3.52 \(\int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac{2 C \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{3 b^4 d}+\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 C \sin (c+d x)}{3 b^3 d \sqrt{b \sec (c+d x)}} \]

[Out]

(6*B*EllipticE[(c + d*x)/2, 2])/(5*b^3*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*C*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^4*d) + (2*B*Sin[c + d*x])/(5*b^2*d*(b*Sec[c + d*x])^(3/2))
+ (2*C*Sin[c + d*x])/(3*b^3*d*Sqrt[b*Sec[c + d*x]])

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Rubi [A]  time = 0.134684, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {4047, 3769, 3771, 2639, 12, 16, 2641} \[ \frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 C \sin (c+d x)}{3 b^3 d \sqrt{b \sec (c+d x)}}+\frac{2 C \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*B*EllipticE[(c + d*x)/2, 2])/(5*b^3*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*C*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^4*d) + (2*B*Sin[c + d*x])/(5*b^2*d*(b*Sec[c + d*x])^(3/2))
+ (2*C*Sin[c + d*x])/(3*b^3*d*Sqrt[b*Sec[c + d*x]])

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx &=\frac{B \int \frac{1}{(b \sec (c+d x))^{5/2}} \, dx}{b}+\int \frac{C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx\\ &=\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{(3 B) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx}{5 b^3}+C \int \frac{\sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx\\ &=\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{C \int \frac{1}{(b \sec (c+d x))^{3/2}} \, dx}{b^2}+\frac{(3 B) \int \sqrt{\cos (c+d x)} \, dx}{5 b^3 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{2 C \sin (c+d x)}{3 b^3 d \sqrt{b \sec (c+d x)}}+\frac{C \int \sqrt{b \sec (c+d x)} \, dx}{3 b^4}\\ &=\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{2 C \sin (c+d x)}{3 b^3 d \sqrt{b \sec (c+d x)}}+\frac{\left (C \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^4}\\ &=\frac{6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 C \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 b^4 d}+\frac{2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac{2 C \sin (c+d x)}{3 b^3 d \sqrt{b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.531016, size = 91, normalized size = 0.62 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)} \left (5 C \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (c+d x) \sqrt{\cos (c+d x)} (3 B \cos (c+d x)+5 C)+9 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]*(9*B*EllipticE[(c + d*x)/2, 2] + 5*C*EllipticF[(c + d*x)/2, 2] + Sq
rt[Cos[c + d*x]]*(5*C + 3*B*Cos[c + d*x])*Sin[c + d*x]))/(15*b^4*d)

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Maple [C]  time = 0.25, size = 482, 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((B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x)

[Out]

-2/15/d*(9*I*B*sin(d*x+c)*cos(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)-9*I*B*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-5*I*C*(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)*cos(d*x+c)*sin(d*x+c)+9*I*B*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))
/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-9*I*B*EllipticF(I*(-1+cos(d*x+c))/si
n(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-5*I*C*sin(d*x+c)*EllipticF(I
*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+3*B*cos(d*x+c)^4+5*C
*cos(d*x+c)^3+6*B*cos(d*x+c)^2-9*B*cos(d*x+c)-5*C*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^4/(b/cos(d*x+c))^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right ) + B\right )} \sqrt{b \sec \left (d x + c\right )}}{b^{4} \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c) + B)*sqrt(b*sec(d*x + c))/(b^4*sec(d*x + c)^3), 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)+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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