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

Optimal. Leaf size=135 \[ -\frac {2 \left (5 a^2+3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2+3 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {4 a b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

-2/5*(5*a^2+3*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+4/3
*a*b*(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+2/5*b^2*sin(d*x+c
)/d/cos(d*x+c)^(5/2)+4/3*a*b*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(5*a^2+3*b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4264, 3788, 3768, 3771, 2641, 4046, 2639} \[ -\frac {2 \left (5 a^2+3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2+3 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {4 a b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*a^2 + 3*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a*b*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*b^2*Sin[c
+ d*x])/(5*d*Cos[c + d*x]^(5/2)) + (4*a*b*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(5*a^2 + 3*b^2)*Sin[c +
d*x])/(5*d*Sqrt[Cos[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]

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 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

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

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Mathematica [A]  time = 0.44, size = 124, normalized size = 0.92 \[ \frac {-6 \left (5 a^2+3 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+15 a^2 \sin (2 (c+d x))+20 a b \sin (c+d x)+20 a b \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+9 b^2 \sin (2 (c+d x))+6 b^2 \tan (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(5*a^2 + 3*b^2)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 20*a*b*Cos[c + d*x]^(3/2)*EllipticF[(c + d*
x)/2, 2] + 20*a*b*Sin[c + d*x] + 15*a^2*Sin[2*(c + d*x)] + 9*b^2*Sin[2*(c + d*x)] + 6*b^2*Tan[c + d*x])/(15*d*
Cos[c + d*x]^(3/2))

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fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 10.71, size = 660, normalized size = 4.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^2/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*a^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)^(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)-2/5*b^2/(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)+4*a*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))))/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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.67, size = 113, normalized size = 0.84 \[ \frac {6\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*b^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeo
m([-1/4, 1/2], 3/4, cos(c + d*x)^2) + 20*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x
)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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