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

Optimal. Leaf size=100 \[ \frac{2 b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac{10 b^4 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{10 b^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{b \cos (c+d x)}} \]

[Out]

(10*b^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (2*b^6*Sin[c + d*x])/(7*d*
(b*Cos[c + d*x])^(7/2)) + (10*b^4*Sin[c + d*x])/(21*d*(b*Cos[c + d*x])^(3/2))

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Rubi [A]  time = 0.0741106, 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, 2636, 2642, 2641} \[ \frac{2 b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac{10 b^4 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{10 b^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*b^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (2*b^6*Sin[c + d*x])/(7*d*
(b*Cos[c + d*x])^(7/2)) + (10*b^4*Sin[c + d*x])/(21*d*(b*Cos[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 2636

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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 (b \cos (c+d x))^{5/2} \sec ^7(c+d x) \, dx &=b^7 \int \frac{1}{(b \cos (c+d x))^{9/2}} \, dx\\ &=\frac{2 b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac{1}{7} \left (5 b^5\right ) \int \frac{1}{(b \cos (c+d x))^{5/2}} \, dx\\ &=\frac{2 b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac{10 b^4 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{1}{21} \left (5 b^3\right ) \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx\\ &=\frac{2 b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac{10 b^4 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac{\left (5 b^3 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 \sqrt{b \cos (c+d x)}}\\ &=\frac{10 b^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{b \cos (c+d x)}}+\frac{2 b^6 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac{10 b^4 \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.177607, size = 69, normalized size = 0.69 \[ \frac{\sec ^5(c+d x) (b \cos (c+d x))^{5/2} \left (5 \sin (2 (c+d x))+6 \tan (c+d x)+10 \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 1.819, size = 398, normalized size = 4. \begin{align*} -{\frac{2\,{b}^{3}}{21\,d} \left ( -40\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+60\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) -30\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +5\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -16\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}{\frac{1}{\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) }}} \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-3} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*(-40*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*sin(1/2*d*x+1/2*c)^6+60*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-40*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-30*(2*sin(1/2*d*x+1/2*c)^2-1)^(
1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+40*sin(1/2*d*x+1/
2*c)^4*cos(1/2*d*x+1/2*c)+5*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))-16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*b^3*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/
2*c)^2)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^3/sin(1/2*d*
x+1/2*c)/(b*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}} \sec \left (d x + c\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*cos(d*x + c))^(5/2)*sec(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 \cos \left (d x + c\right )} b^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{7}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}} \sec \left (d x + c\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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