3.491 \(\int \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=126 \[ \frac{4 a^2 (A+2 B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{4 a^2 (4 A+5 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 (7 A+5 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d} \]

[Out]

(4*a^2*(4*A + 5*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(A + 2*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a
^2*(7*A + 5*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])*Sin[
c + d*x])/(5*d)

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Rubi [A]  time = 0.347456, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2954, 2976, 2968, 3023, 2748, 2641, 2639} \[ \frac{4 a^2 (A+2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^2 (4 A+5 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 (7 A+5 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]

[Out]

(4*a^2*(4*A + 5*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(A + 2*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a
^2*(7*A + 5*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])*Sin[
c + d*x])/(5*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 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

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{5}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\int \frac{(a+a \cos (c+d x))^2 (B+A \cos (c+d x))}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{2}{5} \int \frac{(a+a \cos (c+d x)) \left (\frac{1}{2} a (A+5 B)+\frac{1}{2} a (7 A+5 B) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{2}{5} \int \frac{\frac{1}{2} a^2 (A+5 B)+\left (\frac{1}{2} a^2 (A+5 B)+\frac{1}{2} a^2 (7 A+5 B)\right ) \cos (c+d x)+\frac{1}{2} a^2 (7 A+5 B) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a^2 (7 A+5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{4}{15} \int \frac{\frac{5}{2} a^2 (A+2 B)+\frac{3}{2} a^2 (4 A+5 B) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a^2 (7 A+5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{1}{3} \left (2 a^2 (A+2 B)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (2 a^2 (4 A+5 B)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^2 (4 A+5 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^2 (A+2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 (7 A+5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 6.30096, size = 994, normalized size = 7.89 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]

[Out]

(Cos[c + d*x]^(7/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*(-((4*A + 5*B)*Cot[c])/(5
*d) + ((2*A + B)*Cos[d*x]*Sin[c])/(6*d) + (A*Cos[2*d*x]*Sin[2*c])/(20*d) + ((2*A + B)*Cos[c]*Sin[d*x])/(6*d) +
(A*Cos[2*c]*Sin[2*d*x])/(20*d)))/(B + A*Cos[c + d*x]) - (A*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}
, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*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*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (2*B*Cos[c + d*x]^3
*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c +
d*x])^2*(A + B*Sec[c + d*x])*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*(B + A*Cos[c + d*x])*Sqrt[
1 + Cot[c]^2]) - (2*A*Cos[c + d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*(
(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]]))/(5*d*(B + A*Cos[c + d*x])) - (B*Cos[c + d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^
2*(A + B*Sec[c + d*x])*((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*(B + A*Cos[c + d*x]))

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Maple [B]  time = 1.776, size = 357, normalized size = 2.8 \begin{align*} -{\frac{4\,{a}^{2}}{15\,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 ( -12\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 32\,A+10\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -13\,A-5\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +5\,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 ) -12\,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 ) +10\,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 ) -15\,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)^(5/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)

[Out]

-4/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-12*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^6+(32*A+10*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-13*A-5*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*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))-12*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))+10*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))-15*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(-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)^(5/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{3} +{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A a^{2} \cos \left (d x + c\right )^{2}\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)^(5/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*a^2*cos(d*x + c)^2*sec(d*x + c)^3 + (A + 2*B)*a^2*cos(d*x + c)^2*sec(d*x + c)^2 + (2*A + B)*a^2*co
s(d*x + c)^2*sec(d*x + c) + A*a^2*cos(d*x + c)^2)*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)**(5/2)*(a+a*sec(d*x+c))**2*(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 )}^{2} \cos \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*cos(d*x + c)^(5/2), x)