3.98 \(\int \cos ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=100 \[ \frac{10 b^2 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{21 d}+\frac{2 b^5 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac{10 b^3 \sin (c+d x)}{21 d \sqrt{b \sec (c+d x)}} \]

[Out]

(10*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*d) + (2*b^5*Sin[c + d*x])/(7*d*
(b*Sec[c + d*x])^(5/2)) + (10*b^3*Sin[c + d*x])/(21*d*Sqrt[b*Sec[c + d*x]])

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Rubi [A]  time = 0.0714372, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {16, 3769, 3771, 2641} \[ \frac{2 b^5 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac{10 b^3 \sin (c+d x)}{21 d \sqrt{b \sec (c+d x)}}+\frac{10 b^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{21 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(b*Sec[c + d*x])^(5/2),x]

[Out]

(10*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*d) + (2*b^5*Sin[c + d*x])/(7*d*
(b*Sec[c + d*x])^(5/2)) + (10*b^3*Sin[c + d*x])/(21*d*Sqrt[b*Sec[c + d*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 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 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 ^6(c+d x) (b \sec (c+d x))^{5/2} \, dx &=b^6 \int \frac{1}{(b \sec (c+d x))^{7/2}} \, dx\\ &=\frac{2 b^5 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac{1}{7} \left (5 b^4\right ) \int \frac{1}{(b \sec (c+d x))^{3/2}} \, dx\\ &=\frac{2 b^5 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac{10 b^3 \sin (c+d x)}{21 d \sqrt{b \sec (c+d x)}}+\frac{1}{21} \left (5 b^2\right ) \int \sqrt{b \sec (c+d x)} \, dx\\ &=\frac{2 b^5 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac{10 b^3 \sin (c+d x)}{21 d \sqrt{b \sec (c+d x)}}+\frac{1}{21} \left (5 b^2 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{10 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)}}{21 d}+\frac{2 b^5 \sin (c+d x)}{7 d (b \sec (c+d x))^{5/2}}+\frac{10 b^3 \sin (c+d x)}{21 d \sqrt{b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0732741, size = 66, normalized size = 0.66 \[ \frac{b^2 \sqrt{b \sec (c+d x)} \left (40 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+26 \sin (2 (c+d x))+3 \sin (4 (c+d x))\right )}{84 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(b*Sec[c + d*x])^(5/2),x]

[Out]

(b^2*Sqrt[b*Sec[c + d*x]]*(40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 26*Sin[2*(c + d*x)] + 3*Sin[4*(c
+ d*x)]))/(84*d)

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Maple [C]  time = 0.188, size = 153, normalized size = 1.5 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{21\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}} \left ( -5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) +3\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+5\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-5\,\cos \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x)

[Out]

2/21/d*(cos(d*x+c)+1)^2*(b/cos(d*x+c))^(5/2)*(-1+cos(d*x+c))*cos(d*x+c)^2*(-5*I*(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)*sin(d*x+c)+3*cos(d*x+c)^4-3*cos(d*x+c)^
3+5*cos(d*x+c)^2-5*cos(d*x+c))/sin(d*x+c)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (d x + c\right )\right )^{\frac{5}{2}} \cos \left (d x + c\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^6, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sec \left (d x + c\right )} b^{2} \cos \left (d x + c\right )^{6} \sec \left (d x + c\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c))*b^2*cos(d*x + c)^6*sec(d*x + c)^2, 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)**6*(b*sec(d*x+c))**(5/2),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 )\right )^{\frac{5}{2}} \cos \left (d x + c\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^6, x)