3.1255 \(\int \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=181 \[ \frac{8 a^2 (19 A+21 B+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a (19 A+21 B+35 C) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 (3 A+7 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d} \]

[Out]

(8*a^2*(19*A + 21*B + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(19*A + 2
1*B + 35*C)*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d) + (2*(3*A + 7*B)*Cos[c + d*x]^(3
/2)*(a + a*Sec[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])^(3/2)*Sin[c
+ d*x])/(7*d)

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Rubi [A]  time = 0.600766, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4265, 4086, 4013, 3809, 3804} \[ \frac{8 a^2 (19 A+21 B+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a (19 A+21 B+35 C) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 (3 A+7 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d} \]

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] + C*Sec[c + d*x]^2),x]

[Out]

(8*a^2*(19*A + 21*B + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(19*A + 2
1*B + 35*C)*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d) + (2*(3*A + 7*B)*Cos[c + d*x]^(3
/2)*(a + a*Sec[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])^(3/2)*Sin[c
+ d*x])/(7*d)

Rule 4265

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]

Rule 4086

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

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))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+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) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (3 A+7 B)+\frac{1}{2} a (2 A+7 C) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac{2 (3 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))^{3/2} \sin (c+d x)}{7 d}+\frac{1}{35} \left ((19 A+21 B+35 C) \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{2 a (19 A+21 B+35 C) \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (3 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))^{3/2} \sin (c+d x)}{7 d}+\frac{1}{105} \left (4 a (19 A+21 B+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{8 a^2 (19 A+21 B+35 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 a (19 A+21 B+35 C) \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (3 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))^{3/2} \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.98434, size = 100, normalized size = 0.55 \[ \frac{a \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} ((253 A+28 (9 B+5 C)) \cos (c+d x)+6 (13 A+7 B) \cos (2 (c+d x))+15 A \cos (3 (c+d x))+494 A+546 B+700 C)}{210 d} \]

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] + C*Sec[c + d*x]^2),x]

[Out]

(a*Sqrt[Cos[c + d*x]]*(494*A + 546*B + 700*C + (253*A + 28*(9*B + 5*C))*Cos[c + d*x] + 6*(13*A + 7*B)*Cos[2*(c
+ d*x)] + 15*A*Cos[3*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(210*d)

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Maple [A]  time = 0.319, size = 121, normalized size = 0.7 \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 ) +35\,C\cos \left ( dx+c \right ) +104\,A+126\,B+175\,C \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)+C*sec(d*x+c)^2),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)+35*C*cos(d*x+c)+104*A+126*B+175*C)*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 = 2.37214, size = 693, normalized size = 3.83 \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)+C*sec(d*x+c)^2),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) + 280*(sqrt(2)*a*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 9*sqrt(
2)*a*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(a))/d

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Fricas [A]  time = 0.493011, size = 302, normalized size = 1.67 \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 + 35 \, C\right )} a \cos \left (d x + c\right ) +{\left (104 \, A + 126 \, B + 175 \, C\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)+C*sec(d*x+c)^2),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 + 35*C)*a*cos(d*x + c) + (104*A
+ 126*B + 175*C)*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)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + 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)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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