3.486 \(\int \sqrt{\cos (c+d x)} (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=66 \[ \frac{2 a (A+B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{d}+\frac{2 a (A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a B \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.194115, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2954, 2968, 3021, 2748, 2641, 2639} \[ \frac{2 a (A+B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a (A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a B \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

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

Rule 2954

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 2968

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

Rule 3021

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

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

Mathematica [C]  time = 6.074, size = 252, normalized size = 3.82 \[ \frac{a (\cos (c+d x)+1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{2 (A-B) \sec (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\sec ^2(c)} \sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )}}-4 (A+B) \sin (c) \sqrt{\csc ^2(c)} \cos (c+d x) \sqrt{\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+\frac{\csc (c) \left (-2 \sqrt{\sec ^2(c)} ((A-2 B) \cos (d x)+A \cos (2 c+d x))+3 (A-B) \sec (c) \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )+(A-B) \sec (c) \cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\sec ^2(c)}}\right )}{4 d \sqrt{\cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*((Csc[c]*(3*(A - B)*Cos[c - d*x - ArcTan[Tan[c]]]*Sec[c] + (A - B)*Co
s[c + d*x + ArcTan[Tan[c]]]*Sec[c] - 2*((A - 2*B)*Cos[d*x] + A*Cos[2*c + d*x])*Sqrt[Sec[c]^2]))/Sqrt[Sec[c]^2]
- 4*(A + B)*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}
, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] - (2*(A - B)*HypergeometricPFQ[{-1/2, -1/4}, {
3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sec[c]*Sin[d*x + ArcTan[Tan[c]]])/(Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan
[c]]]^2])))/(4*d*Sqrt[Cos[c + d*x]])

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Maple [B]  time = 2.013, size = 240, normalized size = 3.6 \begin{align*} -2\,{\frac{a \left ( 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 EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -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 ) +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 EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +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 ) -2\,B\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }{\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x)

[Out]

-2*a*(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))-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))+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))+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))-2*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 (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)*sqrt(cos(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c))*cos(d*x+c)**(1/2),x)

[Out]

a*(Integral(A*sqrt(cos(c + d*x)), x) + Integral(A*sqrt(cos(c + d*x))*sec(c + d*x), x) + Integral(B*sqrt(cos(c
+ d*x))*sec(c + d*x), x) + Integral(B*sqrt(cos(c + d*x))*sec(c + d*x)**2, x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)*sqrt(cos(d*x + c)), x)