3.99 \(\int \cos ^7(c+d x) (b \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=100 \[ \frac{2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac{14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac{14 b^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}} \]

[Out]

(14*b^3*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b^6*Sin[c + d*x])/(9*d*
(b*Sec[c + d*x])^(7/2)) + (14*b^4*Sin[c + d*x])/(45*d*(b*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.0740449, 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, 2639} \[ \frac{2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac{14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac{14 b^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(b*Sec[c + d*x])^(5/2),x]

[Out]

(14*b^3*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b^6*Sin[c + d*x])/(9*d*
(b*Sec[c + d*x])^(7/2)) + (14*b^4*Sin[c + d*x])/(45*d*(b*Sec[c + d*x])^(3/2))

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 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 \cos ^7(c+d x) (b \sec (c+d x))^{5/2} \, dx &=b^7 \int \frac{1}{(b \sec (c+d x))^{9/2}} \, dx\\ &=\frac{2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac{1}{9} \left (7 b^5\right ) \int \frac{1}{(b \sec (c+d x))^{5/2}} \, dx\\ &=\frac{2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac{14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac{1}{15} \left (7 b^3\right ) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx\\ &=\frac{2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac{14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}+\frac{\left (7 b^3\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=\frac{14 b^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 b^6 \sin (c+d x)}{9 d (b \sec (c+d x))^{7/2}}+\frac{14 b^4 \sin (c+d x)}{45 d (b \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.166259, size = 74, normalized size = 0.74 \[ \frac{b^2 \sqrt{b \sec (c+d x)} \left ((33 \sin (c+d x)+5 \sin (3 (c+d x))) \cos ^2(c+d x)+84 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{90 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(b*Sec[c + d*x])^(5/2),x]

[Out]

(b^2*Sqrt[b*Sec[c + d*x]]*(84*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + Cos[c + d*x]^2*(33*Sin[c + d*x] +
5*Sin[3*(c + d*x)])))/(90*d)

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Maple [C]  time = 0.186, size = 333, normalized size = 3.3 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{45\,d\sin \left ( dx+c \right ) } \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}} \left ( 21\,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 ) \cos \left ( dx+c \right ) -21\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) -5\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}+21\,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 ) -21\,i\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) -2\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-14\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,\cos \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(b*sec(d*x+c))^(5/2),x)

[Out]

2/45/d*(b/cos(d*x+c))^(5/2)*cos(d*x+c)^2*(21*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elli
pticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)-21*I*cos(d*x+c)*sin(d*x+c)*(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)-5*cos(d*x+c)^6+21*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)-21*I*sin(d
*x+c)*(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)-2*c
os(d*x+c)^4-14*cos(d*x+c)^2+21*cos(d*x+c))/sin(d*x+c)

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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 )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^7, 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 )^{7} \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)^7*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c))*b^2*cos(d*x + c)^7*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)**7*(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 )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(5/2)*cos(d*x + c)^7, x)