3.7.8 \(\int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\) [608]

Optimal. Leaf size=231 \[ -\frac {\sqrt {i a-b} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\left (a^2+8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {i a+b} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d} \]

[Out]

-1/4*(a^2+8*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(3/2)/d-arctan((I*a-b)^(1/2)*tan(d
*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a-b)^(1/2)/d+arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1
/2))*(I*a+b)^(1/2)/d-1/4*a*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/b/d+1/2*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(
3/2)/b/d

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Rubi [A]
time = 1.12, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3647, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} -\frac {\left (a^2+8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}-\frac {\sqrt {-b+i a} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}-\frac {a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {\sqrt {b+i a} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) - ((a^2 + 8*b^2)*ArcT
anh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(4*b^(3/2)*d) + (Sqrt[I*a + b]*ArcTanh[(Sqrt[I*a +
b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(4*b*d)
+ (Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*b*d)

Rule 65

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 95

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 211

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

Rule 212

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

Rule 214

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3736

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] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 1.90, size = 282, normalized size = 1.22 \begin {gather*} \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {-\frac {a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {-2 \sqrt [4]{-1} \sqrt {-a+i b} b \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \sqrt [4]{-1} \sqrt {a+i b} b \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {\sqrt {a} \left (a^2+8 b^2\right ) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 \sqrt {b} \sqrt {a+b \tan (c+d x)}}}{d}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*b*d) + (-1/2*(a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]
)/d + (-2*(-1)^(1/4)*Sqrt[-a + I*b]*b*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c +
d*x]]] - 2*(-1)^(1/4)*Sqrt[a + I*b]*b*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c +
d*x]]] - (Sqrt[a]*(a^2 + 8*b^2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(
2*Sqrt[b]*Sqrt[a + b*Tan[c + d*x]]))/d)/(2*b)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 1.26, size = 1092020, normalized size = 4727.36 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),x)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(5/2)*(a+b*tan(d*x+c))**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))^(1/2),x)

[Out]

int(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))^(1/2), x)

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