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

Optimal. Leaf size=132 \[ \frac{2 a (A+B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{2 a (5 A+3 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (A+B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a (5 A+3 B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

[Out]

(-2*a*(5*A + 3*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(A + B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*B*Si
n[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*a*(A + B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*a*(5*A + 3*B)*
Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.232189, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {2954, 2968, 3021, 2748, 2636, 2641, 2639} \[ \frac{2 a (A+B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 a (5 A+3 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (A+B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a (5 A+3 B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(5*A + 3*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(A + B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*B*Si
n[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*a*(A + B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*a*(5*A + 3*B)*
Sin[c + d*x])/(5*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 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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 \frac{(a+a \sec (c+d x)) (A+B \sec (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\int \frac{(a+a \cos (c+d x)) (B+A \cos (c+d x))}{\cos ^{\frac{7}{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{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{\frac{5}{2} a (A+B)+\frac{1}{2} a (5 A+3 B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+(a (A+B)) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx+\frac{1}{5} (a (5 A+3 B)) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (A+B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a (5 A+3 B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{1}{3} (a (A+B)) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} (a (5 A+3 B)) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 a (5 A+3 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (A+B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a B \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (A+B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a (5 A+3 B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.38727, size = 865, normalized size = 6.55 \[ a \left (\sqrt{\cos (c+d x)} (\cos (c+d x)+1) \left (\frac{B \sec (c) \sin (d x) \sec ^3(c+d x)}{5 d}+\frac{\sec (c) (3 B \sin (c)+5 A \sin (d x)+5 B \sin (d x)) \sec ^2(c+d x)}{15 d}+\frac{\sec (c) (5 A \sin (c)+5 B \sin (c)+15 A \sin (d x)+9 B \sin (d x)) \sec (c+d x)}{15 d}+\frac{(5 A+3 B) \csc (c) \sec (c)}{5 d}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )+\frac{A (\cos (c+d x)+1) \csc (c) \left (\frac{\text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt{\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}} \sqrt{\tan ^2(c)+1}}-\frac{\frac{2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac{\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{\tan ^2(c)+1}}}{\sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}}}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}+\frac{3 B (\cos (c+d x)+1) \csc (c) \left (\frac{\text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt{\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}} \sqrt{\tan ^2(c)+1}}-\frac{\frac{2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac{\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{\tan ^2(c)+1}}}{\sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}}}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{10 d}-\frac{A (\cos (c+d x)+1) \csc (c) \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 ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d \sqrt{\cot ^2(c)+1}}-\frac{B (\cos (c+d x)+1) \csc (c) \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 ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d \sqrt{\cot ^2(c)+1}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(((5*A + 3*B)*Csc[c]*Sec[c])/(5*d) + (B*Sec[c]*S
ec[c + d*x]^3*Sin[d*x])/(5*d) + (Sec[c]*Sec[c + d*x]^2*(3*B*Sin[c] + 5*A*Sin[d*x] + 5*B*Sin[d*x]))/(15*d) + (S
ec[c]*Sec[c + d*x]*(5*A*Sin[c] + 5*B*Sin[c] + 15*A*Sin[d*x] + 9*B*Sin[d*x]))/(15*d)) - (A*(1 + Cos[c + d*x])*C
sc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[
Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt
[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (B*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/
4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x -
ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c
]]]])/(3*d*Sqrt[1 + Cot[c]^2]) + (A*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1/2,
-1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]
]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan
[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[
1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d) + (3*B
*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[
c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c
]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]
]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2
))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(10*d))

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Maple [B]  time = 5.993, size = 661, normalized size = 5. \begin{align*} \text{result too large to display} \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)^(3/2),x)

[Out]

-4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*((1/2*A+1/2*B)*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-
2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2
*c),2^(1/2)))-1/10*B/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2
*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*
d*x+1/2*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/
2)-8*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)+1/2*A*(-(si
n(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)+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*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/(2*sin(1/2*d*x+1/2*c)^2-1))/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(-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+a*sec(d*x+c))*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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

[Out]

integral((B*a*sec(d*x + c)^2 + (A + B)*a*sec(d*x + c) + A*a)/cos(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+a*sec(d*x+c))*(A+B*sec(d*x+c))/cos(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 \frac{{\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((a+a*sec(d*x+c))*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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