3.16 \(\int (b \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=110 \[ \frac {2 b^2 (7 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 b (7 A+5 C) \sin (c+d x) (b \sec (c+d x))^{3/2}}{21 d}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d} \]

[Out]

2/21*b*(7*A+5*C)*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/21*b^2*(7*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x
+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/d+2/7*C*(b*sec(d*x+c))^(5/
2)*tan(d*x+c)/d

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Rubi [A]  time = 0.08, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4046, 3768, 3771, 2641} \[ \frac {2 b^2 (7 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 b (7 A+5 C) \sin (c+d x) (b \sec (c+d x))^{3/2}}{21 d}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

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

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]

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

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Mathematica [A]  time = 1.38, size = 84, normalized size = 0.76 \[ \frac {(b \sec (c+d x))^{7/2} \left (4 (7 A+5 C) \cos ^{\frac {7}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) ((7 A+5 C) \cos (2 (c+d x))+7 A+11 C)\right )}{42 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

((b*Sec[c + d*x])^(7/2)*(4*(7*A + 5*C)*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2] + 2*(7*A + 11*C + (7*A + 5
*C)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(42*b*d)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \sec \left (d x + c\right )^{4} + A b^{2} \sec \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b^2*sec(d*x + c)^4 + A*b^2*sec(d*x + c)^2)*sqrt(b*sec(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^(5/2), x)

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maple [C]  time = 1.45, size = 251, normalized size = 2.28 \[ -\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (7 i A \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 i C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-7 A \left (\cos ^{3}\left (d x +c \right )\right )-5 C \left (\cos ^{3}\left (d x +c \right )\right )+7 A \left (\cos ^{2}\left (d x +c \right )\right )+5 C \left (\cos ^{2}\left (d x +c \right )\right )-3 C \cos \left (d x +c \right )+3 C \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{21 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)

[Out]

-2/21/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(7*I*A*sin(d*x+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)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*C*sin(d*x+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)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-7*A*cos(d*x+c)^3-5*C*cos(d
*x+c)^3+7*A*cos(d*x+c)^2+5*C*cos(d*x+c)^2-3*C*cos(d*x+c)+3*C)*(b/cos(d*x+c))^(5/2)/sin(d*x+c)^3/cos(d*x+c)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(5/2),x)

[Out]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(5/2), 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((b*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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