3.1246 \(\int \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=178 \[ \frac {2 a (24 A+28 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a (A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d} \]

[Out]

2/35*a*(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*(24*A+28*B+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+28*B+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 = 0.55, antiderivative size = 178, normalized size of antiderivative = 1.00, 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, 4015, 3805, 3804} \[ \frac {2 a (24 A+28 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a (A+7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(24*A + 28*B + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(24*A + 28*
B + 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(A + 7*B)*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 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]

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 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 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 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]

Rubi steps

\begin {align*} \int \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \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 {\sqrt {a+a \sec (c+d x)} \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) \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 {1}{2} a (A+7 B)+\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+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 \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+28 B+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+28 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (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 \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+28 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 {4 a (24 A+28 B+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+28 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (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 \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 = 0.38, size = 105, normalized size = 0.59 \[ \frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a (\sec (c+d x)+1)} ((141 A+28 (4 B+5 C)) \cos (c+d x)+6 (6 A+7 B) \cos (2 (c+d x))+15 A \cos (3 (c+d x))+228 A+266 B+280 C)}{210 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(228*A + 266*B + 280*C + (141*A + 28*(4*B + 5*C))*Cos[c + d*x] + 6*(6*A + 7*B)*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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fricas [A]  time = 0.43, size = 104, normalized size = 0.58 \[ \frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{3} + 3 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right ) + 48 \, A + 56 \, B + 70 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+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 + 3*(6*A + 7*B)*cos(d*x + c)^2 + (24*A + 28*B + 35*C)*cos(d*x + c) + 48*A + 56*B +
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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 2.35, size = 120, normalized size = 0.67 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (15 A \left (\cos ^{3}\left (d x +c \right )\right )+18 A \left (\cos ^{2}\left (d x +c \right )\right )+21 B \left (\cos ^{2}\left (d x +c \right )\right )+24 A \cos \left (d x +c \right )+28 B \cos \left (d x +c \right )+35 C \cos \left (d x +c \right )+48 A +56 B +70 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\sqrt {\cos }\left (d x +c \right )\right )}{105 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/d*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^3+18*A*cos(d*x+c)^2+21*B*cos(d*x+c)^2+24*A*cos(d*x+c)+28*B*cos(d*x+c
)+35*C*cos(d*x+c)+48*A+56*B+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]  time = 0.80, size = 508, normalized size = 2.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+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) - 14*sqrt(2)*(5*(6*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - (30*cos(
2*d*x + 2*c) + 5*cos(d*x + c) + 6)*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - 5*sin(3/2*arctan2(sin(d*x +
c), cos(d*x + c))) - 30*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*B*sqrt(a) - 140*(3*sqrt(2)*cos(3/2*arcta
n2(sin(d*x + c), cos(d*x + c)))*sin(d*x + c) - (3*sqrt(2)*cos(d*x + c) + 2*sqrt(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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