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

Optimal. Leaf size=70 \[ \frac{2 a (A+3 B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 a (A+B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a A \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.185505, antiderivative size = 70, 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, 3023, 2748, 2641, 2639} \[ \frac{2 a (A+3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a (A+B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a A \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(3/2)*(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 + 3*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*A*Sqrt[Cos[
c + d*x]]*Sin[c + d*x])/(3*d)

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

Mathematica [C]  time = 5.91764, size = 309, normalized size = 4.41 \[ \frac{a (\cos (c+d x)+1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )} \left (-4 (A+3 B) \sin (c) \sqrt{\csc ^2(c)} \sqrt{\sec ^2(c)} \cos (c+d x) \sqrt{\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+9 (A+B) \csc (c) \sec (c) \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )-12 A \cot (c) \sqrt{\sec ^2(c)} \cos (c+d x)+4 A \sqrt{\sec ^2(c)} \sin (c+d x) \cos (c+d x)+3 A \csc (c) \sec (c) \cos \left (c+\tan ^{-1}(\tan (c))+d x\right )-12 B \cot (c) \sqrt{\sec ^2(c)} \cos (c+d x)+3 B \csc (c) \sec (c) \cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )-6 (A+B) \sec (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )\right )}{12 d \sqrt{\sec ^2(c)} \sqrt{\cos (c+d x)} \sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(3/2)*(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*(-6*(A + B)*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[T
an[c]]]^2]*Sec[c]*Sin[d*x + ArcTan[Tan[c]]] + (9*(A + B)*Cos[c - d*x - ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 3*A*Cos
[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 3*B*Cos[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] - 12*A*Cos[c + d*x]
*Cot[c]*Sqrt[Sec[c]^2] - 12*B*Cos[c + d*x]*Cot[c]*Sqrt[Sec[c]^2] - 4*(A + 3*B)*Cos[c + d*x]*Sqrt[Cos[d*x - Arc
Tan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Sec[c]^2
]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + 4*A*Cos[c + d*x]*Sqrt[Sec[c]^2]*Sin[c + d*x])*Sqrt[Sin[d*x + ArcTan[Tan[c
]]]^2]))/(12*d*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])

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Maple [B]  time = 2.037, size = 321, normalized size = 4.6 \begin{align*} -{\frac{2\,a}{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\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+A\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\,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 ) -2\,A \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/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 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 ) \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+a*sec(d*x+c))*(A+B*sec(d*x+c)),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)*a*(4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+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*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))-2*A*sin(1/2*d*x+
1/2*c)^2*cos(1/2*d*x+1/2*c)+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)*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)))/(-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{\left (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )} \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+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)*cos(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 (B a \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} +{\left (A + B\right )} a \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A a \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+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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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 )} \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+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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