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]

-6/5*a*(a^2+5*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))/d+2*b
*(a^2+b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+8/5*a^2*b*s
in(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*a^2*(a+b*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(5/2)+6/5*a*(a^2+5*b^2)*sin(d*x+
c)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 149, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.96, size = 125, normalized size = 0.84 \[ \frac {3 \left (a^3+5 a b^2\right ) \sin (2 (c+d x))+2 a^3 \tan (c+d x)+10 b \left (a^2+b^2\right ) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-6 a \left (a^2+5 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 a^2 b \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)} \]

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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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

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)

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maple [B]  time = 2.07, size = 738, normalized size = 4.95 \[ \text {result too large to display} \]

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))-
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*Ell
ipticE(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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*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)+6*b^2*a*(-(-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)^(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))+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)+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)/(-1/2+cos(1/2*d*x+1/2*c)^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))))/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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

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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mupad [B]  time = 1.74, size = 156, normalized size = 1.05 \[ \frac {2\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,a\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*a^3*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d
*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)) + (6*a*b^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2
))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*a^2*b*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d
*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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