3.166 \(\int \frac{(a+a \cos (c+d x))^3}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=147 \[ \frac{52 a^3 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{28 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{52 a^3 \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{6 a^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a^3 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{28 a^3 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}} \]

[Out]

(-28*a^3*EllipticE[(c + d*x)/2, 2])/(5*d) + (52*a^3*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a^3*Sin[c + d*x])/(
7*d*Cos[c + d*x]^(7/2)) + (6*a^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (52*a^3*Sin[c + d*x])/(21*d*Cos[c +
d*x]^(3/2)) + (28*a^3*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.14811, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2757, 2636, 2641, 2639} \[ \frac{52 a^3 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{28 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{52 a^3 \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{6 a^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a^3 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{28 a^3 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3/Cos[c + d*x]^(9/2),x]

[Out]

(-28*a^3*EllipticE[(c + d*x)/2, 2])/(5*d) + (52*a^3*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a^3*Sin[c + d*x])/(
7*d*Cos[c + d*x]^(7/2)) + (6*a^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (52*a^3*Sin[c + d*x])/(21*d*Cos[c +
d*x]^(3/2)) + (28*a^3*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

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 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 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]

Rubi steps

\begin{align*} \int \frac{(a+a \cos (c+d x))^3}{\cos ^{\frac{9}{2}}(c+d x)} \, dx &=\int \left (\frac{a^3}{\cos ^{\frac{9}{2}}(c+d x)}+\frac{3 a^3}{\cos ^{\frac{7}{2}}(c+d x)}+\frac{3 a^3}{\cos ^{\frac{5}{2}}(c+d x)}+\frac{a^3}{\cos ^{\frac{3}{2}}(c+d x)}\right ) \, dx\\ &=a^3 \int \frac{1}{\cos ^{\frac{9}{2}}(c+d x)} \, dx+a^3 \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx+\left (3 a^3\right ) \int \frac{1}{\cos ^{\frac{7}{2}}(c+d x)} \, dx+\left (3 a^3\right ) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{6 a^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a^3 \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a^3 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{1}{7} \left (5 a^3\right ) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx+a^3 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-a^3 \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{5} \left (9 a^3\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^3 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^3 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{6 a^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{52 a^3 \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{28 a^3 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{1}{21} \left (5 a^3\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} \left (9 a^3\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{28 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{52 a^3 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a^3 \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{6 a^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{52 a^3 \sin (c+d x)}{21 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{28 a^3 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.23853, size = 515, normalized size = 3.5 \[ \frac{7 \csc (c) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\tan ^2(c)+1} \sqrt{1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt{\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac{\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt{\tan ^2(c)+1}}+\frac{2 \cos ^2(c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{20 d}-\frac{13 \csc (c) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin (c) \left (-\sqrt{\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{42 d \sqrt{\cot ^2(c)+1}}+\sqrt{\cos (c+d x)} \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (\frac{\sec (c) \sin (d x) \sec ^4(c+d x)}{28 d}+\frac{\sec (c) (5 \sin (c)+21 \sin (d x)) \sec ^3(c+d x)}{140 d}+\frac{\sec (c) (63 \sin (c)+130 \sin (d x)) \sec ^2(c+d x)}{420 d}+\frac{\sec (c) (65 \sin (c)+147 \sin (d x)) \sec (c+d x)}{210 d}+\frac{7 \csc (c) \sec (c)}{10 d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Cos[c + d*x])^3/Cos[c + d*x]^(9/2),x]

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*((7*Csc[c]*Sec[c])/(10*d) + (Sec[c]*Sec[c + d*x
]^4*Sin[d*x])/(28*d) + (Sec[c]*Sec[c + d*x]^3*(5*Sin[c] + 21*Sin[d*x]))/(140*d) + (Sec[c]*Sec[c + d*x]^2*(63*S
in[c] + 130*Sin[d*x]))/(420*d) + (Sec[c]*Sec[c + d*x]*(65*Sin[c] + 147*Sin[d*x]))/(210*d)) - (13*(a + a*Cos[c
+ d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*
x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[
c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(42*d*Sqrt[1 + Cot[c]^2]) + (7*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c
/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]
]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTa
n[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] +
(2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[
Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(20*d)

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Maple [B]  time = 3.825, size = 439, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^3/cos(d*x+c)^(9/2),x)

[Out]

-16*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-13/168*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+53/105*(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))-7/10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-
7/20*(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-3/160*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)^3-1/448*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)^4)/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 (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^3/cos(d*x + c)^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}{\cos \left (d x + c\right )^{\frac{9}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

integral((a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x + c) + a^3)/cos(d*x + c)^(9/2), 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+a*cos(d*x+c))**3/cos(d*x+c)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^3/cos(d*x + c)^(9/2), x)