3.207 \(\int \frac{1}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=178 \[ \frac{10 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}+\frac{10 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{7 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}-\frac{7 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{\sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

[Out]

(-7*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + (10*Sqrt[Cos[c + d*x]]*Elliptic
F[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (10*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - (7*Sin[c +
d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) - Sin[c + d*x]/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*
x])^2)

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Rubi [A]  time = 0.224789, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3817, 4020, 3787, 3769, 3771, 2641, 2639} \[ \frac{10 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{7 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}+\frac{10 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{7 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{\sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(-7*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + (10*Sqrt[Cos[c + d*x]]*Elliptic
F[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (10*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - (7*Sin[c +
d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) - Sin[c + d*x]/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*
x])^2)

Rule 3817

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[
e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc
[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d
, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4020

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

Rule 3787

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

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 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{1}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=-\frac{\sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{\int \frac{-\frac{9 a}{2}+\frac{5}{2} a \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{7 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{\sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{\int \frac{-15 a^2+\frac{21}{2} a^2 \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{7 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{\sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{7 \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}+\frac{5 \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{10 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{7 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{\sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{5 \int \sqrt{\sec (c+d x)} \, dx}{3 a^2}-\frac{\left (7 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac{7 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{10 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{7 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{\sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\left (5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^2}\\ &=-\frac{7 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{10 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{7 \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{\sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 1.7424, size = 257, normalized size = 1.44 \[ \frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (7 i e^{-\frac{1}{2} i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^3 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+80 \cos ^3\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+3 \sin \left (\frac{1}{2} (c+d x)\right )+10 \sin \left (\frac{3}{2} (c+d x)\right )+12 \sin \left (\frac{5}{2} (c+d x)\right )+\sin \left (\frac{7}{2} (c+d x)\right )-84 i \cos \left (\frac{1}{2} (c+d x)\right )-63 i \cos \left (\frac{3}{2} (c+d x)\right )-21 i \cos \left (\frac{5}{2} (c+d x)\right )\right )}{6 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(Cos[d*x] + I*Sin[d*x])*((-84*I)*Cos[(c + d*x)/2] - (63*I)*Cos[(3*(c + d*
x))/2] - (21*I)*Cos[(5*(c + d*x))/2] + 80*Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((
7*I)*(1 + E^(I*(c + d*x)))^3*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x
))])/E^((I/2)*(c + d*x)) + 3*Sin[(c + d*x)/2] + 10*Sin[(3*(c + d*x))/2] + 12*Sin[(5*(c + d*x))/2] + Sin[(7*(c
+ d*x))/2]))/(6*a^2*d*E^(I*d*x)*(1 + Sec[c + d*x])^2)

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Maple [A]  time = 1.421, size = 270, normalized size = 1.5 \begin{align*} -{\frac{1}{6\,{a}^{2}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 ( 16\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+12\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+20\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+42\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -48\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+21\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \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 ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3} \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(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x)

[Out]

-1/6/a^2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*cos(1/2*d*x+1/2*c)^8+12*cos(1/2*d*x+1/2*c
)^6+20*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*co
s(1/2*d*x+1/2*c)^3+42*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^3*Elli
pticE(cos(1/2*d*x+1/2*c),2^(1/2))-48*cos(1/2*d*x+1/2*c)^4+21*cos(1/2*d*x+1/2*c)^2-1)/(-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)^3/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 \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(3/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\sec \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{4} + 2 \, a^{2} \sec \left (d x + c\right )^{3} + a^{2} \sec \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(sqrt(sec(d*x + c))/(a^2*sec(d*x + c)^4 + 2*a^2*sec(d*x + c)^3 + a^2*sec(d*x + c)^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sec ^{\frac{7}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac{5}{2}}{\left (c + d x \right )} + \sec ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)**(3/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(1/(sec(c + d*x)**(7/2) + 2*sec(c + d*x)**(5/2) + sec(c + d*x)**(3/2)), x)/a**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(3/2)), x)