3.629 \(\int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=219 \[ \frac {2 \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {(-b+i a)^{5/2} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(b+i a)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]

[Out]

-(I*a-b)^(5/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+(I*a+b)^(5/2)*arctanh((I*a+b)^(
1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+2/15*(15*a^2-23*b^2)*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
-2/5*a^2*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)-22/15*a*b*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)

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Rubi [A]  time = 0.99, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3565, 3649, 3616, 3615, 93, 203, 206} \[ \frac {2 \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {(-b+i a)^{5/2} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(b+i a)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(7/2),x]

[Out]

-(((I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + ((I*a + b)^(5/2)*
ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2*a^2*Sqrt[a + b*Tan[c + d*x]])/(5*
d*Tan[c + d*x]^(5/2)) - (22*a*b*Sqrt[a + b*Tan[c + d*x]])/(15*d*Tan[c + d*x]^(3/2)) + (2*(15*a^2 - 23*b^2)*Sqr
t[a + b*Tan[c + d*x]])/(15*d*Sqrt[Tan[c + d*x]])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3615

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

Rule 3616

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

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\frac {11 a^2 b}{2}-\frac {5}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac {1}{2} b \left (4 a^2-5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 \int \frac {\frac {1}{4} a^2 \left (15 a^2-23 b^2\right )+\frac {15}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {11}{2} a^2 b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}+\frac {8 \int \frac {-\frac {15}{8} a^2 b \left (3 a^2-b^2\right )+\frac {15}{8} a^3 \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {1}{2} (i a-b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx+\frac {1}{2} (i a+b)^3 \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {(i a-b)^3 \operatorname {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(i a+b)^3 \operatorname {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {(i a-b)^3 \operatorname {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^3 \operatorname {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=-\frac {(i a-b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.65, size = 194, normalized size = 0.89 \[ \frac {\frac {2 \sqrt {a+b \tan (c+d x)} \left (\left (15 a^2-23 b^2\right ) \tan ^2(c+d x)-3 a^2-11 a b \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)}+15 \sqrt [4]{-1} (-a+i b)^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+15 \sqrt [4]{-1} (a+i b)^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(7/2),x]

[Out]

(15*(-1)^(1/4)*(-a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]
] + 15*(-1)^(1/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]
]] + (2*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2 - 11*a*b*Tan[c + d*x] + (15*a^2 - 23*b^2)*Tan[c + d*x]^2))/Tan[c + d*
x]^(5/2))/(15*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)/tan(d*x+c)^(7/2),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(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.74, size = 1344685, normalized size = 6140.11 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^(5/2)/tan(c + d*x)^(7/2),x)

[Out]

int((a + b*tan(c + d*x))^(5/2)/tan(c + d*x)^(7/2), x)

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sympy [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)/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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