3.3.62 \(\int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\) [262]

Optimal. Leaf size=217 \[ -\frac {6 b^2 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^3 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^5 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 b^3 B \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}} \]

[Out]

2/7*A*b^6*sin(d*x+c)/d/(b*cos(d*x+c))^(7/2)+2/5*b^5*B*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/21*b^4*(5*A+7*C)*sin
(d*x+c)/d/(b*cos(d*x+c))^(3/2)+6/5*b^3*B*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)+2/21*b^3*(5*A+7*C)*(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))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)-
6/5*b^2*B*(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))*(b*cos(d*x+c))
^(1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {16, 3100, 2827, 2716, 2721, 2719, 2720} \begin {gather*} \frac {2 A b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^4 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {2 b^3 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 b^5 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {6 b^3 B \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}-\frac {6 b^2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(-6*b^2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b^3*(5*A + 7*C)*Sqrt[C
os[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^6*Sin[c + d*x])/(7*d*(b*Cos[c + d
*x])^(7/2)) + (2*b^5*B*Sin[c + d*x])/(5*d*(b*Cos[c + d*x])^(5/2)) + (2*b^4*(5*A + 7*C)*Sin[c + d*x])/(21*d*(b*
Cos[c + d*x])^(3/2)) + (6*b^3*B*Sin[c + d*x])/(5*d*Sqrt[b*Cos[c + d*x]])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2716

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 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 2721

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

Rule 2827

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 3100

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 (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=b^7 \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{9/2}} \, dx\\ &=\frac {2 A b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {1}{7} \left (2 b^4\right ) \int \frac {\frac {7 b^2 B}{2}+\frac {1}{2} b^2 (5 A+7 C) \cos (c+d x)}{(b \cos (c+d x))^{7/2}} \, dx\\ &=\frac {2 A b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\left (b^6 B\right ) \int \frac {1}{(b \cos (c+d x))^{7/2}} \, dx+\frac {1}{7} \left (b^5 (5 A+7 C)\right ) \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx\\ &=\frac {2 A b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^5 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {1}{5} \left (3 b^4 B\right ) \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx+\frac {1}{21} \left (b^3 (5 A+7 C)\right ) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx\\ &=\frac {2 A b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^5 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 b^3 B \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}-\frac {1}{5} \left (3 b^2 B\right ) \int \sqrt {b \cos (c+d x)} \, dx+\frac {\left (b^3 (5 A+7 C) \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {b \cos (c+d x)}}\\ &=\frac {2 b^3 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^5 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 b^3 B \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}-\frac {\left (3 b^2 B \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)}}\\ &=-\frac {6 b^2 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^3 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^5 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 b^3 B \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.82, size = 134, normalized size = 0.62 \begin {gather*} \frac {(b \cos (c+d x))^{5/2} \sec ^6(c+d x) \left (-504 B \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+40 (5 A+7 C) \cos ^{\frac {7}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 (110 A+70 C+273 B \cos (c+d x)+10 (5 A+7 C) \cos (2 (c+d x))+63 B \cos (3 (c+d x))) \sin (c+d x)\right )}{420 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

((b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^6*(-504*B*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2] + 40*(5*A + 7*C)*C
os[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2] + 2*(110*A + 70*C + 273*B*Cos[c + d*x] + 10*(5*A + 7*C)*Cos[2*(c +
d*x)] + 63*B*Cos[3*(c + d*x)])*Sin[c + d*x]))/(420*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(726\) vs. \(2(241)=482\).
time = 1.15, size = 727, normalized size = 3.35

method	result	size
default	\(\text {Expression too large to display}\)	\(727\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x,method=_RETURNVERBOSE)

[Out]

-2*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(1/5*B/b/sin(1/2*d*x+1/2*c)^2/(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)*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*Elli
pticE(cos(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)*sin(1/2*d*x+1/
2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*EllipticE(cos(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)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(s
in(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*sin(1/2
*d*x+1/2*c)^4*b+sin(1/2*d*x+1/2*c)^2*b)^(1/2)+A*(-1/56*cos(1/2*d*x+1/2*c)/b*(-b*(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)^4-5/42*cos(1/2*d*x+1/2*c)/b*(-b*(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+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*
c)^2+1)^(1/2)/(-b*(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)))+
C*(-1/6*cos(1/2*d*x+1/2*c)/b*(-b*(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)/(-b*(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)/(b*(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((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^7, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 243, normalized size = 1.12 \begin {gather*} \frac {-5 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 63 i \, \sqrt {2} B b^{\frac {5}{2}} \cos \left (d x + c\right )^{4} {\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 ) + 63 i \, \sqrt {2} B b^{\frac {5}{2}} \cos \left (d x + c\right )^{4} {\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 ) + 2 \, {\left (63 \, B b^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (5 \, A + 7 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 21 \, B b^{2} \cos \left (d x + c\right ) + 15 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/105*(-5*I*sqrt(2)*(5*A + 7*C)*b^(5/2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
)) + 5*I*sqrt(2)*(5*A + 7*C)*b^(5/2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))
- 63*I*sqrt(2)*B*b^(5/2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin
(d*x + c))) + 63*I*sqrt(2)*B*b^(5/2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) - I*sin(d*x + c))) + 2*(63*B*b^2*cos(d*x + c)^3 + 5*(5*A + 7*C)*b^2*cos(d*x + c)^2 + 21*B*b^2*cos(d*x + c
) + 15*A*b^2)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

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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((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**7,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((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^7, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^7,x)

[Out]

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^7, x)

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