3.1293 \(\int \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=154 \[ \frac{2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (5 a A+7 a C+7 b B)}{21 d}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a B+3 A b+5 b C)}{5 d}+\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} (5 a A+7 a C+7 b B)}{21 d}+\frac{2 (a B+A b) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d} \]

[Out]

(2*(3*A*b + 3*a*B + 5*b*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(5*a*A + 7*b*B + 7*a*C)*EllipticF[(c + d*x)/2
, 2])/(21*d) + (2*(5*a*A + 7*b*B + 7*a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*(A*b + a*B)*Cos[c + d*x
]^(3/2)*Sin[c + d*x])/(5*d) + (2*a*A*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d)

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Rubi [A]  time = 0.269183, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4112, 3033, 3023, 2748, 2639, 2635, 2641} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (5 a A+7 a C+7 b B)}{21 d}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a B+3 A b+5 b C)}{5 d}+\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} (5 a A+7 a C+7 b B)}{21 d}+\frac{2 (a B+A b) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*A*b + 3*a*B + 5*b*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(5*a*A + 7*b*B + 7*a*C)*EllipticF[(c + d*x)/2
, 2])/(21*d) + (2*(5*a*A + 7*b*B + 7*a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*(A*b + a*B)*Cos[c + d*x
]^(3/2)*Sin[c + d*x])/(5*d) + (2*a*A*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d)

Rule 4112

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sqrt{\cos (c+d x)} (b+a \cos (c+d x)) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 a A \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2}{7} \int \sqrt{\cos (c+d x)} \left (\frac{7 b C}{2}+\frac{1}{2} (5 a A+7 b B+7 a C) \cos (c+d x)+\frac{7}{2} (A b+a B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 (A b+a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a A \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{4}{35} \int \sqrt{\cos (c+d x)} \left (\frac{7}{4} (3 A b+3 a B+5 b C)+\frac{5}{4} (5 a A+7 b B+7 a C) \cos (c+d x)\right ) \, dx\\ &=\frac{2 (A b+a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a A \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{7} (5 a A+7 b B+7 a C) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{5} (3 A b+3 a B+5 b C) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 (3 A b+3 a B+5 b C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (5 a A+7 b B+7 a C) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 (A b+a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a A \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{21} (5 a A+7 b B+7 a C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (3 A b+3 a B+5 b C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (5 a A+7 b B+7 a C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 (5 a A+7 b B+7 a C) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 (A b+a B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a A \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.900022, size = 117, normalized size = 0.76 \[ \frac{10 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (5 a A+7 a C+7 b B)+42 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a B+3 A b+5 b C)+\sin (c+d x) \sqrt{\cos (c+d x)} (42 (a B+A b) \cos (c+d x)+15 a A \cos (2 (c+d x))+65 a A+70 a C+70 b B)}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(42*(3*A*b + 3*a*B + 5*b*C)*EllipticE[(c + d*x)/2, 2] + 10*(5*a*A + 7*b*B + 7*a*C)*EllipticF[(c + d*x)/2, 2] +
Sqrt[Cos[c + d*x]]*(65*a*A + 70*b*B + 70*a*C + 42*(A*b + a*B)*Cos[c + d*x] + 15*a*A*Cos[2*(c + d*x)])*Sin[c +
d*x])/(105*d)

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Maple [B]  time = 2.419, size = 515, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*A*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8+(-360*A*a-168*A*b-168*B*a)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*A*a+168*A*b+168*B*a+140*B*b+140*C*a)
*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-80*A*a-42*A*b-42*B*a-70*B*b-70*C*a)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)+25*A*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))-63*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
*b+35*B*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-
63*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a+35*
a*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*C*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)/(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [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+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} +{\left (C a + B b\right )} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + A a \cos \left (d x + c\right )^{3} +{\left (B a + A b\right )} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b*cos(d*x + c)^3*sec(d*x + c)^3 + (C*a + B*b)*cos(d*x + c)^3*sec(d*x + c)^2 + A*a*cos(d*x + c)^3 +
(B*a + A*b)*cos(d*x + c)^3*sec(d*x + c))*sqrt(cos(d*x + c)), x)

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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+b*sec(d*x+c))*(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 (b \sec \left (d x + c\right ) + a\right )} \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+b*sec(d*x+c))*(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)*(b*sec(d*x + c) + a)*cos(d*x + c)^(7/2), x)