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

Optimal. Leaf size=214 \[ \frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}-\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {(a-i b)^{3/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (-B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.62, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3607, 3630, 3528, 3539, 3537, 63, 208} \[ \frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}-\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {(a-i b)^{3/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (-B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3607

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*
f*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !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 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {2 \int (a+b \tan (c+d x))^{3/2} \left (-a B-\frac {7}{2} b B \tan (c+d x)+\frac {1}{2} (7 A b-2 a B) \tan ^2(c+d x)\right ) \, dx}{7 b}\\ &=\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {2 \int (a+b \tan (c+d x))^{3/2} \left (-\frac {7 A b}{2}-\frac {7}{2} b B \tan (c+d x)\right ) \, dx}{7 b}\\ &=-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {2 \int \sqrt {a+b \tan (c+d x)} \left (-\frac {7}{2} b (a A-b B)-\frac {7}{2} b (A b+a B) \tan (c+d x)\right ) \, dx}{7 b}\\ &=-\frac {2 (A b+a B) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {2 \int \frac {-\frac {7}{2} b \left (a^2 A-A b^2-2 a b B\right )-\frac {7}{2} b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx}{7 b}\\ &=-\frac {2 (A b+a B) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {1}{2} \left ((a-i b)^2 (A-i B)\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} \left ((a+i b)^2 (A+i B)\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 (A b+a B) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {\left ((a+i b)^2 (i A-B)\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}-\frac {\left ((a-i b)^2 (i A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}\\ &=-\frac {2 (A b+a B) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {\left ((a-i b)^2 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-\frac {i a}{b}+\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {\left ((a+i b)^2 (A+i B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\frac {i a}{b}-\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=\frac {(a-i b)^{3/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (i A-B) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 (A b+a B) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}\\ \end {align*}

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Mathematica [A]  time = 2.49, size = 252, normalized size = 1.18 \[ \frac {\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{b}+\frac {35}{3} b (A-i B) \left (3 \sqrt {a-i b} (b+i a) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-i \sqrt {a+b \tan (c+d x)} (4 a+b \tan (c+d x)-3 i b)\right )+\frac {35}{3} b (A+i B) \left (i \sqrt {a+b \tan (c+d x)} (4 a+b \tan (c+d x)+3 i b)+3 \sqrt {a+i b} (b-i a) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )\right )+10 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{35 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(7*A*b - 2*a*B)*(a + b*Tan[c + d*x])^(5/2))/b + 10*B*Tan[c + d*x]*(a + b*Tan[c + d*x])^(5/2) + (35*b*(A -
I*B)*(3*Sqrt[a - I*b]*(I*a + b)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - I*Sqrt[a + b*Tan[c + d*x]]*(
4*a - (3*I)*b + b*Tan[c + d*x])))/3 + (35*b*(A + I*B)*(3*Sqrt[a + I*b]*((-I)*a + b)*ArcTanh[Sqrt[a + b*Tan[c +
d*x]]/Sqrt[a + I*b]] + I*Sqrt[a + b*Tan[c + d*x]]*(4*a + (3*I)*b + b*Tan[c + d*x])))/3)/(35*b*d)

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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(tan(d*x+c)^2*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),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(tan(d*x+c)^2*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.37, size = 1729, normalized size = 8.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)
^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a-1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+
c)-a-(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a+1/d*b^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*
arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B-1/d*b^2/(2*(a
^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*
a)^(1/2))*B-1/4/d*b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*A*
(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2/5/d/b^2*B*(a+b*tan(d*x+c))^(5/2)*a+1/4/d*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2
)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)-1/2/d*ln((a
+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(
1/2)*a-1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a
^2+b^2)^(1/2)-2*a)^(1/2))*B*a^2-1/4/d*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(
a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+1/2/d*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)
*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+1/d/(2*(a^2+b^2)^(1/2)-2*a)^
(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*a^2+1/4
/d*b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1
/2)+2*a)^(1/2)+2/7/d/b^2*B*(a+b*tan(d*x+c))^(7/2)-2/d*b*A*(a+b*tan(d*x+c))^(1/2)+2/5/d/b*A*(a+b*tan(d*x+c))^(5
/2)-2/d*B*(a+b*tan(d*x+c))^(1/2)*a-1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1
/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*ta
n(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*(a^2+b^2)^(1/2)+2/d*b/(2*(a^2+
b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^
(1/2))*A*a+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(
2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*(a^2+b^2)^(1/2)*a-1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)
+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*(a^2+b^2)^(1/2)*a-1/d*b/(2*(a^2+b^2)^(1
/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*
A*(a^2+b^2)^(1/2)-2/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)
^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a+1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*t
an(d*x+c)-a-(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-2/3*B*(a+b*tan(d*x+c))^(3/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^2, x)

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mupad [B]  time = 69.45, size = 2993, normalized size = 13.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((16*A^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*
a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b + a*d*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^
2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d + (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b
^2))/d^2)*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2))/2)*((6*A^4*a^2*b
^4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2)/(4*d^4) - (A^2*a^3)/(4*d^2) + (3*A^2*a*b^2)/(4*d^2))^(1/2) - l
og((16*A^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*
a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b - a*d*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b
^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d - (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*
b^2))/d^2)*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2))/2)*(-((6*A^4*a
^2*b^4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/(4*d^4))^(1/2) - log((16*
A^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^
2)/d^4)^(1/2)*(A*b^3 + A*a^2*b - a*d*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d
^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d - (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2
)*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2))/2)*(((6*A^4*a^2*b^4*d^4
- A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/(4*d^4))^(1/2) - ((2*B*(a^2 + b^2))/
(3*b^2*d) - (2*B*a^2)/(3*b^2*d))*(a + b*tan(c + d*x))^(3/2) + log((16*A^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2
*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b + a*d*(-
((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))
/d + (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2)*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1
/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2))/2)*((3*A^2*a*b^2)/(4*d^2) - (A^2*a^3)/(4*d^2) - (6*A^4*a^2*b^
4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2)/(4*d^4))^(1/2) - (a + b*tan(c + d*x))^(1/2)*(2*a*((2*B*(a^2 + b
^2))/(b^2*d) - (2*B*a^2)/(b^2*d)) + (2*B*a^3)/(b^2*d) - (2*B*a*(a^2 + b^2))/(b^2*d)) - log((8*B^3*b^2*(a^2 - b
^2)*(a^2 + b^2)^2)/d^3 - ((((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(
(16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2 + (16*a*b^2*(((-B^4*b^2*d^4*(3*a^2 - b^2)^
2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(B*a^2 + B*b^2 - d*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2)
+ B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d))/2)*(((6*B^4*a^2*b^4*d^4 - B^4*b^
6*d^4 - 9*B^4*a^4*b^2*d^4)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/(4*d^4))^(1/2) - log((8*B^3*b^2*(a^2 - b^2)*
(a^2 + b^2)^2)/d^3 - ((-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2)*((16
*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2 + (16*a*b^2*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)
^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(B*a^2 + B*b^2 - d*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2)
- B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d))/2)*(-((6*B^4*a^2*b^4*d^4 - B^4*b^
6*d^4 - 9*B^4*a^4*b^2*d^4)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/(4*d^4))^(1/2) + log(((((-B^4*b^2*d^4*(3*a^2
- b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2)*((16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^
4 - 6*a^2*b^2))/d^2 - (16*a*b^2*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(
1/2)*(B*a^2 + B*b^2 + d*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(a
+ b*tan(c + d*x))^(1/2)))/d))/2 + (8*B^3*b^2*(a^2 - b^2)*(a^2 + b^2)^2)/d^3)*((6*B^4*a^2*b^4*d^4 - B^4*b^6*d^4
- 9*B^4*a^4*b^2*d^4)^(1/2)/(4*d^4) + (B^2*a^3)/(4*d^2) - (3*B^2*a*b^2)/(4*d^2))^(1/2) + log(((-((-B^4*b^2*d^4
*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2)*((16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a
^4 + b^4 - 6*a^2*b^2))/d^2 - (16*a*b^2*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2
)/d^4)^(1/2)*(B*a^2 + B*b^2 + d*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^
(1/2)*(a + b*tan(c + d*x))^(1/2)))/d))/2 + (8*B^3*b^2*(a^2 - b^2)*(a^2 + b^2)^2)/d^3)*((B^2*a^3)/(4*d^2) - (6*
B^4*a^2*b^4*d^4 - B^4*b^6*d^4 - 9*B^4*a^4*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^2)/(4*d^2))^(1/2) - ((2*A*(a^2 +
b^2))/(b*d) - (2*A*a^2)/(b*d))*(a + b*tan(c + d*x))^(1/2) + (2*A*(a + b*tan(c + d*x))^(5/2))/(5*b*d) + (2*B*(
a + b*tan(c + d*x))^(7/2))/(7*b^2*d) - (2*B*a*(a + b*tan(c + d*x))^(5/2))/(5*b^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+b*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

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

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