3.352 \(\int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=138 \[ \frac {2 \sqrt {2} a^3 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}+\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac {4 a^3 \sqrt {d \tan (e+f x)}}{f}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f} \]

[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^(1/2)/f+4*a^3*(d*tan(f*x
+e))^(1/2)/f+8/5*a^3*(d*tan(f*x+e))^(3/2)/d/f+2/5*(d*tan(f*x+e))^(3/2)*(a^3+a^3*tan(f*x+e))/d/f

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Rubi [A]  time = 0.18, antiderivative size = 138, 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 = {3566, 3630, 3528, 3532, 205} \[ \frac {2 \sqrt {2} a^3 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}+\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac {4 a^3 \sqrt {d \tan (e+f x)}}{f}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3528

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

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 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

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

Rubi steps

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

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Mathematica [C]  time = 1.80, size = 315, normalized size = 2.28 \[ \frac {a^3 \cos (e+f x) (\tan (e+f x)+1)^3 \sqrt {d \tan (e+f x)} \left (3 \left (4 \sin ^2(e+f x) \sqrt {\tan (e+f x)}+10 \sin (2 (e+f x)) \sqrt {\tan (e+f x)}+10 \sqrt {2} \cos ^2(e+f x) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-10 \sqrt {2} \cos ^2(e+f x) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+40 \cos ^2(e+f x) \sqrt {\tan (e+f x)}+5 \sqrt {2} \cos ^2(e+f x) \log \left (\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1\right )-5 \sqrt {2} \cos ^2(e+f x) \log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )\right )-20 \sin (2 (e+f x)) \sqrt {\tan (e+f x)} \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right )\right )}{30 f \sqrt {\tan (e+f x)} (\sin (e+f x)+\cos (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[e + f*x]*(3*(10*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^2 - 10*Sqrt[2]*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^2 + 5*Sqrt[2]*Cos[e + f*x]^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Ta
n[e + f*x]] - 5*Sqrt[2]*Cos[e + f*x]^2*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] + 40*Cos[e + f*x]^2*
Sqrt[Tan[e + f*x]] + 4*Sin[e + f*x]^2*Sqrt[Tan[e + f*x]] + 10*Sin[2*(e + f*x)]*Sqrt[Tan[e + f*x]]) - 20*Hyperg
eometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2]*Sin[2*(e + f*x)]*Sqrt[Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]]*(1 + Tan[e
+ f*x])^3)/(30*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[Tan[e + f*x]])

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fricas [A]  time = 0.59, size = 217, normalized size = 1.57 \[ \left [\frac {5 \, \sqrt {2} a^{3} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + 10 \, a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{5 \, f}, -\frac {2 \, {\left (5 \, \sqrt {2} a^{3} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) - {\left (a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + 10 \, a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{5 \, f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(1/2)*(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.56, size = 344, normalized size = 2.49 \[ -\frac {\sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} - a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, d f} + \frac {\sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} - a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, d f} - \frac {{\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} + \sqrt {2} a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{d f} - \frac {{\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} + \sqrt {2} a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{d f} + \frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right )^{2} + 5 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right ) + 10 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4}\right )}}{5 \, d^{10} f^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(1/2)*(a+a*tan(f*x+e))^3,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*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)*sqrt(
d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d*f) - (sqrt(2)*a^3*d*sqrt(abs(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*f) - (sqrt(2)*a^3*d*sqrt(abs(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*f) + 2/5*(sqrt(d*tan(f*x + e))*a^3*d^10*f^4*tan(f*x + e)^2 + 5*sqrt(d*tan(f*x + e))*a^3*d^10*f^4*tan(f*x +
e) + 10*sqrt(d*tan(f*x + e))*a^3*d^10*f^4)/(d^10*f^5)

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maple [B]  time = 0.26, size = 391, normalized size = 2.83 \[ \frac {2 a^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f \,d^{2}}+\frac {2 a^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{d f}+\frac {4 a^{3} \sqrt {d \tan \left (f x +e \right )}}{f}-\frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f}-\frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f}+\frac {a^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f}-\frac {a^{3} d \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} d \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} d \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \left (d^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(1/2)*(a+a*tan(f*x+e))^3,x)

[Out]

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

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maxima [A]  time = 0.61, size = 144, normalized size = 1.04 \[ -\frac {2 \, {\left (5 \, a^{3} d^{2} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )} - \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} + 5 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d + 10 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{2}}{d}\right )}}{5 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(1/2)*(a+a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.62, size = 137, normalized size = 0.99 \[ \frac {4\,a^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{d\,f}+\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d^2\,f}-\frac {\sqrt {2}\,a^3\,\sqrt {d}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(1/2)*(a + a*tan(e + f*x))^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(1/2)*(a+a*tan(f*x+e))**3,x)

[Out]

a**3*(Integral(sqrt(d*tan(e + f*x)), x) + Integral(3*sqrt(d*tan(e + f*x))*tan(e + f*x), x) + Integral(3*sqrt(d
*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(sqrt(d*tan(e + f*x))*tan(e + f*x)**3, x))

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