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

Optimal. Leaf size=171 \[ \frac{2 A b^2 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{3 d}+\frac{2 A b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}+\frac{6 b^2 B \sin (c+d x) \sqrt{b \sec (c+d x)}}{5 d}-\frac{6 b^3 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 B \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d} \]

[Out]

(-6*b^3*B*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*A*b^2*Sqrt[Cos[c + d*x
]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (6*b^2*B*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(5*d) +
(2*A*b*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + (2*B*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.118789, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3787, 3768, 3771, 2641, 2639} \[ \frac{2 A b^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 d}+\frac{2 A b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}+\frac{6 b^2 B \sin (c+d x) \sqrt{b \sec (c+d x)}}{5 d}-\frac{6 b^3 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 B \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*b^3*B*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*A*b^2*Sqrt[Cos[c + d*x
]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (6*b^2*B*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(5*d) +
(2*A*b*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + (2*B*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d)

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, 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 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 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]

Rubi steps

\begin{align*} \int (b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=A \int (b \sec (c+d x))^{5/2} \, dx+\frac{B \int (b \sec (c+d x))^{7/2} \, dx}{b}\\ &=\frac{2 A b (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 B (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{1}{3} \left (A b^2\right ) \int \sqrt{b \sec (c+d x)} \, dx+\frac{1}{5} (3 b B) \int (b \sec (c+d x))^{3/2} \, dx\\ &=\frac{6 b^2 B \sqrt{b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 A b (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 B (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac{1}{5} \left (3 b^3 B\right ) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx+\frac{1}{3} \left (A b^2 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 A b^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 d}+\frac{6 b^2 B \sqrt{b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 A b (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 B (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac{\left (3 b^3 B\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=-\frac{6 b^3 B 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 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 d}+\frac{6 b^2 B \sqrt{b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 A b (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 B (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.475985, size = 99, normalized size = 0.58 \[ \frac{(b \sec (c+d x))^{5/2} \left (20 A \cos ^{\frac{5}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+10 A \sin (2 (c+d x))+21 B \sin (c+d x)+9 B \sin (3 (c+d x))-36 B \cos ^{\frac{5}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sec[c + d*x])^(5/2)*(-36*B*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 20*A*Cos[c + d*x]^(5/2)*Elliptic
F[(c + d*x)/2, 2] + 21*B*Sin[c + d*x] + 10*A*Sin[2*(c + d*x)] + 9*B*Sin[3*(c + d*x)]))/(30*d)

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Maple [C]  time = 0.275, size = 518, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^2*(5*I*A*sin(d*x+c)*cos(d*x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+9*I*B*sin(d*x+c)*cos(d*x+c)^3*(1/(cos(d*x+c)+1
))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-9*I*B*sin(d*x+c)*cos(d*x+
c)^3*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*
A*sin(d*x+c)*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c
))/sin(d*x+c),I)+9*I*B*sin(d*x+c)*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elli
pticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-9*I*B*sin(d*x+c)*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-5*A*cos(d*x+c)^3-9*B*cos(d*x+c)^3+6*B*cos(d*x+c)^2+
5*A*cos(d*x+c)+3*B)*(b/cos(d*x+c))^(5/2)/sin(d*x+c)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*sec(d*x + c)^3 + A*b^2*sec(d*x + c)^2)*sqrt(b*sec(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((b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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