3.6.25 \(\int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [525]

Optimal. Leaf size=181 \[ \frac {4 a^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (52 A+63 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (8 A+7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d} \]

[Out]

2/35*a^2*(8*A+7*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+4/105*a^2*(52*A+63*B)*sin(d*x+c)/d/cos
(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+2/105*a^2*(52*A+63*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2
)+2/7*a*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 = 0.36, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3034, 4102, 4100, 3890, 3889} \begin {gather*} \frac {2 a^2 (8 A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (52 A+63 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(52*A + 63*B)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(52*A + 63*B)*
Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(8*A + 7*B)*Cos[c + d*x]^(3/2)*Sin[
c + d*x])/(35*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(
7*d)

Rule 3034

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Csc[e + f*x])^m*((
c + d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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 4102

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

Rubi steps

\begin {align*} \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d}+\frac {1}{7} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (8 A+7 B)+\frac {1}{2} a (4 A+7 B) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (8 A+7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a 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 (a (52 A+63 B) \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^2 (52 A+63 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (8 A+7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a 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 a (52 A+63 B) \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^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (52 A+63 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (8 A+7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a 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 = 0.32, size = 100, normalized size = 0.55 \begin {gather*} \frac {2 a \sqrt {\cos (c+d x)} \left (2 (52 A+63 B)+(52 A+63 B) \cos (c+d x)+3 (13 A+7 B) \cos ^2(c+d x)+15 A \cos ^3(c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{105 d (1+\cos (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[Cos[c + d*x]]*(2*(52*A + 63*B) + (52*A + 63*B)*Cos[c + d*x] + 3*(13*A + 7*B)*Cos[c + d*x]^2 + 15*A*C
os[c + d*x]^3)*Sqrt[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(105*d*(1 + Cos[c + d*x]))

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Maple [A]
time = 11.85, size = 109, normalized size = 0.60

method	result	size
default	\(-\frac {2 a \left (-1+\cos \left (d x +c \right )\right ) \left (15 A \left (\cos ^{3}\left (d x +c \right )\right )+39 A \left (\cos ^{2}\left (d x +c \right )\right )+21 B \left (\cos ^{2}\left (d x +c \right )\right )+52 A \cos \left (d x +c \right )+63 B \cos \left (d x +c \right )+104 A +126 B \right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{105 d \sin \left (d x +c \right )}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/d*a*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^3+39*A*cos(d*x+c)^2+21*B*cos(d*x+c)^2+52*A*cos(d*x+c)+63*B*cos(d*x
+c)+104*A+126*B)*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c))/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. 451 vs. \(2 (157) = 314\).
time = 0.90, size = 451, normalized size = 2.49 \begin {gather*} \frac {\sqrt {2} {\left (735 \, a \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 ) + 175 \, a \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 ) + 63 \, a \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 ) - 735 \, a \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 ) - 175 \, a \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 ) - 63 \, a \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 ) + 30 \, a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, a \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 ) + 175 \, a \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 ) + 735 \, a \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} - 84 \, {\left (10 \, \sqrt {2} a \cos \left (\frac {5}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) \sin \left (2 \, d x + 2 \, c\right ) - 5 \, \sqrt {2} a \sin \left (\frac {3}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - 10 \, \sqrt {2} a \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - {\left (10 \, \sqrt {2} a \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} a\right )} \sin \left (\frac {5}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )\right )} B \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+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/840*(sqrt(2)*(735*a*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) + 175*
a*cos(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) + 63*a*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) - 735*a*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(si
n(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 175*a*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))) - 63*a*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))
) + 30*a*sin(7/2*d*x + 7/2*c) + 63*a*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*a*sin(
3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 735*a*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2
*d*x + 7/2*c))))*A*sqrt(a) - 84*(10*sqrt(2)*a*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x +
2*c) - 5*sqrt(2)*a*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 10*sqrt(2)*a*sin(1/4*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c))) - (10*sqrt(2)*a*cos(2*d*x + 2*c) + sqrt(2)*a)*sin(5/4*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c))))*B*sqrt(a))/d

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Fricas [A]
time = 2.53, size = 105, normalized size = 0.58 \begin {gather*} \frac {2 \, {\left (15 \, A a \cos \left (d x + c\right )^{3} + 3 \, {\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right ) + 2 \, {\left (52 \, A + 63 \, B\right )} a\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+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/105*(15*A*a*cos(d*x + c)^3 + 3*(13*A + 7*B)*a*cos(d*x + c)^2 + (52*A + 63*B)*a*cos(d*x + c) + 2*(52*A + 63*B
)*a)*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+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)),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))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(3/2)*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 {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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