3.1454 \(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=147 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a B+3 A b+b C)}{3 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (b B-a (A-C))}{d}+\frac{2 a A \sin (c+d x) \sqrt{\sec (c+d x)}}{d}+\frac{2 b C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}} \]

[Out]

(2*(b*B - a*(A - C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*(3*A*b + 3*a*B +
b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b*C*Sin[c + d*x])/(3*d*Sqrt[S
ec[c + d*x]]) + (2*a*A*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

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Rubi [A]  time = 0.301595, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4221, 3031, 3023, 2748, 2641, 2639} \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a B+3 A b+b C)}{3 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (b B-a (A-C))}{d}+\frac{2 a A \sin (c+d x) \sqrt{\sec (c+d x)}}{d}+\frac{2 b C \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*B - a*(A - C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*(3*A*b + 3*a*B +
b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b*C*Sin[c + d*x])/(3*d*Sqrt[S
ec[c + d*x]]) + (2*a*A*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

Rule 4221

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

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, 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]

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

Mathematica [A]  time = 0.713809, size = 109, normalized size = 0.74 \[ \frac{\sqrt{\sec (c+d x)} \left (2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a B+3 A b+b C)+6 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (a (C-A)+b B)+2 \sin (c+d x) (3 a A+b C \cos (c+d x))\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(6*(b*B + a*(-A + C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 2*(3*A*b + 3*a*B + b*
C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(3*a*A + b*C*Cos[c + d*x])*Sin[c + d*x]))/(3*d)

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Maple [B]  time = 1.436, size = 388, normalized size = 2.6 \begin{align*} -{\frac{2}{3\,d} \left ( 4\,Cb\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+3\,Ab\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 ) +3\,A\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-6\,Aa\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+3\,Ba\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 ) -3\,B\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+Cb\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\,C\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\,Cb\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) \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))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x)

[Out]

-2/3*(4*C*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*A*b*(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*A*(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-6*A*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+3*B*a*(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*(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+C*b*(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*C*(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*C*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^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{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b*cos(d*x + c)^3 + (C*a + B*b)*cos(d*x + c)^2 + A*a + (B*a + A*b)*cos(d*x + c))*sec(d*x + c)^(3/2)
, 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))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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