3.356 \(\int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=141 \[ \frac {2 \sqrt {2} a^3 \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac {4 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}} \]

[Out]

2*a^3*arctan(1/2*(d^(1/2)-d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/d^(7/2)/f-4*a^3/d^3/f/(d*t
an(f*x+e))^(1/2)-8/5*a^3/d^2/f/(d*tan(f*x+e))^(3/2)-2/5*(a^3+a^3*tan(f*x+e))/d/f/(d*tan(f*x+e))^(5/2)

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Rubi [A]  time = 0.22, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3565, 3628, 3529, 3532, 205} \[ \frac {2 \sqrt {2} a^3 \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac {4 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]

[Out]

(2*Sqrt[2]*a^3*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(7/2)*f) - (8*a^3)/
(5*d^2*f*(d*Tan[e + f*x])^(3/2)) - (4*a^3)/(d^3*f*Sqrt[d*Tan[e + f*x]]) - (2*(a^3 + a^3*Tan[e + f*x]))/(5*d*f*
(d*Tan[e + f*x])^(5/2))

Rule 205

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

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3532

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*d^2)/f,
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 \int \frac {6 a^3 d^2+5 a^3 d^2 \tan (e+f x)+a^3 d^2 \tan ^2(e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{5 d^3}\\ &=-\frac {8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 \int \frac {5 a^3 d^3-5 a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{5 d^5}\\ &=-\frac {8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {4 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 \int \frac {-5 a^3 d^4-5 a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{5 d^7}\\ &=-\frac {8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {4 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\left (20 a^6 d\right ) \operatorname {Subst}\left (\int \frac {1}{50 a^6 d^8+d x^2} \, dx,x,\frac {-5 a^3 d^4+5 a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=\frac {2 \sqrt {2} a^3 \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac {8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {4 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 2.18, size = 271, normalized size = 1.92 \[ -\frac {a^3 (\cot (e+f x)+1)^3 \left (8 \sin (e+f x) \cos ^2(e+f x) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\tan ^2(e+f x)\right )+5 \left (24 \sin ^3(e+f x) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\tan ^2(e+f x)\right )+8 \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(e+f x)\right )+\sqrt {2} \cos ^3(e+f x) \tan ^{\frac {7}{2}}(e+f x) \left (2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-2 \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+\log \left (\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1\right )-\log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )\right )\right )\right )}{20 d^3 f \sqrt {d \tan (e+f x)} (\sin (e+f x)+\cos (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]

[Out]

-1/20*(a^3*(1 + Cot[e + f*x])^3*(8*Cos[e + f*x]^2*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[e + f*x]^2]*Sin[e + f*
x] + 5*(8*Cos[e + f*x]*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[e + f*x]^2]*Sin[e + f*x]^2 + 24*Hypergeometric2F1[
-1/4, 1, 3/4, -Tan[e + f*x]^2]*Sin[e + f*x]^3 + Sqrt[2]*Cos[e + f*x]^3*(2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]
]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - Log[1 + S
qrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])*Tan[e + f*x]^(7/2))))/(d^3*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d
*Tan[e + f*x]])

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fricas [A]  time = 0.57, size = 257, normalized size = 1.82 \[ \left [\frac {5 \, \sqrt {2} a^{3} d \sqrt {-\frac {1}{d}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-\frac {1}{d}} {\left (\tan \left (f x + e\right ) - 1\right )} - \tan \left (f x + e\right )^{2} + 4 \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} - 2 \, {\left (10 \, a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{5 \, d^{4} f \tan \left (f x + e\right )^{3}}, -\frac {2 \, {\left (5 \, \sqrt {2} a^{3} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{3} + {\left (10 \, a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{5 \, d^{4} f \tan \left (f x + e\right )^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

[1/5*(5*sqrt(2)*a^3*d*sqrt(-1/d)*log(-(2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-1/d)*(tan(f*x + e) - 1) - tan(f*x
+ e)^2 + 4*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1))*tan(f*x + e)^3 - 2*(10*a^3*tan(f*x + e)^2 + 5*a^3*tan(f*x +
e) + a^3)*sqrt(d*tan(f*x + e)))/(d^4*f*tan(f*x + e)^3), -2/5*(5*sqrt(2)*a^3*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d
*tan(f*x + e))*(tan(f*x + e) - 1)/(sqrt(d)*tan(f*x + e)))*tan(f*x + e)^3 + (10*a^3*tan(f*x + e)^2 + 5*a^3*tan(
f*x + e) + a^3)*sqrt(d*tan(f*x + e)))/(d^4*f*tan(f*x + e)^3)]

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giac [B]  time = 1.74, size = 322, normalized size = 2.28 \[ -\frac {\sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} - a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, d^{5} f} + \frac {\sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} - a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, d^{5} f} - \frac {{\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} + \sqrt {2} a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{d^{5} f} - \frac {{\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} + \sqrt {2} a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{d^{5} f} - \frac {2 \, {\left (10 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} + 5 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}\right )}}{5 \, \sqrt {d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) - a^3*abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(ab
s(d)) + abs(d))/(d^5*f) + 1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) - a^3*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqr
t(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d^5*f) - (sqrt(2)*a^3*d*sqrt(abs(d)) + sqrt(2)*a^3*abs(d)^(3/2))*arc
tan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d^5*f) - (sqrt(2)*a^3*d*sqrt(ab
s(d)) + sqrt(2)*a^3*abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs
(d)))/(d^5*f) - 2/5*(10*a^3*d^2*tan(f*x + e)^2 + 5*a^3*d^2*tan(f*x + e) + a^3*d^2)/(sqrt(d*tan(f*x + e))*d^5*f
*tan(f*x + e)^2)

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maple [B]  time = 0.19, size = 409, normalized size = 2.90 \[ -\frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \,d^{4}}-\frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{4}}+\frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{4}}-\frac {a^{3} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \,d^{3} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{3} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{3} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2 a^{3}}{5 f d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {2 a^{3}}{d^{2} f \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {4 a^{3}}{d^{3} f \sqrt {d \tan \left (f x +e \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x)

[Out]

-1/2/f*a^3/d^4*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*t
an(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/f*a^3/d^4*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2
)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/f*a^3/d^4*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e
))^(1/2)+1)-1/2/f*a^3/d^3*2^(1/2)/(d^2)^(1/4)*ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^
(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/f*a^3/d^3*2^(1/2)/(d^2)^(1/4)*ar
ctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/f*a^3/d^3*2^(1/2)/(d^2)^(1/4)*arctan(-2^(1/2)/(d^2)^(1/4)*(
d*tan(f*x+e))^(1/2)+1)-2/5/f*a^3/d/(d*tan(f*x+e))^(5/2)-2*a^3/d^2/f/(d*tan(f*x+e))^(3/2)-4*a^3/d^3/f/(d*tan(f*
x+e))^(1/2)

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maxima [A]  time = 0.58, size = 142, normalized size = 1.01 \[ -\frac {2 \, {\left (\frac {5 \, a^{3} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{d^{2}} + \frac {10 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} + 5 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}\right )}}{5 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-2/5*(5*a^3*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*
arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/d^2 + (10*a^3*d^2*tan(f*x + e
)^2 + 5*a^3*d^2*tan(f*x + e) + a^3*d^2)/((d*tan(f*x + e))^(5/2)*d^2))/(d*f)

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mupad [B]  time = 5.08, size = 128, normalized size = 0.91 \[ -\frac {4\,d\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,d\,a^3\,\mathrm {tan}\left (e+f\,x\right )+\frac {2\,d\,a^3}{5}}{d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}-\frac {\sqrt {2}\,a^3\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{d^{7/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x))^3/(d*tan(e + f*x))^(7/2),x)

[Out]

- ((2*a^3*d)/5 + 4*a^3*d*tan(e + f*x)^2 + 2*a^3*d*tan(e + f*x))/(d^2*f*(d*tan(e + f*x))^(5/2)) - (2^(1/2)*a^3*
(2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2))) + 2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2)) +
(2^(1/2)*(d*tan(e + f*x))^(3/2))/(2*d^(3/2)))))/(d^(7/2)*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {3 \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))**3/(d*tan(f*x+e))**(7/2),x)

[Out]

a**3*(Integral((d*tan(e + f*x))**(-7/2), x) + Integral(3*tan(e + f*x)/(d*tan(e + f*x))**(7/2), x) + Integral(3
*tan(e + f*x)**2/(d*tan(e + f*x))**(7/2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))**(7/2), x))

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