{6 a \left (a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 b \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)} \]

[Out]

(-6*a*(a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*b*(a^2 + b^2)*EllipticF[(c + d*x)/2, 2])/d + (8*a^2*
b*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2)) + (6*a*(a^2 + 5*b^2)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*a^2*
(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))

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Rubi [A]  time = 0.210559, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2792, 3021, 2748, 2636, 2639, 2641} \[ \frac{2 b \left (a^2+b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}-\frac{6 a \left (a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{6 a \left (a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 b \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*(a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*b*(a^2 + b^2)*EllipticF[(c + d*x)/2, 2])/d + (8*a^2*
b*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2)) + (6*a*(a^2 + 5*b^2)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*a^2*
(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2636

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+b \cos (c+d x))^3}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{6 a^2 b+\frac{3}{2} a \left (a^2+5 b^2\right ) \cos (c+d x)+\frac{1}{2} b \left (a^2+5 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{8 a^2 b \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4}{15} \int \frac{\frac{9}{4} a \left (a^2+5 b^2\right )+\frac{15}{4} b \left (a^2+b^2\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{8 a^2 b \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\left (b \left (a^2+b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (3 a \left (a^2+5 b^2\right )\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b \left (a^2+b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{8 a^2 b \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{6 a \left (a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{1}{5} \left (3 a \left (a^2+5 b^2\right )\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{6 a \left (a^2+5 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b \left (a^2+b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{8 a^2 b \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{6 a \left (a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.860373, size = 125, normalized size = 0.84 \[ \frac{3 \left (a^3+5 a b^2\right ) \sin (2 (c+d x))+10 b \left (a^2+b^2\right ) \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 a \left (a^2+5 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+10 a^2 b \sin (c+d x)+2 a^3 \tan (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*(a^2 + 5*b^2)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*b*(a^2 + b^2)*Cos[c + d*x]^(3/2)*Ellipti
cF[(c + d*x)/2, 2] + 10*a^2*b*Sin[c + d*x] + 3*(a^3 + 5*a*b^2)*Sin[2*(c + d*x)] + 2*a^3*Tan[c + d*x])/(5*d*Cos
[c + d*x]^(3/2))

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Maple [B]  time = 8.242, size = 738, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+
6*a^2*b*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/
2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+6*a*b^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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c
)^2/(2*sin(1/2*d*x+1/2*c)^2-1)-2/5*a^3/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-
1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1/2*d*x+1/2
*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x
+1/2*c)^4*cos(1/2*d*x+1/2*c)+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))-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)
^(1/2))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}{\cos \left (d x + c\right )^{\frac{7}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3/cos(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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