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

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

[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)

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Rubi [A]  time = 0.616517, antiderivative size = 181, normalized size of antiderivative = 1., 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 = {2955, 4017, 4015, 3805, 3804} \[ \frac{2 a^2 (8 A+7 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{35 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (52 A+63 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{4 a^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{7 d} \]

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 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 4017

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]

Rule 4015

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 3805

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 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))^{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*}

Mathematica [A]  time = 0.316271, size = 100, normalized size = 0.55 \[ \frac{2 a \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)} \left (3 (13 A+7 B) \cos ^2(c+d x)+(52 A+63 B) \cos (c+d x)+2 (52 A+63 B)+15 A \cos ^3(c+d x)\right )}{105 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(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 = 0.273, size = 109, normalized size = 0.6 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 15\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+39\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+52\,A\cos \left ( dx+c \right ) +63\,B\cos \left ( dx+c \right ) +104\,A+126\,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))^(3/2)*(A+B*sec(d*x+c)),x)

[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*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.06328, size = 609, normalized size = 3.36 \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))^(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 = 0.48177, size = 282, normalized size = 1.56 \begin{align*} \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{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))^(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)]  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))**(3/2)*(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{3}{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))^(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)