3.334 \(\int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=188 \[ \frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a B+A b) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {(a-i b)^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} (-B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a B+A b) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {(a-i b)^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} (-B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.08, size = 233, normalized size = 1.24 \[ \frac {i \left ((A-i B) \left (\frac {2}{5} (a+b \tan (c+d x))^{5/2}+\frac {2}{3} (a-i b) \left (\sqrt {a+b \tan (c+d x)} (4 a+b \tan (c+d x)-3 i b)-3 (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )\right )\right )-(A+i B) \left (\frac {2}{5} (a+b \tan (c+d x))^{5/2}+\frac {2}{3} (a+i b) \left (\sqrt {a+b \tan (c+d x)} (4 a+b \tan (c+d x)+3 i b)-3 (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )\right )\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

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

[Out]

Timed out

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maple [B]  time = 0.26, size = 2405, normalized size = 12.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)

[Out]

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

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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(d*x+c))^(5/2)*(A+B*tan(d*x+c)),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(b-a>0)', see `assume?` for mor
e details)Is b-a positive, negative or zero?

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mupad [B]  time = 34.45, size = 3863, normalized size = 20.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2),x)

[Out]

log(- ((((((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*
d^2)/d^4)^(1/2)*(32*B*b^6 - 32*B*a^4*b^2 + 32*a*b^2*d*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^
2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) - (16*B^2*b^2*
(a + b*tan(c + d*x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b
^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2))/2 - (8*B^3*a*b^2*(a^2 - 3*b^2)*
(a^2 + b^2)^3)/d^3)*((20*B^4*a^2*b^8*d^4 - B^4*b^10*d^4 - 110*B^4*a^4*b^6*d^4 + 100*B^4*a^6*b^4*d^4 - 25*B^4*a
^8*b^2*d^4)^(1/2)/(4*d^4) + (B^2*a^5)/(4*d^2) - (5*B^2*a^3*b^2)/(2*d^2) + (5*B^2*a*b^4)/(4*d^2))^(1/2) - log((
(((((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^
4)^(1/2)*(32*B*a^4*b^2 - 32*B*b^6 + 32*a*b^2*d*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d
^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) - (16*B^2*b^2*(a + b*
tan(c + d*x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^
(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2))/2 - (8*B^3*a*b^2*(a^2 - 3*b^2)*(a^2 +
b^2)^3)/d^3)*(((20*B^4*a^2*b^8*d^4 - B^4*b^10*d^4 - 110*B^4*a^4*b^6*d^4 + 100*B^4*a^6*b^4*d^4 - 25*B^4*a^8*b^2
*d^4)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/(4*d^4))^(1/2) - log(((((-((-B^4*b^2*d^4*(5*
a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - B^2*a^5*d^2 + 10*B^2*a^3*b^2*d^2 - 5*B^2*a*b^4*d^2)/d^4)^(1/2)*(32*B*a^4*b^
2 - 32*B*b^6 + 32*a*b^2*d*(-((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - B^2*a^5*d^2 + 10*B^2*a^3*b^2*
d^2 - 5*B^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) - (16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*
(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(-((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - B^2*a^5*d^2
+ 10*B^2*a^3*b^2*d^2 - 5*B^2*a*b^4*d^2)/d^4)^(1/2))/2 - (8*B^3*a*b^2*(a^2 - 3*b^2)*(a^2 + b^2)^3)/d^3)*(-((20
*B^4*a^2*b^8*d^4 - B^4*b^10*d^4 - 110*B^4*a^4*b^6*d^4 + 100*B^4*a^6*b^4*d^4 - 25*B^4*a^8*b^2*d^4)^(1/2) - B^2*
a^5*d^2 + 10*B^2*a^3*b^2*d^2 - 5*B^2*a*b^4*d^2)/(4*d^4))^(1/2) + log(- ((((-((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a
^2*b^2)^2)^(1/2) - B^2*a^5*d^2 + 10*B^2*a^3*b^2*d^2 - 5*B^2*a*b^4*d^2)/d^4)^(1/2)*(32*B*b^6 - 32*B*a^4*b^2 + 3
2*a*b^2*d*(-((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - B^2*a^5*d^2 + 10*B^2*a^3*b^2*d^2 - 5*B^2*a*b^
4*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) - (16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^6 - b^6 + 15*
a^2*b^4 - 15*a^4*b^2))/d^2)*(-((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - B^2*a^5*d^2 + 10*B^2*a^3*b^
2*d^2 - 5*B^2*a*b^4*d^2)/d^4)^(1/2))/2 - (8*B^3*a*b^2*(a^2 - 3*b^2)*(a^2 + b^2)^3)/d^3)*((B^2*a^5)/(4*d^2) - (
20*B^4*a^2*b^8*d^4 - B^4*b^10*d^4 - 110*B^4*a^4*b^6*d^4 + 100*B^4*a^6*b^4*d^4 - 25*B^4*a^8*b^2*d^4)^(1/2)/(4*d
^4) - (5*B^2*a^3*b^2)/(2*d^2) + (5*B^2*a*b^4)/(4*d^2))^(1/2) + ((4*B*a^2)/d - (2*B*(a^2 + b^2))/d)*(a + b*tan(
c + d*x))^(1/2) - log(((-((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2
+ 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(((-((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a
^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(64*A*a^3*b^3 + 64*A*a*b^5 + 32*a*b^2*d*(-((-A^4*b^2*d^4*(5*a^4 + b^4
- 10*a^2*b^2)^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^
(1/2)))/(2*d) + (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2))/2 - (8*A^3
*b^3*(3*a^2 - b^2)*(a^2 + b^2)^3)/d^3)*(-((20*A^4*a^2*b^8*d^4 - A^4*b^10*d^4 - 110*A^4*a^4*b^6*d^4 + 100*A^4*a
^6*b^4*d^4 - 25*A^4*a^8*b^2*d^4)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/(4*d^4))^(1/2) -
log(((((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - A^2*a^5*d^2 + 10*A^2*a^3*b^2*d^2 - 5*A^2*a*b^4*d^2)
/d^4)^(1/2)*(((((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - A^2*a^5*d^2 + 10*A^2*a^3*b^2*d^2 - 5*A^2*a
*b^4*d^2)/d^4)^(1/2)*(64*A*a^3*b^3 + 64*A*a*b^5 + 32*a*b^2*d*(((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/
2) - A^2*a^5*d^2 + 10*A^2*a^3*b^2*d^2 - 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) + (16*A
^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2))/2 - (8*A^3*b^3*(3*a^2 - b^2)*(a
^2 + b^2)^3)/d^3)*(((20*A^4*a^2*b^8*d^4 - A^4*b^10*d^4 - 110*A^4*a^4*b^6*d^4 + 100*A^4*a^6*b^4*d^4 - 25*A^4*a^
8*b^2*d^4)^(1/2) - A^2*a^5*d^2 + 10*A^2*a^3*b^2*d^2 - 5*A^2*a*b^4*d^2)/(4*d^4))^(1/2) + log(((((-A^4*b^2*d^4*(
5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - A^2*a^5*d^2 + 10*A^2*a^3*b^2*d^2 - 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(((((-A^4*
b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - A^2*a^5*d^2 + 10*A^2*a^3*b^2*d^2 - 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(
64*A*a^3*b^3 + 64*A*a*b^5 - 32*a*b^2*d*(((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - A^2*a^5*d^2 + 10*
A^2*a^3*b^2*d^2 - 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) - (16*A^2*b^2*(a + b*tan(c +
d*x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2))/2 - (8*A^3*b^3*(3*a^2 - b^2)*(a^2 + b^2)^3)/d^3)*((20
*A^4*a^2*b^8*d^4 - A^4*b^10*d^4 - 110*A^4*a^4*b^6*d^4 + 100*A^4*a^6*b^4*d^4 - 25*A^4*a^8*b^2*d^4)^(1/2)/(4*d^4
) - (A^2*a^5)/(4*d^2) + (5*A^2*a^3*b^2)/(2*d^2) - (5*A^2*a*b^4)/(4*d^2))^(1/2) + log(((-((-A^4*b^2*d^4*(5*a^4
+ b^4 - 10*a^2*b^2)^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(((-((-A^4*b^2*d
^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(64*A*
a^3*b^3 + 64*A*a*b^5 - 32*a*b^2*d*(-((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*
a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) - (16*A^2*b^2*(a + b*tan(c + d*x)
)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2))/2 - (8*A^3*b^3*(3*a^2 - b^2)*(a^2 + b^2)^3)/d^3)*((5*A^2*
a^3*b^2)/(2*d^2) - (A^2*a^5)/(4*d^2) - (20*A^4*a^2*b^8*d^4 - A^4*b^10*d^4 - 110*A^4*a^4*b^6*d^4 + 100*A^4*a^6*
b^4*d^4 - 25*A^4*a^8*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a*b^4)/(4*d^2))^(1/2) + (2*B*(a + b*tan(c + d*x))^(5/2))/
(5*d) + (2*A*b*(a + b*tan(c + d*x))^(3/2))/(3*d) + (2*B*a*(a + b*tan(c + d*x))^(3/2))/(3*d) + (4*A*a*b*(a + b*
tan(c + d*x))^(1/2))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(5/2), x)

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