3.4.68 \(\int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx\) [368]

Optimal. Leaf size=121 \[ \frac {28 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {52 a^3 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {52 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {6 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^3 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

28/5*a^3*(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+52/21*a^3*(co
s(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+6/5*a^3*cos(d*x+c)^(3/2)*
sin(d*x+c)/d+2/7*a^3*cos(d*x+c)^(5/2)*sin(d*x+c)/d+52/21*a^3*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.17, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4349, 3876, 3854, 3856, 2720, 2719} \begin {gather*} \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 {2 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {6 a^3 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {52 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2720

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

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3876

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

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.11, size = 516, normalized size = 4.26 \begin {gather*} \cos ^{\frac {7}{2}}(c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \left (-\frac {7 \cot (c)}{10 d}+\frac {107 \cos (d x) \sin (c)}{336 d}+\frac {3 \cos (2 d x) \sin (2 c)}{40 d}+\frac {\cos (3 d x) \sin (3 c)}{112 d}+\frac {107 \cos (c) \sin (d x)}{336 d}+\frac {3 \cos (2 c) \sin (2 d x)}{40 d}+\frac {\cos (3 c) \sin (3 d x)}{112 d}\right )-\frac {13 \cos ^3(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{42 d \sqrt {1+\cot ^2(c)}}-\frac {7 \cos ^3(c+d x) \csc (c) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^3 \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{20 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Cos[c + d*x]^(7/2)*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*((-7*Cot[c])/(10*d) + (107*Cos[d*x]*Sin[c])/(33
6*d) + (3*Cos[2*d*x]*Sin[2*c])/(40*d) + (Cos[3*d*x]*Sin[3*c])/(112*d) + (107*Cos[c]*Sin[d*x])/(336*d) + (3*Cos
[2*c]*Sin[2*d*x])/(40*d) + (Cos[3*c]*Sin[3*d*x])/(112*d)) - (13*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*(a + a*Sec[c + d*x])^3*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 + Si
n[d*x - ArcTan[Cot[c]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (7*Cos[c + d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec
[c + d*x])^3*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*T
an[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[T
an[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 [A]
time = 0.11, size = 272, normalized size = 2.25

method	result	size
default	\(-\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{3} \left (120 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-432 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+602 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-208 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+65 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-147 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(120*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8-432*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+602*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-208*sin(1/2*d*x+1/2*
c)^2*cos(1/2*d*x+1/2*c)+65*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x
+1/2*c),2^(1/2))-147*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2)))/(-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 = 0.00, 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(cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((a*sec(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 = 0.95, size = 162, normalized size = 1.34 \begin {gather*} -\frac {2 \, {\left (65 i \, \sqrt {2} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 65 i \, \sqrt {2} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 147 i \, \sqrt {2} a^{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 ) + 147 i \, \sqrt {2} a^{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 ) - {\left (15 \, a^{3} \cos \left (d x + c\right )^{2} + 63 \, a^{3} \cos \left (d x + c\right ) + 130 \, a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(65*I*sqrt(2)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 65*I*sqrt(2)*a^3*weierstr
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 147*I*sqrt(2)*a^3*weierstrassZeta(-4, 0, weierstrassPInver
se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 147*I*sqrt(2)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0
, cos(d*x + c) - I*sin(d*x + c))) - (15*a^3*cos(d*x + c)^2 + 63*a^3*cos(d*x + c) + 130*a^3)*sqrt(cos(d*x + c))
*sin(d*x + c))/d

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

[Out]

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

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Mupad [B]
time = 1.03, size = 143, normalized size = 1.18 \begin {gather*} \frac {2\,\left (a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}-\frac {6\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a^3*ellipticE(c/2 + (d*x)/2, 2) + a^3*ellipticF(c/2 + (d*x)/2, 2) + a^3*cos(c + d*x)^(1/2)*sin(c + d*x)))/
d - (6*a^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^
(1/2)) - (2*a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x
)^2)^(1/2))

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