3.805 \(\int \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=72 \[ \frac{2 \left (a^2+3 b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 a^2 \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{4 a b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

[Out]

(4*a*b*EllipticE[(c + d*x)/2, 2])/d + (2*(a^2 + 3*b^2)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*Sqrt[Cos[c +
d*x]]*Sin[c + d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.139091, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4264, 3788, 3771, 2639, 4045, 2641} \[ \frac{2 \left (a^2+3 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{4 a b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2,x]

[Out]

(4*a*b*EllipticE[(c + d*x)/2, 2])/d + (2*(a^2 + 3*b^2)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*Sqrt[Cos[c +
d*x]]*Sin[c + d*x])/(3*d)

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

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

Mathematica [A]  time = 0.155578, size = 64, normalized size = 0.89 \[ \frac{2 \left (\left (a^2+3 b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+a^2 \sin (c+d x) \sqrt{\cos (c+d x)}+6 a b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2,x]

[Out]

(2*(6*a*b*EllipticE[(c + d*x)/2, 2] + (a^2 + 3*b^2)*EllipticF[(c + d*x)/2, 2] + a^2*Sqrt[Cos[c + d*x]]*Sin[c +
d*x]))/(3*d)

________________________________________________________________________________________

Maple [B]  time = 1.731, size = 283, normalized size = 3.9 \begin{align*} -{\frac{2}{3\,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 ( 4\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+{a}^{2}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +3\,{b}^{2}\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 EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -6\,\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 ) ab-2\,{a}^{2} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \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(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2,x)

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+a^
2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(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))+3*b^2*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(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))-6*(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*b-2*a^2*sin(1/2*d*
x+1/2*c)^2*cos(1/2*d*x+1/2*c))/(-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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^2*cos(d*x + c)^(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)*sec(d*x + c)^2 + 2*a*b*cos(d*x + c)*sec(d*x + c) + a^2*cos(d*x + c))*sqrt(cos(d*x +
c)), x)

________________________________________________________________________________________

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)**(3/2)*(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^2*cos(d*x + c)^(3/2), x)