3.1.53 \(\int (c+d x)^{5/2} \sinh ^3(a+b x) \, dx\) [53]

Optimal. Leaf size=381 \[ -\frac {45 d^2 \sqrt {c+d x} \cosh (a+b x)}{16 b^3}-\frac {2 (c+d x)^{5/2} \cosh (a+b x)}{3 b}+\frac {5 d^2 \sqrt {c+d x} \cosh (3 a+3 b x)}{144 b^3}+\frac {45 d^{5/2} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{7/2}}-\frac {5 d^{5/2} e^{-3 a+\frac {3 b c}{d}} \sqrt {\frac {\pi }{3}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{576 b^{7/2}}+\frac {45 d^{5/2} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{7/2}}-\frac {5 d^{5/2} e^{3 a-\frac {3 b c}{d}} \sqrt {\frac {\pi }{3}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{576 b^{7/2}}+\frac {5 d (c+d x)^{3/2} \sinh (a+b x)}{3 b^2}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2} \]

[Out]

-2/3*(d*x+c)^(5/2)*cosh(b*x+a)/b+5/3*d*(d*x+c)^(3/2)*sinh(b*x+a)/b^2+1/3*(d*x+c)^(5/2)*cosh(b*x+a)*sinh(b*x+a)
^2/b-5/18*d*(d*x+c)^(3/2)*sinh(b*x+a)^3/b^2-5/1728*d^(5/2)*exp(-3*a+3*b*c/d)*erf(3^(1/2)*b^(1/2)*(d*x+c)^(1/2)
/d^(1/2))*3^(1/2)*Pi^(1/2)/b^(7/2)-5/1728*d^(5/2)*exp(3*a-3*b*c/d)*erfi(3^(1/2)*b^(1/2)*(d*x+c)^(1/2)/d^(1/2))
*3^(1/2)*Pi^(1/2)/b^(7/2)+45/64*d^(5/2)*exp(-a+b*c/d)*erf(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/b^(7/2)+45/6
4*d^(5/2)*exp(a-b*c/d)*erfi(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/b^(7/2)-45/16*d^2*cosh(b*x+a)*(d*x+c)^(1/2
)/b^3+5/144*d^2*cosh(3*b*x+3*a)*(d*x+c)^(1/2)/b^3

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Rubi [A]
time = 0.80, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3392, 3377, 3388, 2211, 2235, 2236, 3393} \begin {gather*} \frac {45 \sqrt {\pi } d^{5/2} e^{\frac {b c}{d}-a} \text {Erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{7/2}}-\frac {5 \sqrt {\frac {\pi }{3}} d^{5/2} e^{\frac {3 b c}{d}-3 a} \text {Erf}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{576 b^{7/2}}+\frac {45 \sqrt {\pi } d^{5/2} e^{a-\frac {b c}{d}} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{7/2}}-\frac {5 \sqrt {\frac {\pi }{3}} d^{5/2} e^{3 a-\frac {3 b c}{d}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{576 b^{7/2}}-\frac {45 d^2 \sqrt {c+d x} \cosh (a+b x)}{16 b^3}+\frac {5 d^2 \sqrt {c+d x} \cosh (3 a+3 b x)}{144 b^3}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2}+\frac {5 d (c+d x)^{3/2} \sinh (a+b x)}{3 b^2}-\frac {2 (c+d x)^{5/2} \cosh (a+b x)}{3 b}+\frac {(c+d x)^{5/2} \sinh ^2(a+b x) \cosh (a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)*Sinh[a + b*x]^3,x]

[Out]

(-45*d^2*Sqrt[c + d*x]*Cosh[a + b*x])/(16*b^3) - (2*(c + d*x)^(5/2)*Cosh[a + b*x])/(3*b) + (5*d^2*Sqrt[c + d*x
]*Cosh[3*a + 3*b*x])/(144*b^3) + (45*d^(5/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(
64*b^(7/2)) - (5*d^(5/2)*E^(-3*a + (3*b*c)/d)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(576*b^
(7/2)) + (45*d^(5/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(64*b^(7/2)) - (5*d^(5/2)
*E^(3*a - (3*b*c)/d)*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(576*b^(7/2)) + (5*d*(c + d*x)^
(3/2)*Sinh[a + b*x])/(3*b^2) + ((c + d*x)^(5/2)*Cosh[a + b*x]*Sinh[a + b*x]^2)/(3*b) - (5*d*(c + d*x)^(3/2)*Si
nh[a + b*x]^3)/(18*b^2)

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3392

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*m*(c + d*x)^(m - 1)*((
b*Sin[e + f*x])^n/(f^2*n^2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[d^2*m*((m - 1)/(f^2*n^2)), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[b*(c + d*x)^m*Cos[e + f
*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin {align*} \int (c+d x)^{5/2} \sinh ^3(a+b x) \, dx &=\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2}-\frac {2}{3} \int (c+d x)^{5/2} \sinh (a+b x) \, dx+\frac {\left (5 d^2\right ) \int \sqrt {c+d x} \sinh ^3(a+b x) \, dx}{12 b^2}\\ &=-\frac {2 (c+d x)^{5/2} \cosh (a+b x)}{3 b}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2}+\frac {(5 d) \int (c+d x)^{3/2} \cosh (a+b x) \, dx}{3 b}+\frac {\left (5 i d^2\right ) \int \left (\frac {3}{4} i \sqrt {c+d x} \sinh (a+b x)-\frac {1}{4} i \sqrt {c+d x} \sinh (3 a+3 b x)\right ) \, dx}{12 b^2}\\ &=-\frac {2 (c+d x)^{5/2} \cosh (a+b x)}{3 b}+\frac {5 d (c+d x)^{3/2} \sinh (a+b x)}{3 b^2}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2}+\frac {\left (5 d^2\right ) \int \sqrt {c+d x} \sinh (3 a+3 b x) \, dx}{48 b^2}-\frac {\left (5 d^2\right ) \int \sqrt {c+d x} \sinh (a+b x) \, dx}{16 b^2}-\frac {\left (5 d^2\right ) \int \sqrt {c+d x} \sinh (a+b x) \, dx}{2 b^2}\\ &=-\frac {45 d^2 \sqrt {c+d x} \cosh (a+b x)}{16 b^3}-\frac {2 (c+d x)^{5/2} \cosh (a+b x)}{3 b}+\frac {5 d^2 \sqrt {c+d x} \cosh (3 a+3 b x)}{144 b^3}+\frac {5 d (c+d x)^{3/2} \sinh (a+b x)}{3 b^2}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2}-\frac {\left (5 d^3\right ) \int \frac {\cosh (3 a+3 b x)}{\sqrt {c+d x}} \, dx}{288 b^3}+\frac {\left (5 d^3\right ) \int \frac {\cosh (a+b x)}{\sqrt {c+d x}} \, dx}{32 b^3}+\frac {\left (5 d^3\right ) \int \frac {\cosh (a+b x)}{\sqrt {c+d x}} \, dx}{4 b^3}\\ &=-\frac {45 d^2 \sqrt {c+d x} \cosh (a+b x)}{16 b^3}-\frac {2 (c+d x)^{5/2} \cosh (a+b x)}{3 b}+\frac {5 d^2 \sqrt {c+d x} \cosh (3 a+3 b x)}{144 b^3}+\frac {5 d (c+d x)^{3/2} \sinh (a+b x)}{3 b^2}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2}-\frac {\left (5 d^3\right ) \int \frac {e^{-i (3 i a+3 i b x)}}{\sqrt {c+d x}} \, dx}{576 b^3}-\frac {\left (5 d^3\right ) \int \frac {e^{i (3 i a+3 i b x)}}{\sqrt {c+d x}} \, dx}{576 b^3}+\frac {\left (5 d^3\right ) \int \frac {e^{-i (i a+i b x)}}{\sqrt {c+d x}} \, dx}{64 b^3}+\frac {\left (5 d^3\right ) \int \frac {e^{i (i a+i b x)}}{\sqrt {c+d x}} \, dx}{64 b^3}+\frac {\left (5 d^3\right ) \int \frac {e^{-i (i a+i b x)}}{\sqrt {c+d x}} \, dx}{8 b^3}+\frac {\left (5 d^3\right ) \int \frac {e^{i (i a+i b x)}}{\sqrt {c+d x}} \, dx}{8 b^3}\\ &=-\frac {45 d^2 \sqrt {c+d x} \cosh (a+b x)}{16 b^3}-\frac {2 (c+d x)^{5/2} \cosh (a+b x)}{3 b}+\frac {5 d^2 \sqrt {c+d x} \cosh (3 a+3 b x)}{144 b^3}+\frac {5 d (c+d x)^{3/2} \sinh (a+b x)}{3 b^2}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2}-\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{i \left (3 i a-\frac {3 i b c}{d}\right )-\frac {3 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{288 b^3}-\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{-i \left (3 i a-\frac {3 i b c}{d}\right )+\frac {3 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{288 b^3}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{i \left (i a-\frac {i b c}{d}\right )-\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{32 b^3}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{-i \left (i a-\frac {i b c}{d}\right )+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{32 b^3}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{i \left (i a-\frac {i b c}{d}\right )-\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 b^3}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{-i \left (i a-\frac {i b c}{d}\right )+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 b^3}\\ &=-\frac {45 d^2 \sqrt {c+d x} \cosh (a+b x)}{16 b^3}-\frac {2 (c+d x)^{5/2} \cosh (a+b x)}{3 b}+\frac {5 d^2 \sqrt {c+d x} \cosh (3 a+3 b x)}{144 b^3}+\frac {45 d^{5/2} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{7/2}}-\frac {5 d^{5/2} e^{-3 a+\frac {3 b c}{d}} \sqrt {\frac {\pi }{3}} \text {erf}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{576 b^{7/2}}+\frac {45 d^{5/2} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{7/2}}-\frac {5 d^{5/2} e^{3 a-\frac {3 b c}{d}} \sqrt {\frac {\pi }{3}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{576 b^{7/2}}+\frac {5 d (c+d x)^{3/2} \sinh (a+b x)}{3 b^2}+\frac {(c+d x)^{5/2} \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac {5 d (c+d x)^{3/2} \sinh ^3(a+b x)}{18 b^2}\\ \end {align*}

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Mathematica [A]
time = 6.68, size = 243, normalized size = 0.64 \begin {gather*} -\frac {d^3 \left (\sqrt {3} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {3 b (c+d x)}{d}\right ) \left (\cosh \left (3 a-\frac {3 b c}{d}\right )+\sinh \left (3 a-\frac {3 b c}{d}\right )\right )+\left (\sqrt {\frac {b (c+d x)}{d}} \left (-243 \Gamma \left (\frac {7}{2},\frac {b (c+d x)}{d}\right )+\sqrt {3} \Gamma \left (\frac {7}{2},\frac {3 b (c+d x)}{d}\right ) \left (\cosh \left (2 a-\frac {2 b c}{d}\right )-\sinh \left (2 a-\frac {2 b c}{d}\right )\right )\right )+243 \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {b (c+d x)}{d}\right ) \left (\cosh \left (2 a-\frac {2 b c}{d}\right )+\sinh \left (2 a-\frac {2 b c}{d}\right )\right )\right ) \left (-\cosh \left (a-\frac {b c}{d}\right )+\sinh \left (a-\frac {b c}{d}\right )\right )\right )}{648 b^4 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)*Sinh[a + b*x]^3,x]

[Out]

-1/648*(d^3*(Sqrt[3]*Sqrt[-((b*(c + d*x))/d)]*Gamma[7/2, (-3*b*(c + d*x))/d]*(Cosh[3*a - (3*b*c)/d] + Sinh[3*a
- (3*b*c)/d]) + (Sqrt[(b*(c + d*x))/d]*(-243*Gamma[7/2, (b*(c + d*x))/d] + Sqrt[3]*Gamma[7/2, (3*b*(c + d*x))
/d]*(Cosh[2*a - (2*b*c)/d] - Sinh[2*a - (2*b*c)/d])) + 243*Sqrt[-((b*(c + d*x))/d)]*Gamma[7/2, -((b*(c + d*x))
/d)]*(Cosh[2*a - (2*b*c)/d] + Sinh[2*a - (2*b*c)/d]))*(-Cosh[a - (b*c)/d] + Sinh[a - (b*c)/d])))/(b^4*Sqrt[c +
d*x])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{\frac {5}{2}} \left (\sinh ^{3}\left (b x +a \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)*sinh(b*x+a)^3,x)

[Out]

int((d*x+c)^(5/2)*sinh(b*x+a)^3,x)

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Maxima [A]
time = 0.51, size = 513, normalized size = 1.35 \begin {gather*} -\frac {\frac {5 \, \sqrt {3} \sqrt {\pi } d^{3} \operatorname {erf}\left (\sqrt {3} \sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (3 \, a - \frac {3 \, b c}{d}\right )}}{b^{3} \sqrt {-\frac {b}{d}}} + \frac {5 \, \sqrt {3} \sqrt {\pi } d^{3} \operatorname {erf}\left (\sqrt {3} \sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-3 \, a + \frac {3 \, b c}{d}\right )}}{b^{3} \sqrt {\frac {b}{d}}} - \frac {1215 \, \sqrt {\pi } d^{3} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (a - \frac {b c}{d}\right )}}{b^{3} \sqrt {-\frac {b}{d}}} - \frac {1215 \, \sqrt {\pi } d^{3} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-a + \frac {b c}{d}\right )}}{b^{3} \sqrt {\frac {b}{d}}} + \frac {162 \, {\left (4 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d e^{\left (\frac {b c}{d}\right )} + 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} e^{\left (\frac {b c}{d}\right )} + 15 \, \sqrt {d x + c} d^{3} e^{\left (\frac {b c}{d}\right )}\right )} e^{\left (-a - \frac {{\left (d x + c\right )} b}{d}\right )}}{b^{3}} - \frac {6 \, {\left (12 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d e^{\left (\frac {3 \, b c}{d}\right )} + 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} e^{\left (\frac {3 \, b c}{d}\right )} + 5 \, \sqrt {d x + c} d^{3} e^{\left (\frac {3 \, b c}{d}\right )}\right )} e^{\left (-3 \, a - \frac {3 \, {\left (d x + c\right )} b}{d}\right )}}{b^{3}} - \frac {6 \, {\left (12 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d e^{\left (3 \, a\right )} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} e^{\left (3 \, a\right )} + 5 \, \sqrt {d x + c} d^{3} e^{\left (3 \, a\right )}\right )} e^{\left (\frac {3 \, {\left (d x + c\right )} b}{d} - \frac {3 \, b c}{d}\right )}}{b^{3}} + \frac {162 \, {\left (4 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d e^{a} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} e^{a} + 15 \, \sqrt {d x + c} d^{3} e^{a}\right )} e^{\left (\frac {{\left (d x + c\right )} b}{d} - \frac {b c}{d}\right )}}{b^{3}}}{1728 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/1728*(5*sqrt(3)*sqrt(pi)*d^3*erf(sqrt(3)*sqrt(d*x + c)*sqrt(-b/d))*e^(3*a - 3*b*c/d)/(b^3*sqrt(-b/d)) + 5*s
qrt(3)*sqrt(pi)*d^3*erf(sqrt(3)*sqrt(d*x + c)*sqrt(b/d))*e^(-3*a + 3*b*c/d)/(b^3*sqrt(b/d)) - 1215*sqrt(pi)*d^
3*erf(sqrt(d*x + c)*sqrt(-b/d))*e^(a - b*c/d)/(b^3*sqrt(-b/d)) - 1215*sqrt(pi)*d^3*erf(sqrt(d*x + c)*sqrt(b/d)
)*e^(-a + b*c/d)/(b^3*sqrt(b/d)) + 162*(4*(d*x + c)^(5/2)*b^2*d*e^(b*c/d) + 10*(d*x + c)^(3/2)*b*d^2*e^(b*c/d)
+ 15*sqrt(d*x + c)*d^3*e^(b*c/d))*e^(-a - (d*x + c)*b/d)/b^3 - 6*(12*(d*x + c)^(5/2)*b^2*d*e^(3*b*c/d) + 10*(
d*x + c)^(3/2)*b*d^2*e^(3*b*c/d) + 5*sqrt(d*x + c)*d^3*e^(3*b*c/d))*e^(-3*a - 3*(d*x + c)*b/d)/b^3 - 6*(12*(d*
x + c)^(5/2)*b^2*d*e^(3*a) - 10*(d*x + c)^(3/2)*b*d^2*e^(3*a) + 5*sqrt(d*x + c)*d^3*e^(3*a))*e^(3*(d*x + c)*b/
d - 3*b*c/d)/b^3 + 162*(4*(d*x + c)^(5/2)*b^2*d*e^a - 10*(d*x + c)^(3/2)*b*d^2*e^a + 15*sqrt(d*x + c)*d^3*e^a)
*e^((d*x + c)*b/d - b*c/d)/b^3)/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2090 vs. \(2 (291) = 582\).
time = 0.46, size = 2090, normalized size = 5.49 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/1728*(5*sqrt(3)*sqrt(pi)*(d^3*cosh(b*x + a)^3*cosh(-3*(b*c - a*d)/d) - d^3*cosh(b*x + a)^3*sinh(-3*(b*c - a
*d)/d) + (d^3*cosh(-3*(b*c - a*d)/d) - d^3*sinh(-3*(b*c - a*d)/d))*sinh(b*x + a)^3 + 3*(d^3*cosh(b*x + a)*cosh
(-3*(b*c - a*d)/d) - d^3*cosh(b*x + a)*sinh(-3*(b*c - a*d)/d))*sinh(b*x + a)^2 + 3*(d^3*cosh(b*x + a)^2*cosh(-
3*(b*c - a*d)/d) - d^3*cosh(b*x + a)^2*sinh(-3*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(3)*sqrt(d*x +
c)*sqrt(b/d)) - 5*sqrt(3)*sqrt(pi)*(d^3*cosh(b*x + a)^3*cosh(-3*(b*c - a*d)/d) + d^3*cosh(b*x + a)^3*sinh(-3*
(b*c - a*d)/d) + (d^3*cosh(-3*(b*c - a*d)/d) + d^3*sinh(-3*(b*c - a*d)/d))*sinh(b*x + a)^3 + 3*(d^3*cosh(b*x +
a)*cosh(-3*(b*c - a*d)/d) + d^3*cosh(b*x + a)*sinh(-3*(b*c - a*d)/d))*sinh(b*x + a)^2 + 3*(d^3*cosh(b*x + a)^
2*cosh(-3*(b*c - a*d)/d) + d^3*cosh(b*x + a)^2*sinh(-3*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(3)*s
qrt(d*x + c)*sqrt(-b/d)) - 1215*sqrt(pi)*(d^3*cosh(b*x + a)^3*cosh(-(b*c - a*d)/d) - d^3*cosh(b*x + a)^3*sinh(
-(b*c - a*d)/d) + (d^3*cosh(-(b*c - a*d)/d) - d^3*sinh(-(b*c - a*d)/d))*sinh(b*x + a)^3 + 3*(d^3*cosh(b*x + a)
*cosh(-(b*c - a*d)/d) - d^3*cosh(b*x + a)*sinh(-(b*c - a*d)/d))*sinh(b*x + a)^2 + 3*(d^3*cosh(b*x + a)^2*cosh(
-(b*c - a*d)/d) - d^3*cosh(b*x + a)^2*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(d*x + c)*sqrt(b/
d)) + 1215*sqrt(pi)*(d^3*cosh(b*x + a)^3*cosh(-(b*c - a*d)/d) + d^3*cosh(b*x + a)^3*sinh(-(b*c - a*d)/d) + (d^
3*cosh(-(b*c - a*d)/d) + d^3*sinh(-(b*c - a*d)/d))*sinh(b*x + a)^3 + 3*(d^3*cosh(b*x + a)*cosh(-(b*c - a*d)/d)
+ d^3*cosh(b*x + a)*sinh(-(b*c - a*d)/d))*sinh(b*x + a)^2 + 3*(d^3*cosh(b*x + a)^2*cosh(-(b*c - a*d)/d) + d^3
*cosh(b*x + a)^2*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(d*x + c)*sqrt(-b/d)) - 6*(12*b^3*d^2
*x^2 + (12*b^3*d^2*x^2 + 12*b^3*c^2 - 10*b^2*c*d + 5*b*d^2 + 2*(12*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a)^6 + 6
*(12*b^3*d^2*x^2 + 12*b^3*c^2 - 10*b^2*c*d + 5*b*d^2 + 2*(12*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a)*sinh(b*x +
a)^5 + (12*b^3*d^2*x^2 + 12*b^3*c^2 - 10*b^2*c*d + 5*b*d^2 + 2*(12*b^3*c*d - 5*b^2*d^2)*x)*sinh(b*x + a)^6 + 1
2*b^3*c^2 - 27*(4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a)
^4 - 3*(36*b^3*d^2*x^2 + 36*b^3*c^2 - 90*b^2*c*d + 135*b*d^2 - 5*(12*b^3*d^2*x^2 + 12*b^3*c^2 - 10*b^2*c*d + 5
*b*d^2 + 2*(12*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a)^2 + 18*(4*b^3*c*d - 5*b^2*d^2)*x)*sinh(b*x + a)^4 + 10*b^
2*c*d + 4*(5*(12*b^3*d^2*x^2 + 12*b^3*c^2 - 10*b^2*c*d + 5*b*d^2 + 2*(12*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a)
^3 - 27*(4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a))*sinh(
b*x + a)^3 + 5*b*d^2 - 27*(4*b^3*d^2*x^2 + 4*b^3*c^2 + 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d + 5*b^2*d^2)*x)*co
sh(b*x + a)^2 - 3*(36*b^3*d^2*x^2 + 36*b^3*c^2 - 5*(12*b^3*d^2*x^2 + 12*b^3*c^2 - 10*b^2*c*d + 5*b*d^2 + 2*(12
*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a)^4 + 90*b^2*c*d + 135*b*d^2 + 54*(4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d
+ 15*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a)^2 + 18*(4*b^3*c*d + 5*b^2*d^2)*x)*sinh(b*x + a)^2 + 2
*(12*b^3*c*d + 5*b^2*d^2)*x + 6*((12*b^3*d^2*x^2 + 12*b^3*c^2 - 10*b^2*c*d + 5*b*d^2 + 2*(12*b^3*c*d - 5*b^2*d
^2)*x)*cosh(b*x + a)^5 - 18*(4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2)*x)*
cosh(b*x + a)^3 - 9*(4*b^3*d^2*x^2 + 4*b^3*c^2 + 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d + 5*b^2*d^2)*x)*cosh(b*x
+ a))*sinh(b*x + a))*sqrt(d*x + c))/(b^4*cosh(b*x + a)^3 + 3*b^4*cosh(b*x + a)^2*sinh(b*x + a) + 3*b^4*cosh(b
*x + a)*sinh(b*x + a)^2 + b^4*sinh(b*x + a)^3)

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Sympy [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((d*x+c)**(5/2)*sinh(b*x+a)**3,x)

[Out]

Timed out

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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((d*x+c)^(5/2)*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/2)*sinh(b*x + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {sinh}\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^3*(c + d*x)^(5/2),x)

[Out]

int(sinh(a + b*x)^3*(c + d*x)^(5/2), x)

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