3.155 \(\int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=128 \[ \frac{2 a^4 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}+\frac{2 a^3 \tan ^3(c+d x)}{d (a \sec (c+d x)+a)^{3/2}}-\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a^2 \tan (c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(-2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*Tan[c + d*x])/(d*Sqrt[a + a*Se
c[c + d*x]]) + (2*a^3*Tan[c + d*x]^3)/(d*(a + a*Sec[c + d*x])^(3/2)) + (2*a^4*Tan[c + d*x]^5)/(5*d*(a + a*Sec[
c + d*x])^(5/2))

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Rubi [A]  time = 0.0927509, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3887, 461, 203} \[ \frac{2 a^4 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}+\frac{2 a^3 \tan ^3(c+d x)}{d (a \sec (c+d x)+a)^{3/2}}-\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a^2 \tan (c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*Tan[c + d*x])/(d*Sqrt[a + a*Se
c[c + d*x]]) + (2*a^3*Tan[c + d*x]^3)/(d*(a + a*Sec[c + d*x])^(3/2)) + (2*a^4*Tan[c + d*x]^5)/(5*d*(a + a*Sec[
c + d*x])^(5/2))

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx &=-\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (2+a x^2\right )^2}{1+a x^2} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{a}+3 x^2+a x^4-\frac{1}{a \left (1+a x^2\right )}\right ) \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{2 a^2 \tan (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 \tan ^3(c+d x)}{d (a+a \sec (c+d x))^{3/2}}+\frac{2 a^4 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}+\frac{2 a^2 \tan (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 \tan ^3(c+d x)}{d (a+a \sec (c+d x))^{3/2}}+\frac{2 a^4 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 5.53601, size = 97, normalized size = 0.76 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (5 \sin \left (\frac{3}{2} (c+d x)\right )+\sin \left (\frac{5}{2} (c+d x)\right )-10 \sqrt{2} \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{5}{2}}(c+d x)\right )}{10 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[(c + d*x)/2]*Sec[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(-10*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Co
s[c + d*x]^(5/2) + 5*Sin[(3*(c + d*x))/2] + Sin[(5*(c + d*x))/2]))/(10*d)

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Maple [B]  time = 0.183, size = 300, normalized size = 2.3 \begin{align*}{\frac{a}{20\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 5\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sqrt{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+10\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sqrt{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +5\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sin \left ( dx+c \right ) -8\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-16\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+16\,\cos \left ( dx+c \right ) +8 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(5*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d
*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2+10*arctanh(1/2*2^(1/2)*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)*sin(d
*x+c)*cos(d*x+c)+5*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)-8*cos(d*x+c)^3-16*cos(d*x+c)^2+16*cos(d*x+c)+8)/sin(d*x+c)/cos(d*
x+c)^2

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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))^(3/2)*tan(d*x+c)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.74454, size = 832, normalized size = 6.5 \begin{align*} \left [\frac{5 \,{\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \,{\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac{2 \,{\left (5 \,{\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) +{\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{5 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/5*(5*(a*cos(d*x + c)^3 + a*cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(a*cos(d*x + c)^
2 + 3*a*cos(d*x + c) + a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x
+ c)^2), 2/5*(5*(a*cos(d*x + c)^3 + a*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*c
os(d*x + c)/(sqrt(a)*sin(d*x + c))) + (a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + a)*sqrt((a*cos(d*x + c) + a)/cos(
d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError