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

Optimal. Leaf size=273 \[ \frac {2 (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}-\frac {(b+i a) (c-i d)^{3/2} (A-i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(-b+i a) (c+i d)^{3/2} (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}-\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f} \]

[Out]

-(I*a+b)*(A-I*B-C)*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+(I*a-b)*(A+I*B-C)*(c+I*d)^(3/
2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(A*a*d+A*b*c+B*a*c-B*b*d-C*a*d-C*b*c)*(c+d*tan(f*x+e))^(1
/2)/f+2/3*(A*b+B*a-C*b)*(c+d*tan(f*x+e))^(3/2)/f-2/35*(-7*B*b*d-7*C*a*d+2*C*b*c)*(c+d*tan(f*x+e))^(5/2)/d^2/f+
2/7*b*C*tan(f*x+e)*(c+d*tan(f*x+e))^(5/2)/d/f

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Rubi [A]  time = 0.88, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3637, 3630, 3528, 3539, 3537, 63, 208} \[ \frac {2 (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}-\frac {(b+i a) (c-i d)^{3/2} (A-i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(-b+i a) (c+i d)^{3/2} (A+i B-C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}-\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*a + b)*(A - I*B - C)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f) + ((I*a - b)*(A
+ I*B - C)*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*(A*b*c + a*B*c - b*c*C + a
*A*d - b*B*d - a*C*d)*Sqrt[c + d*Tan[e + f*x]])/f + (2*(A*b + a*B - b*C)*(c + d*Tan[e + f*x])^(3/2))/(3*f) - (
2*(2*b*c*C - 7*b*B*d - 7*a*C*d)*(c + d*Tan[e + f*x])^(5/2))/(35*d^2*f) + (2*b*C*Tan[e + f*x]*(c + d*Tan[e + f*
x])^(5/2))/(7*d*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]

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]

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {2 \int (c+d \tan (e+f x))^{3/2} \left (\frac {1}{2} (2 b c C-7 a A d)-\frac {7}{2} (A b+a B-b C) d \tan (e+f x)+\frac {1}{2} (2 b c C-7 b B d-7 a C d) \tan ^2(e+f x)\right ) \, dx}{7 d}\\ &=-\frac {2 (2 b c C-7 b B d-7 a C d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {2 \int (c+d \tan (e+f x))^{3/2} \left (\frac {7}{2} (b B-a (A-C)) d-\frac {7}{2} (A b+a B-b C) d \tan (e+f x)\right ) \, dx}{7 d}\\ &=\frac {2 (A b+a B-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 (2 b c C-7 b B d-7 a C d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {2 \int \sqrt {c+d \tan (e+f x)} \left (\frac {7}{2} d (b B c+b (A-C) d-a (A c-c C-B d))-\frac {7}{2} d (A b c+a B c-b c C+a A d-b B d-a C d) \tan (e+f x)\right ) \, dx}{7 d}\\ &=\frac {2 (A b c+a B c-b c C+a A d-b B d-a C d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (A b+a B-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 (2 b c C-7 b B d-7 a C d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {2 \int \frac {\frac {7}{2} d \left (a \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )-\frac {7}{2} d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (c^2 C+2 B c d-C d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{7 d}\\ &=\frac {2 (A b c+a B c-b c C+a A d-b B d-a C d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (A b+a B-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 (2 b c C-7 b B d-7 a C d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}+\frac {1}{2} \left ((a-i b) (A-i B-C) (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) (A+i B-C) (c+i d)^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {2 (A b c+a B c-b c C+a A d-b B d-a C d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (A b+a B-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 (2 b c C-7 b B d-7 a C d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}+\frac {\left ((i a+b) (A-i B-C) (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) (A+i B-C) (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 (A b c+a B c-b c C+a A d-b B d-a C d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (A b+a B-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 (2 b c C-7 b B d-7 a C d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\left ((a-i b) (A-i B-C) (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) (A+i B-C) (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 {(a-i b) (i A+B-i C) (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) (A+i B-C) (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 (A b c+a B c-b c C+a A d-b B d-a C d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (A b+a B-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 (2 b c C-7 b B d-7 a C d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}\\ \end {align*}

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Mathematica [A]  time = 4.44, size = 260, normalized size = 0.95 \[ \frac {\frac {35}{3} d (b+i a) (A-i B-C) \left (\sqrt {c+d \tan (e+f x)} (4 c+d \tan (e+f x)-3 i d)-3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )\right )+\frac {35}{3} d (b-i a) (A+i B-C) \left (\sqrt {c+d \tan (e+f x)} (4 c+d \tan (e+f x)+3 i d)-3 (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )\right )+\frac {2 (7 a C d+7 b B d-2 b c C) (c+d \tan (e+f x))^{5/2}}{d}+10 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{35 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(-2*b*c*C + 7*b*B*d + 7*a*C*d)*(c + d*Tan[e + f*x])^(5/2))/d + 10*b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])^(5
/2) + (35*(I*a + b)*(A - I*B - C)*d*(-3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sqrt
[c + d*Tan[e + f*x]]*(4*c - (3*I)*d + d*Tan[e + f*x])))/3 + (35*((-I)*a + b)*(A + I*B - C)*d*(-3*(c + I*d)^(3/
2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])
))/3)/(35*d*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)*(A+B*tan(f*x+e)+C*tan(f*x+e)^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)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.55, size = 5149, normalized size = 18.86 \[ \text {output too large to display} \]

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)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

[Out]

result too large to display

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maxima [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)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

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)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2),x)

[Out]

\text{Hanged}

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

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)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

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

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