3.172 \(\int \frac{(a+a \cos (c+d x))^4}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=121 \[ \frac{32 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{56 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{66 a^4 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}} \]

[Out]

(-56*a^4*EllipticE[(c + d*x)/2, 2])/(5*d) + (32*a^4*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^4*Sin[c + d*x])/(5
*d*Cos[c + d*x]^(5/2)) + (8*a^4*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (66*a^4*Sin[c + d*x])/(5*d*Sqrt[Cos[c
+ d*x]])

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Rubi [A]  time = 0.146417, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2757, 2636, 2639, 2641} \[ \frac{32 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{56 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{66 a^4 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4/Cos[c + d*x]^(7/2),x]

[Out]

(-56*a^4*EllipticE[(c + d*x)/2, 2])/(5*d) + (32*a^4*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^4*Sin[c + d*x])/(5
*d*Cos[c + d*x]^(5/2)) + (8*a^4*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (66*a^4*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 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+a \cos (c+d x))^4}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\int \left (\frac{a^4}{\cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^4}{\cos ^{\frac{5}{2}}(c+d x)}+\frac{6 a^4}{\cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^4}{\sqrt{\cos (c+d x)}}+a^4 \sqrt{\cos (c+d x)}\right ) \, dx\\ &=a^4 \int \frac{1}{\cos ^{\frac{7}{2}}(c+d x)} \, dx+a^4 \int \sqrt{\cos (c+d x)} \, dx+\left (4 a^4\right ) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx+\left (4 a^4\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\left (6 a^4\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{8 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{12 a^4 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{1}{5} \left (3 a^4\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{3} \left (4 a^4\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\left (6 a^4\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{10 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{32 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{66 a^4 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}-\frac{1}{5} \left (3 a^4\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{56 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{32 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{66 a^4 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.34984, size = 283, normalized size = 2.34 \[ \frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{168 \sec (c) \cos ^2(c+d x) \left (\csc (c) \sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )} \left (3 \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )+\cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )-2 \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )\right )}{\sqrt{\sec ^2(c)} \sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )}}-640 \sin (c) \sqrt{\csc ^2(c)} \cos ^3(c+d x) \sqrt{\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+\csc (c) (141 \cos (2 c+d x)+40 \cos (c+2 d x)-40 \cos (3 c+2 d x)+183 \cos (2 c+3 d x)-15 \cos (4 c+3 d x)+363 \cos (d x))\right )}{960 d \cos ^{\frac{5}{2}}(c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Cos[c + d*x])^4/Cos[c + d*x]^(7/2),x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*((363*Cos[d*x] + 141*Cos[2*c + d*x] + 40*Cos[c + 2*d*x] - 40*Cos[
3*c + 2*d*x] + 183*Cos[2*c + 3*d*x] - 15*Cos[4*c + 3*d*x])*Csc[c] - 640*Cos[c + d*x]^3*Sqrt[Cos[d*x - ArcTan[C
ot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[C
ot[c]]]*Sin[c] - (168*Cos[c + d*x]^2*Sec[c]*(-2*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]
]]^2]*Sin[d*x + ArcTan[Tan[c]]] + (3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Csc[c]*Sqr
t[Sin[d*x + ArcTan[Tan[c]]]^2]))/(Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])))/(960*d*Cos[c + d*x]^(5/2
))

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Maple [B]  time = 4.058, size = 386, normalized size = 3.2 \begin{align*} -32\,{\frac{\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{a}^{4}}{\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}d} \left ( -{\frac{7\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ({\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) }{20\,\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}+{\frac{41\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) }{60\,\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}-1/24\,{\frac{\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{ \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{2}}}-{\frac{33\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) }{40\,\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}-{\frac{\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{320\, \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^4/cos(d*x+c)^(7/2),x)

[Out]

-32*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(-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)))+41/60*(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))-1/24*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)^2-33/40*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)-1/320*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)/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 )}^{4}}{\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+a*cos(d*x+c))^4/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^4/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{a^{4} \cos \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right ) + a^{4}}{\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+a*cos(d*x+c))^4/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((a^4*cos(d*x + c)^4 + 4*a^4*cos(d*x + c)^3 + 6*a^4*cos(d*x + c)^2 + 4*a^4*cos(d*x + c) + a^4)/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+a*cos(d*x+c))**4/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 (a \cos \left (d x + c\right ) + a\right )}^{4}}{\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+a*cos(d*x+c))^4/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^4/cos(d*x + c)^(7/2), x)