3.1237 \(\int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=150 \[ \frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}-\frac {(b+i a) (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(-b+i a) (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f} \]

[Out]

-(I*a+b)*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+(I*a-b)*(c+I*d)^(3/2)*arctanh((c+d*tan(
f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(a*d+b*c)*(c+d*tan(f*x+e))^(1/2)/f+2/3*b*(c+d*tan(f*x+e))^(3/2)/f

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Rubi [A]  time = 0.32, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3528, 3539, 3537, 63, 208} \[ \frac {2 (a d+b c) \sqrt {c+d \tan (e+f x)}}{f}-\frac {(b+i a) (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(-b+i a) (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b (c+d \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*a + b)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f) + ((I*a - b)*(c + I*d)^(3/2)*
ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*(b*c + a*d)*Sqrt[c + d*Tan[e + f*x]])/f + (2*b*(c + d*
Tan[e + f*x])^(3/2))/(3*f)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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 3537

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

Rule 3539

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

Rubi steps

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

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Mathematica [A]  time = 0.54, size = 140, normalized size = 0.93 \[ \frac {2 \sqrt {c+d \tan (e+f x)} (3 a d+4 b c+b d \tan (e+f x))-3 i (a-i b) (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+3 i (a+i b) (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*(a - I*b)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + (3*I)*(a + I*b)*(c + I*d)^
(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + 2*Sqrt[c + d*Tan[e + f*x]]*(4*b*c + 3*a*d + b*d*Tan[e
+ f*x]))/(3*f)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.24, size = 1665, normalized size = 11.10 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/f*d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^
(1/2)+2*c)^(1/2)*a+1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^
(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*b*c+1/4/f*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c
^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b+1/4/f/d*ln(d*tan(f*x
+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^
2-1/2/f*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(
1/2)+2*c)^(1/2)*b*c-1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*
c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a+2/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*
tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a*c+1/f/(2*(c^2+d^2)^(1/2)-2*c
)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*c^2-1
/4/f*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2
)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b-1/4/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-
(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^2+1/2/f*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(
1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*c+1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arc
tan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a-
2/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^
2)^(1/2)-2*c)^(1/2))*a*c-1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*
x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*c^2+1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)
+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*d^2-1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*a
rctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b*d^2+1/4/f*d*ln
((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^
(1/2)*a+2/f*a*(c+d*tan(f*x+e))^(1/2)*d+2/f*c*b*(c+d*tan(f*x+e))^(1/2)-1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan
((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*b*c-1
/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1
/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*c+1/4/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e
)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*c+2/3*b*(c+d*tan(f*x+e))^(3/2)/f

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d-c>0)', see `assume?` for mor
e details)Is d-c positive, negative or zero?

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mupad [B]  time = 17.41, size = 2823, normalized size = 18.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(((((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*((16*c*d^2*(((-b^4*d^2
*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(b*c^2 + b*d^2 - f*(((-b^4*d^2*f^4*(3*
c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*b^2*d^2*(
c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (8*b^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*((6*b^
4*c^2*d^4*f^4 - b^4*d^6*f^4 - 9*b^4*c^4*d^2*f^4)^(1/2)/(4*f^4) + (b^2*c^3)/(4*f^2) - (3*b^2*c*d^2)/(4*f^2))^(1
/2) - log(((-((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)*((16*c*d^2*(-((
-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(b*c^2 + b*d^2 + f*(-((-b^4*d^
2*f^4*(3*c^2 - d^2)^2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*
b^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (8*b^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^
3)*(-((6*b^4*c^2*d^4*f^4 - b^4*d^6*f^4 - 9*b^4*c^4*d^2*f^4)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/(4*f^4))^(1
/2) - log(((((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*((16*c*d^2*(((-b
^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(b*c^2 + b*d^2 + f*(((-b^4*d^2*f
^4*(3*c^2 - d^2)^2)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*b^2
*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (8*b^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*
(((6*b^4*c^2*d^4*f^4 - b^4*d^6*f^4 - 9*b^4*c^4*d^2*f^4)^(1/2) + b^2*c^3*f^2 - 3*b^2*c*d^2*f^2)/(4*f^4))^(1/2)
+ log(((-((-b^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)*((16*c*d^2*(-((-b^4
*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(b*c^2 + b*d^2 - f*(-((-b^4*d^2*f^
4*(3*c^2 - d^2)^2)^(1/2) - b^2*c^3*f^2 + 3*b^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*b^2*
d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (8*b^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*(
(b^2*c^3)/(4*f^2) - (6*b^4*c^2*d^4*f^4 - b^4*d^6*f^4 - 9*b^4*c^4*d^2*f^4)^(1/2)/(4*f^4) - (3*b^2*c*d^2)/(4*f^2
))^(1/2) - log(((-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*d^2*(
-((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*d^3 + a*c^2*d + c*f*(-((
-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f
+ (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (16*a^3*c*d^3*(c^2 + d^2)^2)/f^3)
*(-((6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/(4*f^4))^(1/2
) - log(((((-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*d^2*(((-a^4*d
^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*d^3 + a*c^2*d + c*f*(((-a^4*d^2*f
^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*a^2
*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (16*a^3*c*d^3*(c^2 + d^2)^2)/f^3)*(((6*a^4*
c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/(4*f^4))^(1/2) + log((((
(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*d^2*(((-a^4*d^2*f^4*(3*c
^2 - d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*d^3 + a*c^2*d - c*f*(((-a^4*d^2*f^4*(3*c^2 -
d^2)^2)^(1/2) - a^2*c^3*f^2 + 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*a^2*d^2*(c + d
*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (16*a^3*c*d^3*(c^2 + d^2)^2)/f^3)*((6*a^4*c^2*d^4*f^4
- a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2)/(4*f^4) - (a^2*c^3)/(4*f^2) + (3*a^2*c*d^2)/(4*f^2))^(1/2) + log(((-(
(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*((16*d^2*(-((-a^4*d^2*f^4*(3*
c^2 - d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(a*d^3 + a*c^2*d - c*f*(-((-a^4*d^2*f^4*(3*c^2
- d^2)^2)^(1/2) + a^2*c^3*f^2 - 3*a^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*a^2*d^2*(c +
d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2))/2 - (16*a^3*c*d^3*(c^2 + d^2)^2)/f^3)*((3*a^2*c*d^2)/(4*
f^2) - (a^2*c^3)/(4*f^2) - (6*a^4*c^2*d^4*f^4 - a^4*d^6*f^4 - 9*a^4*c^4*d^2*f^4)^(1/2)/(4*f^4))^(1/2) + (2*b*(
c + d*tan(e + f*x))^(3/2))/(3*f) + (2*a*d*(c + d*tan(e + f*x))^(1/2))/f + (2*b*c*(c + d*tan(e + f*x))^(1/2))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(e + f*x))*(c + d*tan(e + f*x))**(3/2), x)

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