3.27 \(\int (A+C \cos ^2(c+d x)) (b \sec (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=115 \[ -\frac {2 b^4 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^3 (3 A+5 C) \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 d}+\frac {2 A b^2 \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d} \]

[Out]

-2/5*b^4*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos
(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/5*b^3*(3*A+5*C)*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d+2/5*A*b^2*(b*sec(d*x+c)
)^(3/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3238, 4046, 3768, 3771, 2639} \[ \frac {2 b^3 (3 A+5 C) \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 d}-\frac {2 b^4 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A b^2 \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^4*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b^3*(3*A + 5
*C)*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*A*b^2*(b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

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 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3768

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 79, normalized size = 0.69 \[ -\frac {b^2 (b \sec (c+d x))^{3/2} \left (-(3 A+5 C) \sin (2 (c+d x))+2 (3 A+5 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-2 A \tan (c+d x)\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(b^2*(b*Sec[c + d*x])^(3/2)*(2*(3*A + 5*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] - (3*A + 5*C)*Sin
[2*(c + d*x)] - 2*A*Tan[c + d*x]))/d

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{3} \cos \left (d x + c\right )^{2} + A b^{3}\right )} \sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.40, size = 668, normalized size = 5.81 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right )^{2} \left (3 i A \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-3 i A \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 i C \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-5 i C \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 i A \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i A \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 i C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-5 i C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 A \left (\cos ^{3}\left (d x +c \right )\right )+5 C \left (\cos ^{3}\left (d x +c \right )\right )-2 A \left (\cos ^{2}\left (d x +c \right )\right )-5 C \left (\cos ^{2}\left (d x +c \right )\right )-A \right ) \cos \left (d x +c \right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}}}{5 d \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/d*(-1+cos(d*x+c))^2*(3*I*A*cos(d*x+c)^3*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-3*I*A*cos(d*x+c)^3*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*C*cos(d*x+c)^3*(1/(1+cos(d*x+c)))^(1/2
)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-5*I*C*cos(d*x+c)^3*(1
/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+
3*I*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*EllipticF(I*(-1+cos(d
*x+c))/sin(d*x+c),I)-3*I*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*
EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*C*cos(d*x+c)^2*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-5*I*C*cos(d*x+c)^2*sin(d*x+c)*(1/(1+cos(d*x+c))
)^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+3*A*cos(d*x+c)^3+5*C*cos(d
*x+c)^3-2*A*cos(d*x+c)^2-5*C*cos(d*x+c)^2-A)*cos(d*x+c)*(1+cos(d*x+c))^2*(b/cos(d*x+c))^(7/2)/sin(d*x+c)^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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