3.704 \(\int \frac{(a+b \cos (c+d x))^2}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=200 \[ \frac{2 \left (9 a^2+7 b^2\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (9 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a b \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{20 a b \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{20 a b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 b^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

[Out]

(2*(9*a^2 + 7*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (20*a*b*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2))
+ (4*a*b*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*(9*a^2 + 7*b^2)*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2))
+ (20*a*b*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.174146, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3238, 3788, 3769, 3771, 2641, 4045, 2639} \[ \frac{2 \left (9 a^2+7 b^2\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (9 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a b \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{20 a b \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{20 a b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 b^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(9*a^2 + 7*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (20*a*b*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2))
+ (4*a*b*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*(9*a^2 + 7*b^2)*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2))
+ (20*a*b*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, 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 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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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 \frac{(a+b \cos (c+d x))^2}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{(b+a \sec (c+d x))^2}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\\ &=(2 a b) \int \frac{1}{\sec ^{\frac{7}{2}}(c+d x)} \, dx+\int \frac{b^2+a^2 \sec ^2(c+d x)}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{4 a b \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{7} (10 a b) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx-\frac{1}{9} \left (-9 a^2-7 b^2\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{4 a b \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (9 a^2+7 b^2\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{20 a b \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{1}{21} (10 a b) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{15} \left (-9 a^2-7 b^2\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 b^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{4 a b \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (9 a^2+7 b^2\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{20 a b \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{1}{21} \left (10 a b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{15} \left (\left (-9 a^2-7 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (9 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{20 a b \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 b^2 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{4 a b \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (9 a^2+7 b^2\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{20 a b \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.02458, size = 135, normalized size = 0.68 \[ \frac{\sqrt{\sec (c+d x)} \left (\sin (2 (c+d x)) \left (7 \left (36 a^2+43 b^2\right ) \cos (c+d x)+5 b (36 a \cos (2 (c+d x))+156 a+7 b \cos (3 (c+d x)))\right )+168 \left (9 a^2+7 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+1200 a b \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(168*(9*a^2 + 7*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 1200*a*b*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2] + (7*(36*a^2 + 43*b^2)*Cos[c + d*x] + 5*b*(156*a + 36*a*Cos[2*(c + d*x)] + 7*b
*Cos[3*(c + d*x)]))*Sin[2*(c + d*x)]))/(1260*d)

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Maple [A]  time = 3.279, size = 398, normalized size = 2. \begin{align*} -{\frac{2}{315\,d}\sqrt{ \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}} \left ( -1120\,{b}^{2} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}\cos \left ( 1/2\,dx+c/2 \right ) + \left ( 1440\,ab+2240\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{8}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -504\,{a}^{2}-2160\,ab-2072\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 504\,{a}^{2}+1680\,ab+952\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -126\,{a}^{2}-480\,ab-168\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +150\,\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 ) ab-189\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{2}-147\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){b}^{2} \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*b^2*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/
2*c)+(1440*a*b+2240*b^2)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-504*a^2-2160*a*b-2072*b^2)*sin(1/2*d*x+1/2*
c)^6*cos(1/2*d*x+1/2*c)+(504*a^2+1680*a*b+952*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-126*a^2-480*a*b-1
68*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+150*(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))*a*b-189*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-147*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/
2*c)/(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 \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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