3.12.28 \(\int \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\) [1128]

Optimal. Leaf size=168 \[ \frac {4 a (24 A+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a (24 A+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d} \]

[Out]

2/35*a*A*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+4/105*a*(24*A+35*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2
)/(a+a*sec(d*x+c))^(1/2)+2/105*a*(24*A+35*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/7*A*cos(d*
x+c)^(5/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.33, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4350, 4172, 4100, 3890, 3889} \begin {gather*} \frac {2 a (24 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a (24 A+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(24*A + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(24*A + 35*C)*Sqrt
[Cos[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(35*d*
Sqrt[a + a*Sec[c + d*x]]) + (2*A*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(7*d)

Rule 3889

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[-2*a*(Co
t[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rule 3890

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e
+ f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[a*((2*n + 1)/(2*b*d*n)), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rule 4100

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]

Rule 4172

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rule 4350

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[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) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {a A}{2}+\frac {1}{2} a (4 A+7 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {1}{35} \left ((24 A+35 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a (24 A+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {1}{105} \left (2 (24 A+35 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {4 a (24 A+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a (24 A+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 90, normalized size = 0.54 \begin {gather*} \frac {\sqrt {\cos (c+d x)} (228 A+280 C+(141 A+140 C) \cos (c+d x)+36 A \cos (2 (c+d x))+15 A \cos (3 (c+d x))) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{210 d (1+\cos (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(228*A + 280*C + (141*A + 140*C)*Cos[c + d*x] + 36*A*Cos[2*(c + d*x)] + 15*A*Cos[3*(c + d*
x)])*Sqrt[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(210*d*(1 + Cos[c + d*x]))

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Maple [A]
time = 0.20, size = 97, normalized size = 0.58

method	result	size
default	\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (15 A \left (\cos ^{3}\left (d x +c \right )\right )+18 A \left (\cos ^{2}\left (d x +c \right )\right )+24 A \cos \left (d x +c \right )+35 C \cos \left (d x +c \right )+48 A +70 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\sqrt {\cos }\left (d x +c \right )\right )}{105 d \sin \left (d x +c \right )}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(7/2)*(A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/d*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^3+18*A*cos(d*x+c)^2+24*A*cos(d*x+c)+35*C*cos(d*x+c)+48*A+70*C)*(a*(1
+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)/sin(d*x+c)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (144) = 288\).
time = 0.64, size = 387, normalized size = 2.30 \begin {gather*} \frac {3 \, \sqrt {2} {\left (105 \, \cos \left (\frac {6}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 35 \, \cos \left (\frac {4}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \cos \left (\frac {2}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 105 \, \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {6}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 35 \, \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {4}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 7 \, \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {2}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 10 \, \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 105 \, \sin \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )\right )} A \sqrt {a} - 140 \, {\left (3 \, \sqrt {2} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) \sin \left (d x + c\right ) - {\left (3 \, \sqrt {2} \cos \left (d x + c\right ) + 2 \, \sqrt {2}\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right )\right )\right )\right )} C \sqrt {a}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/840*(3*sqrt(2)*(105*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 35*c
os(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 7*cos(2/7*arctan2(sin(7/2*d
*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 105*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(sin(7/2*d*
x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 35*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x
+ 7/2*c))) - 7*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 10*sin(7/2
*d*x + 7/2*c) + 7*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 35*sin(3/7*arctan2(sin(7/2*d*
x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 105*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(
a) - 140*(3*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(d*x + c) - (3*sqrt(2)*cos(d*x + c) + 2*sq
rt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) - 3*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*
C*sqrt(a))/d

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Fricas [A]
time = 3.30, size = 92, normalized size = 0.55 \begin {gather*} \frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{3} + 18 \, A \cos \left (d x + c\right )^{2} + {\left (24 \, A + 35 \, C\right )} \cos \left (d x + c\right ) + 48 \, A + 70 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/105*(15*A*cos(d*x + c)^3 + 18*A*cos(d*x + c)^2 + (24*A + 35*C)*cos(d*x + c) + 48*A + 70*C)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(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+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),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+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^{7/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(7/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^(7/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(1/2), x)

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