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

Optimal. Leaf size=194 \[ \frac {2 b \left (21 a^2+5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 a \left (5 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2+9 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {32 a b^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b^2 \sin (c+d x) (a+b \sec (c+d x))}{7 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

-2/5*a*(5*a^2+9*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2
/21*b*(21*a^2+5*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+3
2/35*a*b^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/21*b*(21*a^2+5*b^2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/7*b^2*(a+b*sec(
d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/5*a*(5*a^2+9*b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4264, 3842, 4047, 3768, 3771, 2641, 4046, 2639} \[ \frac {2 b \left (21 a^2+5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 a \left (5 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2+9 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {32 a b^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b^2 \sin (c+d x) (a+b \sec (c+d x))}{7 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(5*a^2 + 9*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*b*(21*a^2 + 5*b^2)*EllipticF[(c + d*x)/2, 2])/(21*
d) + (32*a*b^2*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2)) + (2*b*(21*a^2 + 5*b^2)*Sin[c + d*x])/(21*d*Cos[c + d*x
]^(3/2)) + (2*a*(5*a^2 + 9*b^2)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b^2*(a + b*Sec[c + d*x])*Sin[c + d
*x])/(7*d*Cos[c + d*x]^(5/2))

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]

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 3768

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3842

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[1/(d*(m + n - 1)), In
t[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2) +
3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Integ
erQ[m])

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4264

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

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^3}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx\\ &=\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a \left (7 a^2+3 b^2\right )+\frac {1}{2} b \left (21 a^2+5 b^2\right ) \sec (c+d x)+8 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a \left (7 a^2+3 b^2\right )+8 a b^2 \sec ^2(c+d x)\right ) \, dx+\frac {1}{7} \left (b \left (21 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx\\ &=\frac {32 a b^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{21} \left (b \left (21 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (a \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {32 a b^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2+9 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{21} \left (b \left (21 a^2+5 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (a \left (5 a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b \left (21 a^2+5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {32 a b^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2+9 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} \left (a \left (5 a^2+9 b^2\right )\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 a \left (5 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {32 a b^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2+9 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.94, size = 177, normalized size = 0.91 \[ \frac {210 a^3 \sin (c+d x) \cos ^2(c+d x)+10 b \left (21 a^2+5 b^2\right ) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-42 a \left (5 a^2+9 b^2\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+105 a^2 b \sin (2 (c+d x))+126 a b^2 \sin (c+d x)+378 a b^2 \sin (c+d x) \cos ^2(c+d x)+25 b^3 \sin (2 (c+d x))+30 b^3 \tan (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-42*a*(5*a^2 + 9*b^2)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 10*b*(21*a^2 + 5*b^2)*Cos[c + d*x]^(5/2)
*EllipticF[(c + d*x)/2, 2] + 126*a*b^2*Sin[c + d*x] + 210*a^3*Cos[c + d*x]^2*Sin[c + d*x] + 378*a*b^2*Cos[c +
d*x]^2*Sin[c + d*x] + 105*a^2*b*Sin[2*(c + d*x)] + 25*b^3*Sin[2*(c + d*x)] + 30*b^3*Tan[c + d*x])/(105*d*Cos[c
+ d*x]^(5/2))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 12.60, size = 847, normalized size = 4.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3/cos(d*x+c)^(3/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^3*(-1/56*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+cos(1/2*d*x+1/2*c)^2)^4-5/42*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+cos(1/2*d*x+1/2*c)^2)^2+5/21*(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))
)+6*a^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)/(-1/2+cos(1/2*d*x+1/2*
c)^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)))-6/5*b^2*a/(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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-
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)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(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))-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)+2*a^3*(-(-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)^(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*(-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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [B]  time = 2.24, size = 147, normalized size = 0.76 \[ \frac {\frac {2\,b^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7}+2\,a^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+\frac {6\,a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5}+2\,a^2\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^3/cos(c + d*x)^(3/2),x)

[Out]

((2*b^3*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/7 + 2*a^3*cos(c + d*x)^3*sin(c + d*x)*hyper
geom([-1/4, 1/2], 3/4, cos(c + d*x)^2) + (6*a*b^2*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c
+ d*x)^2))/5 + 2*a^2*b*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*
x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{3}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(c + d*x))**3/cos(c + d*x)**(3/2), x)

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