3.534 \(\int \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=178 \[ \frac{64 a^3 (5 A+7 B) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (5 A+7 B) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (5 A+7 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

[Out]

(64*a^3*(5*A + 7*B)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(5*A + 7*B)*Sq
rt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d) + (2*a*(5*A + 7*B)*Cos[c + d*x]^(3/2)*(a + a*S
ec[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*A*Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(7*
d)

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Rubi [A]  time = 0.458249, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {2955, 4013, 3809, 3804} \[ \frac{64 a^3 (5 A+7 B) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (5 A+7 B) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (5 A+7 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(5*A + 7*B)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(5*A + 7*B)*Sq
rt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d) + (2*a*(5*A + 7*B)*Cos[c + d*x]^(3/2)*(a + a*S
ec[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*A*Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(7*
d)

Rule 2955

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 4013

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

Rule 3809

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

Rule 3804

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]

Rubi steps

\begin{align*} \int \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (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))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 A \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{1}{7} \left ((5 A+7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a (5 A+7 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 A \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{1}{35} \left (8 a (5 A+7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{16 a^2 (5 A+7 B) \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 a (5 A+7 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 A \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{1}{105} \left (32 a^2 (5 A+7 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{64 a^3 (5 A+7 B) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 (5 A+7 B) \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 a (5 A+7 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 A \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.350622, size = 99, normalized size = 0.56 \[ \frac{2 a^2 \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)} \left (3 (20 A+7 B) \cos ^2(c+d x)+(115 A+98 B) \cos (c+d x)+15 A \cos ^3(c+d x)+230 A+301 B\right )}{105 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sqrt[Cos[c + d*x]]*(230*A + 301*B + (115*A + 98*B)*Cos[c + d*x] + 3*(20*A + 7*B)*Cos[c + d*x]^2 + 15*A*
Cos[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 = 0.274, size = 111, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 15\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+60\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+115\,A\cos \left ( dx+c \right ) +98\,B\cos \left ( dx+c \right ) +230\,A+301\,B \right ) }{105\,d\sin \left ( dx+c \right ) }\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/d*a^2*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^3+60*A*cos(d*x+c)^2+21*B*cos(d*x+c)^2+115*A*cos(d*x+c)+98*B*cos(
d*x+c)+230*A+301*B)*cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.06843, size = 651, normalized size = 3.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(5*sqrt(2)*(315*a^2*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) +
77*a^2*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) + 21*a^2*cos(2/7*arct
an2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315*a^2*cos(7/2*d*x + 7/2*c)*sin(6/7*a
rctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 77*a^2*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))) - 21*a^2*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))) + 6*a^2*sin(7/2*d*x + 7/2*c) + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)
)) + 77*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a^2*sin(1/7*arctan2(sin(7/2*d*x
+ 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(a) - 28*(75*sqrt(2)*a^2*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c)))*sin(2*d*x + 2*c) - 25*sqrt(2)*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 75*sqrt(2)*a
^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(25*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*si
n(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 = 0.481874, size = 294, normalized size = 1.65 \begin{align*} \frac{2 \,{\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (115 \, A + 98 \, B\right )} a^{2} \cos \left (d x + c\right ) +{\left (230 \, A + 301 \, B\right )} a^{2}\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*A*a^2*cos(d*x + c)^3 + 3*(20*A + 7*B)*a^2*cos(d*x + c)^2 + (115*A + 98*B)*a^2*cos(d*x + c) + (230*A
+ 301*B)*a^2)*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)]  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(cos(d*x+c)**(7/2)*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \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(cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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